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301 | https://artofproblemsolving.com/wiki/index.php/2017_AMC_8_Problems/Problem_9 | 2 | All of Marcy's marbles are blue, red, green, or yellow. One third of her marbles are blue, one fourth of them are red, and six of them are green. What is the smallest number of yellow marbles that Macy could have?
$\textbf{(A) }1\qquad\textbf{(B) }2\qquad\textbf{(C) }3\qquad\textbf{(D) }4\qquad\textbf{(E) }5$ | Since $\frac{1}{3}$ of the marbles are blue and $\frac{1}{4}$ are red, it is clear that the total number of marbles must be divisible by $12$ . If there are $12$ marbles, then $4$ are blue, $3$ are red, and $6$ are green, meaning that there are $-1$ yellow marbles. This is impossible. Trying the next multiple of $12$ $24$ , we find that $8$ are green, $6$ are red, and $6$ are green, meaning that the minimum number of yellow marbles is $\boxed{4}$ | 4 |
302 | https://artofproblemsolving.com/wiki/index.php/2017_AMC_8_Problems/Problem_11 | 1 | A square-shaped floor is covered with congruent square tiles. If the total number of tiles that lie on the two diagonals is 37, how many tiles cover the floor?
$\textbf{(A) }148\qquad\textbf{(B) }324\qquad\textbf{(C) }361\qquad\textbf{(D) }1296\qquad\textbf{(E) }1369$ | Since the number of tiles lying on both diagonals is $37$ , counting one tile twice, there are $37=2x-1\implies x=19$ tiles on each side. Therefore, our answer is $19^2=361=\boxed{361}$ | 361 |
303 | https://artofproblemsolving.com/wiki/index.php/2017_AMC_8_Problems/Problem_11 | 2 | A square-shaped floor is covered with congruent square tiles. If the total number of tiles that lie on the two diagonals is 37, how many tiles cover the floor?
$\textbf{(A) }148\qquad\textbf{(B) }324\qquad\textbf{(C) }361\qquad\textbf{(D) }1296\qquad\textbf{(E) }1369$ | Visualize it as 4 separate diagonals connecting to one square in the middle. Each square on the diagonal corresponds to one square of horizontal/vertical distance (because it's a square). So, we figure out the length of each separate diagonal, multiply by two, and then add 1. (Realize that we can just join two of the separate diagonals on opposite sides together to save some time in calculations.) Therefore, the edge length is: \[\frac{37-1}{4} \cdot 2 + 1 = 19\] Thus, our solution is $19^2 = 361 = \boxed{361}$ | 361 |
304 | https://artofproblemsolving.com/wiki/index.php/2017_AMC_8_Problems/Problem_14 | 1 | Chloe and Zoe are both students in Ms. Demeanor's math class. Last night, they each solved half of the problems in their homework assignment alone and then solved the other half together. Chloe had correct answers to only $80\%$ of the problems she solved alone, but overall $88\%$ of her answers were correct. Zoe had correct answers to $90\%$ of the problems she solved alone. What was Zoe's overall percentage of correct answers?
$\textbf{(A) }89\qquad\textbf{(B) }92\qquad\textbf{(C) }93\qquad\textbf{(D) }96\qquad\textbf{(E) }98$ | Let the number of questions that they solved alone be $x$ . Let the percentage of problems they correctly solve together be $a$ %.
As given, \[\frac{80x}{100} + \frac{ax}{100} = \frac{2 \cdot 88x}{100}\]
Hence, $a = 96$
Zoe got $\frac{90x}{100} + \frac{ax}{100} = \frac{186x}{100}$ problems right out of $2x$ . Therefore, Zoe got $\frac{\frac{186x}{100}}{2x} = \frac{93}{100} = \boxed{93}$ percent of the problems correct. | 93 |
305 | https://artofproblemsolving.com/wiki/index.php/2017_AMC_8_Problems/Problem_14 | 2 | Chloe and Zoe are both students in Ms. Demeanor's math class. Last night, they each solved half of the problems in their homework assignment alone and then solved the other half together. Chloe had correct answers to only $80\%$ of the problems she solved alone, but overall $88\%$ of her answers were correct. Zoe had correct answers to $90\%$ of the problems she solved alone. What was Zoe's overall percentage of correct answers?
$\textbf{(A) }89\qquad\textbf{(B) }92\qquad\textbf{(C) }93\qquad\textbf{(D) }96\qquad\textbf{(E) }98$ | Assume the total amount of problems is $100$ per half homework assignment since we are dealing with percentages, not values. Then, we know that Chloe got $80$ problems correct by herself and got $176$ problems correct overall. We also know that Zoe had $90$ problems she did correctly alone. We can see that the total amount of correct problems Chloe and Zoe did together was $176-80=96$ . Therefore, Zoe did $96+90=186$ problems out of $200$ problems correctly. This is $\boxed{93}$ percent. | 93 |
306 | https://artofproblemsolving.com/wiki/index.php/2017_AMC_8_Problems/Problem_14 | 3 | Chloe and Zoe are both students in Ms. Demeanor's math class. Last night, they each solved half of the problems in their homework assignment alone and then solved the other half together. Chloe had correct answers to only $80\%$ of the problems she solved alone, but overall $88\%$ of her answers were correct. Zoe had correct answers to $90\%$ of the problems she solved alone. What was Zoe's overall percentage of correct answers?
$\textbf{(A) }89\qquad\textbf{(B) }92\qquad\textbf{(C) }93\qquad\textbf{(D) }96\qquad\textbf{(E) }98$ | In the problem, we can see that Chloe solved 80% of the problems she solved alone, but 88% of her answers are correct. If 80 and another number's average is 88, the other number must be 96. Then, Zoe solved 90% of the problems she did alone, but 96% of her answers were correct. Then, the average of 90 and 96 is $\boxed{93}$ | 93 |
307 | https://artofproblemsolving.com/wiki/index.php/2017_AMC_8_Problems/Problem_14 | 4 | Chloe and Zoe are both students in Ms. Demeanor's math class. Last night, they each solved half of the problems in their homework assignment alone and then solved the other half together. Chloe had correct answers to only $80\%$ of the problems she solved alone, but overall $88\%$ of her answers were correct. Zoe had correct answers to $90\%$ of the problems she solved alone. What was Zoe's overall percentage of correct answers?
$\textbf{(A) }89\qquad\textbf{(B) }92\qquad\textbf{(C) }93\qquad\textbf{(D) }96\qquad\textbf{(E) }98$ | (Slightly different Solution)
Suppose we said that there were $100$ problems in their assignment. Then, Chloe had $40$ correct and $10$ incorrect on her portion, and $48$ correct and $2$ incorrect on the portion she and Zoe solved. Zoe has $45$ correct and $5$ incorrect on her portion, and $48$ correct and $2$ incorrect on the portion that she and Chloe solved. Then, Zoe has $48 + 45 = 93$ correct answers out of $100$ , so the answer is $\boxed{93}$ | 93 |
308 | https://artofproblemsolving.com/wiki/index.php/2017_AMC_8_Problems/Problem_14 | 5 | Chloe and Zoe are both students in Ms. Demeanor's math class. Last night, they each solved half of the problems in their homework assignment alone and then solved the other half together. Chloe had correct answers to only $80\%$ of the problems she solved alone, but overall $88\%$ of her answers were correct. Zoe had correct answers to $90\%$ of the problems she solved alone. What was Zoe's overall percentage of correct answers?
$\textbf{(A) }89\qquad\textbf{(B) }92\qquad\textbf{(C) }93\qquad\textbf{(D) }96\qquad\textbf{(E) }98$ | Let the total number of problems be $t$ . Let the percentage of the number of problems that Chloe and Zoe did together and got right be $x$ . As we can see, Chloe got $80$ % of $\frac {1}{2}$ of the total problems right, hence, ${0.80 \cdot \frac{1}{2}t}$ . We also know that Chloe got $88$ % of $t$ problems right altogether, making it ${0.88 \cdot t}$ total problems right. If we add $x$ to the percentage of correct problems that Chloe solved alone, then that should be equal to the total number of problems that Chloe got right, making the equation: ${0.80 \cdot \frac{1}{2}t} + x = {0.88 \cdot t}$ . Solving that, we get $x = 0.48t$ . We also know that Zoe got $90$ % of $\frac {1}{2}$ of the total problems right, making it ${0.90 \cdot \frac{1}{2}t}$ . We now add that amount to the percentage of problems that Chloe and Zoe got right together, making ${0.90 \cdot \frac{1}{2}t}+ 0.48t$ . Solving that, we get $0.93t$ , which is equal to $93$ %; hence, $\boxed{93}$ | 93 |
309 | https://artofproblemsolving.com/wiki/index.php/2017_AMC_8_Problems/Problem_15 | 1 | In the arrangement of letters and numerals below, by how many different paths can one spell AMC8? Beginning at the A in the middle, a path only allows moves from one letter to an adjacent (above, below, left, or right, but not diagonal) letter. One example of such a path is traced in the picture.
[asy] fill((0.5, 4.5)--(1.5,4.5)--(1.5,2.5)--(0.5,2.5)--cycle,lightgray); fill((1.5,3.5)--(2.5,3.5)--(2.5,1.5)--(1.5,1.5)--cycle,lightgray); label("$8$", (1, 0)); label("$C$", (2, 0)); label("$8$", (3, 0)); label("$8$", (0, 1)); label("$C$", (1, 1)); label("$M$", (2, 1)); label("$C$", (3, 1)); label("$8$", (4, 1)); label("$C$", (0, 2)); label("$M$", (1, 2)); label("$A$", (2, 2)); label("$M$", (3, 2)); label("$C$", (4, 2)); label("$8$", (0, 3)); label("$C$", (1, 3)); label("$M$", (2, 3)); label("$C$", (3, 3)); label("$8$", (4, 3)); label("$8$", (1, 4)); label("$C$", (2, 4)); label("$8$", (3, 4));[/asy]
$\textbf{(A) }8\qquad\textbf{(B) }9\qquad\textbf{(C) }12\qquad\textbf{(D) }24\qquad\textbf{(E) }36$ | Notice that the upper-most section contains a 3 by 3 square that looks like:
[asy]label("$8$", (1, 2)); label("$C$", (2, 2)); label("$8$", (3, 2)); label("$C$", (1, 1)); label("$M$", (2, 1)); label("$C$", (3, 1)); label("$M$", (1, 0)); label("$A$", (2, 0)); label("$M$", (3, 0));[/asy]
It has 6 paths in which you can spell out AMC8. You will find four identical copies of this square in the figure, so multiply ${6 \cdot 4}$ to get $\boxed{24}$ total paths. | 24 |
310 | https://artofproblemsolving.com/wiki/index.php/2017_AMC_8_Problems/Problem_15 | 2 | In the arrangement of letters and numerals below, by how many different paths can one spell AMC8? Beginning at the A in the middle, a path only allows moves from one letter to an adjacent (above, below, left, or right, but not diagonal) letter. One example of such a path is traced in the picture.
[asy] fill((0.5, 4.5)--(1.5,4.5)--(1.5,2.5)--(0.5,2.5)--cycle,lightgray); fill((1.5,3.5)--(2.5,3.5)--(2.5,1.5)--(1.5,1.5)--cycle,lightgray); label("$8$", (1, 0)); label("$C$", (2, 0)); label("$8$", (3, 0)); label("$8$", (0, 1)); label("$C$", (1, 1)); label("$M$", (2, 1)); label("$C$", (3, 1)); label("$8$", (4, 1)); label("$C$", (0, 2)); label("$M$", (1, 2)); label("$A$", (2, 2)); label("$M$", (3, 2)); label("$C$", (4, 2)); label("$8$", (0, 3)); label("$C$", (1, 3)); label("$M$", (2, 3)); label("$C$", (3, 3)); label("$8$", (4, 3)); label("$8$", (1, 4)); label("$C$", (2, 4)); label("$8$", (3, 4));[/asy]
$\textbf{(A) }8\qquad\textbf{(B) }9\qquad\textbf{(C) }12\qquad\textbf{(D) }24\qquad\textbf{(E) }36$ | There are three different kinds of paths that are on this diagram. The first kind is when you directly count $A$ $M$ $C$ in a straight line. The second is when you count $A$ , turn left or right to get $M$ , then go up or down to count $8$ and $C$ . The third is the one where you start with $A$ , move up or down to count $M$ , turn left or right to count $C$ , then move straight again to get $8$
There are 8 paths for each kind of path, making for $8 \cdot 3=\boxed{24}$ paths. | 24 |
311 | https://artofproblemsolving.com/wiki/index.php/2017_AMC_8_Problems/Problem_15 | 3 | In the arrangement of letters and numerals below, by how many different paths can one spell AMC8? Beginning at the A in the middle, a path only allows moves from one letter to an adjacent (above, below, left, or right, but not diagonal) letter. One example of such a path is traced in the picture.
[asy] fill((0.5, 4.5)--(1.5,4.5)--(1.5,2.5)--(0.5,2.5)--cycle,lightgray); fill((1.5,3.5)--(2.5,3.5)--(2.5,1.5)--(1.5,1.5)--cycle,lightgray); label("$8$", (1, 0)); label("$C$", (2, 0)); label("$8$", (3, 0)); label("$8$", (0, 1)); label("$C$", (1, 1)); label("$M$", (2, 1)); label("$C$", (3, 1)); label("$8$", (4, 1)); label("$C$", (0, 2)); label("$M$", (1, 2)); label("$A$", (2, 2)); label("$M$", (3, 2)); label("$C$", (4, 2)); label("$8$", (0, 3)); label("$C$", (1, 3)); label("$M$", (2, 3)); label("$C$", (3, 3)); label("$8$", (4, 3)); label("$8$", (1, 4)); label("$C$", (2, 4)); label("$8$", (3, 4));[/asy]
$\textbf{(A) }8\qquad\textbf{(B) }9\qquad\textbf{(C) }12\qquad\textbf{(D) }24\qquad\textbf{(E) }36$ | Notice that the $A$ is adjacent to $4$ $M$ s, each $M$ is adjacent to $3$ $C$ s, and each $C$ is adjacent to $2$ $8$ 's. So for each $A$ , there are $4$ $M$ s, and for each $M$ , there are $3$ $C$ s, and for each $C$ , there are $2$ $8$ s. Thus, the answer is $1\cdot 4\cdot 3\cdot 2 = \boxed{24}.$ | 24 |
312 | https://artofproblemsolving.com/wiki/index.php/2017_AMC_8_Problems/Problem_15 | 4 | In the arrangement of letters and numerals below, by how many different paths can one spell AMC8? Beginning at the A in the middle, a path only allows moves from one letter to an adjacent (above, below, left, or right, but not diagonal) letter. One example of such a path is traced in the picture.
[asy] fill((0.5, 4.5)--(1.5,4.5)--(1.5,2.5)--(0.5,2.5)--cycle,lightgray); fill((1.5,3.5)--(2.5,3.5)--(2.5,1.5)--(1.5,1.5)--cycle,lightgray); label("$8$", (1, 0)); label("$C$", (2, 0)); label("$8$", (3, 0)); label("$8$", (0, 1)); label("$C$", (1, 1)); label("$M$", (2, 1)); label("$C$", (3, 1)); label("$8$", (4, 1)); label("$C$", (0, 2)); label("$M$", (1, 2)); label("$A$", (2, 2)); label("$M$", (3, 2)); label("$C$", (4, 2)); label("$8$", (0, 3)); label("$C$", (1, 3)); label("$M$", (2, 3)); label("$C$", (3, 3)); label("$8$", (4, 3)); label("$8$", (1, 4)); label("$C$", (2, 4)); label("$8$", (3, 4));[/asy]
$\textbf{(A) }8\qquad\textbf{(B) }9\qquad\textbf{(C) }12\qquad\textbf{(D) }24\qquad\textbf{(E) }36$ | We can do this problem by computing how many ways there are to get to each letter (in order). There is $1$ way to get to the $A$ in the center. We can only get to each of the other $M$ s by going there from the $A$ , so there is $1$ way to get to each of the four $M$ s. For the $C$ s, we notice that four $C$ s are surrounded by one $M$ , and four $C$ s are surrounded by two $M$ s. Finally, each of the $8$ s is surrounded by one $C$ with one way to get there and one $C$ with two ways to get there. Therefore, there are three paths to any of the $8$ s. Since there are eight $8$ s, the answer is $3 \cdot 8 = \boxed{24}.$ | 24 |
313 | https://artofproblemsolving.com/wiki/index.php/2017_AMC_8_Problems/Problem_17 | 2 | Starting with some gold coins and some empty treasure chests, I tried to put $9$ gold coins in each treasure chest, but that left $2$ treasure chests empty. So instead I put $6$ gold coins in each treasure chest, but then I had $3$ gold coins left over. How many gold coins did I have?
$\textbf{(A) }9\qquad\textbf{(B) }27\qquad\textbf{(C) }45\qquad\textbf{(D) }63\qquad\textbf{(E) }81$ | With $9$ coins, there are $\frac{9}{9}+2=1+2=3$ chests, by the first condition. These don't fit in with the second condition, so we move onto $27$ coins. By the same first condition, there are $5$ chests( $\frac{27}{9}+2$ ). This also doesn't fit with the second condition. So, onto $45$ coins. The first condition implies that there are $\frac{45}{9}+2=7$ chests, which DOES fit with the second condition, since $6\cdot7+3=42+3=45$ . Thus, the desired value is $\boxed{45}$ | 45 |
314 | https://artofproblemsolving.com/wiki/index.php/2017_AMC_8_Problems/Problem_18 | 1 | In the non-convex quadrilateral $ABCD$ shown below, $\angle BCD$ is a right angle, $AB=12$ $BC=4$ $CD=3$ , and $AD=13$ . What is the area of quadrilateral $ABCD$
[asy]draw((0,0)--(2.4,3.6)--(0,5)--(12,0)--(0,0)); label("$B$", (0, 0), SW); label("$A$", (12, 0), ESE); label("$C$", (2.4, 3.6), SE); label("$D$", (0, 5), N);[/asy]
$\textbf{(A) }12 \qquad \textbf{(B) }24 \qquad \textbf{(C) }26 \qquad \textbf{(D) }30 \qquad \textbf{(E) }36$ | We first connect point $B$ with point $D$
[asy]draw((0,0)--(2.4,3.6)--(0,5)--(12,0)--(0,0)); draw((0,0)--(0,5)); label("$B$", (0, 0), SW); label("$A$", (12, 0), ESE); label("$C$", (2.4, 3.6), SE); label("$D$", (0, 5), N);[/asy]
We can see that $\triangle BCD$ is a 3-4-5 right triangle. We can also see that $\triangle BDA$ is a right triangle, by the 5-12-13 Pythagorean triple. With these lengths, we can solve the problem. The area of $\triangle BDA$ is $\frac{5\cdot 12}{2}$ , and the area of $\triangle BCD$ is $\frac{3\cdot 4}{2}$ . Thus, the area of quadrilateral $ABCD$ is $30-6 = \boxed{24}.$ | 24 |
315 | https://artofproblemsolving.com/wiki/index.php/2017_AMC_8_Problems/Problem_18 | 2 | In the non-convex quadrilateral $ABCD$ shown below, $\angle BCD$ is a right angle, $AB=12$ $BC=4$ $CD=3$ , and $AD=13$ . What is the area of quadrilateral $ABCD$
[asy]draw((0,0)--(2.4,3.6)--(0,5)--(12,0)--(0,0)); label("$B$", (0, 0), SW); label("$A$", (12, 0), ESE); label("$C$", (2.4, 3.6), SE); label("$D$", (0, 5), N);[/asy]
$\textbf{(A) }12 \qquad \textbf{(B) }24 \qquad \textbf{(C) }26 \qquad \textbf{(D) }30 \qquad \textbf{(E) }36$ | $\triangle BCD$ is a 3-4-5 right triangle. So the area of $\triangle BCD$ is 6. Then we can use Heron's formula to compute the area of $\triangle ABD$ whose sides have lengths 5,12,and 13. The area of $\triangle ABD$ $\sqrt{s(s-5)(s-12)(s-13)}$ , where s is the semi-perimeter of the triangle, that is $s=(5+12+13)/2=15.$ Thus, the area of $\triangle ABD$ is 30, so the area of $ABCD$ is $30-6 = \boxed{24}.$ ---LarryFlora | 24 |
316 | https://artofproblemsolving.com/wiki/index.php/2017_AMC_8_Problems/Problem_19 | 2 | For any positive integer $M$ , the notation $M!$ denotes the product of the integers $1$ through $M$ . What is the largest integer $n$ for which $5^n$ is a factor of the sum $98!+99!+100!$
$\textbf{(A) }23\qquad\textbf{(B) }24\qquad\textbf{(C) }25\qquad\textbf{(D) }26\qquad\textbf{(E) }27$ | Also, keep in mind that the number of $5$ ’s in $98! (10,000)$ is the same as the number of trailing zeros. The number of zeros is $98!$ , which means we need pairs of $5$ ’s and $2$ ’s; we know there will be many more $2$ ’s, so we seek to find the number of $5$ ’s in $98!$ , which the solution tells us. And, that is $22$ factors of $5$ $10,000$ has $4$ trailing zeros, so it has $4$ factors of $5$ and $22 + 4 = \boxed{26}$ | 26 |
317 | https://artofproblemsolving.com/wiki/index.php/2017_AMC_8_Problems/Problem_19 | 3 | For any positive integer $M$ , the notation $M!$ denotes the product of the integers $1$ through $M$ . What is the largest integer $n$ for which $5^n$ is a factor of the sum $98!+99!+100!$
$\textbf{(A) }23\qquad\textbf{(B) }24\qquad\textbf{(C) }25\qquad\textbf{(D) }26\qquad\textbf{(E) }27$ | We can first factor a $98!$ out of the $98! + 99! + 100!$ to get $98! ( 1 + 99 + 100*99 ),$ Simplify to get $98! (10,000)$
Let's first find how many factors of $5 10,000$ has. $10,000$ is $(2*5)^4$ because $10,000$ is $(10)^4$ . After we remove the brackets, we get $2^4$ , and $5^4$ . We only care about the latter (second one), because the problem only ask's for the power of $5$ . We get $4$
Next, we can look at the multiples of 5 in $98!$ $98/5 = 19$ so there is 19 multiples of 5. We get $19$
But we cannot forget the multiples of $5$ with $2$ fives in it. Multiples of $25$ . How many multiples of $25$ are between $1$ and $98$ $3$ $25,50,75,$ and that's it. We get $3$
Finally, we add all of the numbers (powers of $5$ ) up. That is $4 + 19 + 3$ , which is just $26$
So the answer is $26$ . Which is answer choice D $\boxed{26}$ | 26 |
318 | https://artofproblemsolving.com/wiki/index.php/2017_AMC_8_Problems/Problem_21 | 1 | Suppose $a$ $b$ , and $c$ are nonzero real numbers, and $a+b+c=0$ . What are the possible value(s) for $\frac{a}{|a|}+\frac{b}{|b|}+\frac{c}{|c|}+\frac{abc}{|abc|}$
$\textbf{(A) }0\qquad\textbf{(B) }1\text{ and }-1\qquad\textbf{(C) }2\text{ and }-2\qquad\textbf{(D) }0,2,\text{ and }-2\qquad\textbf{(E) }0,1,\text{ and }-1$ | There are $2$ cases to consider:
Case $1$ $2$ of $a$ $b$ , and $c$ are positive and the other is negative. Without loss of generality (WLOG), we can assume that $a$ and $b$ are positive and $c$ is negative. In this case, we have that \[\frac{a}{|a|}+\frac{b}{|b|}+\frac{c}{|c|}+\frac{abc}{|abc|}=1+1-1-1=0.\]
Case $2$ $2$ of $a$ $b$ , and $c$ are negative and the other is positive. WLOG, we can assume that $a$ and $b$ are negative and $c$ is positive. In this case, we have that \[\frac{a}{|a|}+\frac{b}{|b|}+\frac{c}{|c|}+\frac{abc}{|abc|}=-1-1+1+1=0.\]
Note these are the only valid cases, for neither $3$ negatives nor $3$ positives would work as they cannot sum up to $0$ . In both cases, we get that the given expression equals $\boxed{0}$ | 0 |
319 | https://artofproblemsolving.com/wiki/index.php/2017_AMC_8_Problems/Problem_23 | 1 | Each day for four days, Linda traveled for one hour at a speed that resulted in her traveling one mile in an integer number of minutes. Each day after the first, her speed decreased so that the number of minutes to travel one mile increased by $5$ minutes over the preceding day. Each of the four days, her distance traveled was also an integer number of miles. What was the total number of miles for the four trips?
$\textbf{(A) }10\qquad\textbf{(B) }15\qquad\textbf{(C) }25\qquad\textbf{(D) }50\qquad\textbf{(E) }82$ | It is well known that $\text{Distance}=\text{Speed} \cdot \text{Time}$ . In the question, we want distance. From the question, we have that the time is $60$ minutes or $1$ hour. By the equation derived from $\text{Distance}=\text{Speed} \cdot \text{Time}$ , we have $\text{Speed}=\frac{\text{Distance}}{\text{Time}}$ , so the speed is $1$ mile per $x$ minutes. Because we want the distance, we multiply the time and speed together yielding $60\text{ mins}\cdot \frac{1\text{ mile}}{x\text{ mins}}$ . The minutes cancel out, so now we have $\dfrac{60}{x}$ as our distance for the first day. The distance for the following days are: \[\dfrac{60}{x},\dfrac{60}{x+5},\dfrac{60}{x+10},\dfrac{60}{x+15}.\] We know that $x,x+5,x+10,x+15$ are all factors of $60$ , therefore, $x=5$ because the factors have to be in an arithmetic sequence with the common difference being $5$ and $x=5$ is the only solution. \[\dfrac{60}{5}+\dfrac{60}{10}+\dfrac{60}{15}+\dfrac{60}{20}=12+6+4+3=\boxed{25}.\] | 25 |
320 | https://artofproblemsolving.com/wiki/index.php/2017_AMC_8_Problems/Problem_24 | 1 | Mrs. Sanders has three grandchildren, who call her regularly. One calls her every three days, one calls her every four days, and one calls her every five days. All three called her on December 31, 2016. On how many days during the next year did she not receive a phone call from any of her grandchildren?
$\textbf{(A) }78\qquad\textbf{(B) }80\qquad\textbf{(C) }144\qquad\textbf{(D) }146\qquad\textbf{(E) }152$ | We use Principle of Inclusion-Exclusion. There are $365$ days in the year, and we subtract the days that she gets at least $1$ phone call, which is \[\left \lfloor \frac{365}{3} \right \rfloor + \left \lfloor \frac{365}{4} \right \rfloor + \left \lfloor \frac{365}{5} \right \rfloor.\]
To this result we add the number of days where she gets at least $2$ phone calls in a day because we double subtracted these days, which is \[\left \lfloor \frac{365}{12} \right \rfloor + \left \lfloor \frac{365}{15} \right \rfloor + \left \lfloor \frac{365}{20} \right \rfloor.\]
We now subtract the number of days where she gets three phone calls, which is $\left \lfloor \frac{365}{60} \right \rfloor.$ Therefore, our answer is \[365 - \left( \left \lfloor \frac{365}{3} \right \rfloor + \left \lfloor \frac{365}{4} \right \rfloor + \left \lfloor \frac{365}{5} \right \rfloor \right) + \left( \left \lfloor \frac{365}{12} \right \rfloor + \left \lfloor \frac{365}{15} \right \rfloor + \left \lfloor \frac{365}{20} \right \rfloor \right) - \left \lfloor \frac{365}{60} \right \rfloor = 365 - 285+72 - 6 = \boxed{146}.\] | 146 |
321 | https://artofproblemsolving.com/wiki/index.php/2017_AMC_8_Problems/Problem_24 | 2 | Mrs. Sanders has three grandchildren, who call her regularly. One calls her every three days, one calls her every four days, and one calls her every five days. All three called her on December 31, 2016. On how many days during the next year did she not receive a phone call from any of her grandchildren?
$\textbf{(A) }78\qquad\textbf{(B) }80\qquad\textbf{(C) }144\qquad\textbf{(D) }146\qquad\textbf{(E) }152$ | Note that $\operatorname{lcm}(3,4,5)=60,$ so there is a cycle every $60$ days.
As shown below, all days in a cycle that Mrs. Sanders receives a phone call from any of her grandchildren are colored in red, yellow, or green. [asy] /* Made by MRENTHUSIASM */ size(7cm); fill((2,6)--(3,6)--(3,5)--(2,5)--cycle,red); fill((5,6)--(6,6)--(6,5)--(5,5)--cycle,red); fill((8,6)--(9,6)--(9,5)--(8,5)--cycle,red); fill((1,5)--(2,5)--(2,4.5)--(1,4.5)--cycle,red); fill((4,5)--(5,5)--(5,4.5)--(4,4.5)--cycle,red); fill((7,5)--(8,5)--(8,4)--(7,4)--cycle,red); fill((0,4)--(1,4)--(1,3)--(0,3)--cycle,red); fill((3,4)--(4,4)--(4,3.5)--(3,3.5)--cycle,red); fill((6,4)--(7,4)--(7,3)--(6,3)--cycle,red); fill((9,4)--(10,4)--(10,3.5)--(9,3.5)--cycle,red); fill((2,3)--(3,3)--(3,2)--(2,2)--cycle,red); fill((5,3)--(6,3)--(6,2.5)--(5,2.5)--cycle,red); fill((8,3)--(9,3)--(9,2)--(8,2)--cycle,red); fill((1,2)--(2,2)--(2,1)--(1,1)--cycle,red); fill((4,2)--(5,2)--(5,1.5)--(4,1.5)--cycle,red); fill((7,2)--(8,2)--(8,1.5)--(7,1.5)--cycle,red); fill((0,1)--(1,1)--(1,0)--(0,0)--cycle,red); fill((3,1)--(4,1)--(4,0)--(3,0)--cycle,red); fill((6,1)--(7,1)--(7,0)--(6,0)--cycle,red); fill((9,1)--(10,1)--(10,2/3)--(9,2/3)--cycle,red); fill((3,6)--(4,6)--(4,5)--(3,5)--cycle,yellow); fill((7,6)--(8,6)--(8,5)--(7,5)--cycle,yellow); fill((1,4.5)--(2,4.5)--(2,4)--(1,4)--cycle,yellow); fill((5,5)--(6,5)--(6,4)--(5,4)--cycle,yellow); fill((9,5)--(10,5)--(10,4.5)--(9,4.5)--cycle,yellow); fill((3,3.5)--(4,3.5)--(4,3)--(3,3)--cycle,yellow); fill((7,4)--(8,4)--(8,3)--(7,3)--cycle,yellow); fill((1,3)--(2,3)--(2,2)--(1,2)--cycle,yellow); fill((5,2.5)--(6,2.5)--(6,2)--(5,2)--cycle,yellow); fill((9,3)--(10,3)--(10,2.5)--(9,2.5)--cycle,yellow); fill((3,2)--(4,2)--(4,1)--(3,1)--cycle,yellow); fill((7,1.5)--(8,1.5)--(8,1)--(7,1)--cycle,yellow); fill((1,1)--(2,1)--(2,0)--(1,0)--cycle,yellow); fill((5,1)--(6,1)--(6,0)--(5,0)--cycle,yellow); fill((9,2/3)--(10,2/3)--(10,1/3)--(9,1/3)--cycle,yellow); fill((4,6)--(5,6)--(5,5)--(4,5)--cycle,green); fill((9,6)--(10,6)--(10,5)--(9,5)--cycle,green); fill((4,4.5)--(5,4.5)--(5,4)--(4,4)--cycle,green); fill((9,4.5)--(10,4.5)--(10,4)--(9,4)--cycle,green); fill((4,4)--(5,4)--(5,3)--(4,3)--cycle,green); fill((9,3.5)--(10,3.5)--(10,3)--(9,3)--cycle,green); fill((4,3)--(5,3)--(5,2)--(4,2)--cycle,green); fill((9,2.5)--(10,2.5)--(10,2)--(9,2)--cycle,green); fill((4,1.5)--(5,1.5)--(5,1)--(4,1)--cycle,green); fill((9,2)--(10,2)--(10,1)--(9,1)--cycle,green); fill((4,1)--(5,1)--(5,0)--(4,0)--cycle,green); fill((9,1/3)--(10,1/3)--(10,0)--(9,0)--cycle,green); real cur = 1; for (real i=6; i>0; --i) { for (real j=0; j<10; ++j) { label("$"+string(cur)+"$",(j+0.5,i-0.5)); ++cur; } } add(grid(10,6,linewidth(1.25))); [/asy] The year 2017 has $365$ days, or $6$ cycles and $5$ days.
Together, the answer is $24\cdot6+2=\boxed{146}.$ | 146 |
322 | https://artofproblemsolving.com/wiki/index.php/2017_AMC_8_Problems/Problem_24 | 3 | Mrs. Sanders has three grandchildren, who call her regularly. One calls her every three days, one calls her every four days, and one calls her every five days. All three called her on December 31, 2016. On how many days during the next year did she not receive a phone call from any of her grandchildren?
$\textbf{(A) }78\qquad\textbf{(B) }80\qquad\textbf{(C) }144\qquad\textbf{(D) }146\qquad\textbf{(E) }152$ | For any randomly chosen day, there is a $\frac{2}{3}$ chance the first child does not call her, a $\frac{3}{4}$ chance the second child does not call her and a $\frac{4}{5}$ chance the third child does not call her. So, in a randomly chosen day, there is a $\frac{2}{3} \times \frac{3}{4} \times \frac{4}{5} = \frac{2}{5}$ chance no child calls her.
In particular, this "unrigorous" reasoning becomes rigorous, by linearity of expectation, when we take a sample of $360$ days; over the first $360$ days, $360 \times \frac{2}{5} = 144$ days will go without a phone call. Now, we simply check the remaining days; on day $361$ and $362$ , nobody calls her; on day $363$ , the child that calls every three days will call her; on day $364$ , the child that calls every four days will call her; on day $365$ , the child that calls every five days will call her. Thus, we add two more days to $144$ , to get our answer of $\boxed{146}$ | 146 |
323 | https://artofproblemsolving.com/wiki/index.php/2016_AMC_8_Problems/Problem_1 | 1 | The longest professional tennis match ever played lasted a total of $11$ hours and $5$ minutes. How many minutes was this?
$\textbf{(A) }605\qquad\textbf{(B) }655\qquad\textbf{(C) }665\qquad\textbf{(D) }1005\qquad \textbf{(E) }1105$ | It is best to split 11 hours and 5 minutes into 2 parts, one of 11 hours and another of 5 minutes. We know that there is $60$ minutes in a hour. Therefore, there are $11 \cdot 60 = 660$ minutes in 11 hours. Adding the second part(the 5 minutes) we get $660 + 5 = \boxed{665}$ | 665 |
324 | https://artofproblemsolving.com/wiki/index.php/2016_AMC_8_Problems/Problem_1 | 2 | The longest professional tennis match ever played lasted a total of $11$ hours and $5$ minutes. How many minutes was this?
$\textbf{(A) }605\qquad\textbf{(B) }655\qquad\textbf{(C) }665\qquad\textbf{(D) }1005\qquad \textbf{(E) }1105$ | The best method comes when you remember your multiplication tables. Thus trivial, we get our answer of $\boxed{665}$ | 665 |
325 | https://artofproblemsolving.com/wiki/index.php/2016_AMC_8_Problems/Problem_1 | 3 | The longest professional tennis match ever played lasted a total of $11$ hours and $5$ minutes. How many minutes was this?
$\textbf{(A) }605\qquad\textbf{(B) }655\qquad\textbf{(C) }665\qquad\textbf{(D) }1005\qquad \textbf{(E) }1105$ | 11 hours 5 min = $(11 \cdot 60) + 5 \text{min} = 665 \text{min}$ , therefore $\boxed{665}$ | 665 |
326 | https://artofproblemsolving.com/wiki/index.php/2016_AMC_8_Problems/Problem_2 | 1 | In rectangle $ABCD$ $AB=6$ and $AD=8$ . Point $M$ is the midpoint of $\overline{AD}$ . What is the area of $\triangle AMC$
[asy]draw((0,4)--(0,0)--(6,0)--(6,8)--(0,8)--(0,4)--(6,8)--(0,0)); label("$A$", (0,0), SW); label("$B$", (6, 0), SE); label("$C$", (6,8), NE); label("$D$", (0, 8), NW); label("$M$", (0, 4), W); label("$4$", (0, 2), W); label("$6$", (3, 0), S);[/asy]
$\textbf{(A) }12\qquad\textbf{(B) }15\qquad\textbf{(C) }18\qquad\textbf{(D) }20\qquad \textbf{(E) }24$ | Using the triangle area formula for triangles: $A = \frac{bh}{2},$ where $A$ is the area, $b$ is the base, and $h$ is the height. This equation gives us $A = \frac{4 \cdot 6}{2} = \frac{24}{2} =\boxed{12}$ | 12 |
327 | https://artofproblemsolving.com/wiki/index.php/2016_AMC_8_Problems/Problem_2 | 2 | In rectangle $ABCD$ $AB=6$ and $AD=8$ . Point $M$ is the midpoint of $\overline{AD}$ . What is the area of $\triangle AMC$
[asy]draw((0,4)--(0,0)--(6,0)--(6,8)--(0,8)--(0,4)--(6,8)--(0,0)); label("$A$", (0,0), SW); label("$B$", (6, 0), SE); label("$C$", (6,8), NE); label("$D$", (0, 8), NW); label("$M$", (0, 4), W); label("$4$", (0, 2), W); label("$6$", (3, 0), S);[/asy]
$\textbf{(A) }12\qquad\textbf{(B) }15\qquad\textbf{(C) }18\qquad\textbf{(D) }20\qquad \textbf{(E) }24$ | A triangle with the same height and base as a rectangle is half of the rectangle's area. This means that a triangle with half of the base of the rectangle and also the same height means its area is one quarter of the rectangle's area. Therefore, we get $\frac{48}{4} =\boxed{12}$ | 12 |
328 | https://artofproblemsolving.com/wiki/index.php/2016_AMC_8_Problems/Problem_2 | 3 | In rectangle $ABCD$ $AB=6$ and $AD=8$ . Point $M$ is the midpoint of $\overline{AD}$ . What is the area of $\triangle AMC$
[asy]draw((0,4)--(0,0)--(6,0)--(6,8)--(0,8)--(0,4)--(6,8)--(0,0)); label("$A$", (0,0), SW); label("$B$", (6, 0), SE); label("$C$", (6,8), NE); label("$D$", (0, 8), NW); label("$M$", (0, 4), W); label("$4$", (0, 2), W); label("$6$", (3, 0), S);[/asy]
$\textbf{(A) }12\qquad\textbf{(B) }15\qquad\textbf{(C) }18\qquad\textbf{(D) }20\qquad \textbf{(E) }24$ | We can find the area of the entire rectangle, DCBA to be $8 \cdot 6=48$ and find DCM area to be $\frac{6 \cdot 4}{2} = 12$ and BCA to be $\frac{6 \cdot 8}{2}=24$ $48-12-24=$ $\boxed{12}$ | 12 |
329 | https://artofproblemsolving.com/wiki/index.php/2016_AMC_8_Problems/Problem_2 | 4 | In rectangle $ABCD$ $AB=6$ and $AD=8$ . Point $M$ is the midpoint of $\overline{AD}$ . What is the area of $\triangle AMC$
[asy]draw((0,4)--(0,0)--(6,0)--(6,8)--(0,8)--(0,4)--(6,8)--(0,0)); label("$A$", (0,0), SW); label("$B$", (6, 0), SE); label("$C$", (6,8), NE); label("$D$", (0, 8), NW); label("$M$", (0, 4), W); label("$4$", (0, 2), W); label("$6$", (3, 0), S);[/asy]
$\textbf{(A) }12\qquad\textbf{(B) }15\qquad\textbf{(C) }18\qquad\textbf{(D) }20\qquad \textbf{(E) }24$ | A triangle is half of a rectangle. So since M is the midpoint of DA, we can see that triangle DAC is half of the whole rectangle. And it is also true that triangle MAC is half of triangle DAC, so triangle MAC would just be 1/4 of the whole rectangle. Since the rectangle's area is 48, 1/4 of 48 would be 12. Which gives us the answer $\boxed{12}$ | 12 |
330 | https://artofproblemsolving.com/wiki/index.php/2016_AMC_8_Problems/Problem_2 | 5 | In rectangle $ABCD$ $AB=6$ and $AD=8$ . Point $M$ is the midpoint of $\overline{AD}$ . What is the area of $\triangle AMC$
[asy]draw((0,4)--(0,0)--(6,0)--(6,8)--(0,8)--(0,4)--(6,8)--(0,0)); label("$A$", (0,0), SW); label("$B$", (6, 0), SE); label("$C$", (6,8), NE); label("$D$", (0, 8), NW); label("$M$", (0, 4), W); label("$4$", (0, 2), W); label("$6$", (3, 0), S);[/asy]
$\textbf{(A) }12\qquad\textbf{(B) }15\qquad\textbf{(C) }18\qquad\textbf{(D) }20\qquad \textbf{(E) }24$ | We can subtract the total areas of triangles DCM and ABC from the rectangle ABCD. For triangle DCM, the base is 4 and the height is 6, so we multiply 4 and 6, then divide by 2 to get 12. For triangle ABC, the base is 4 and the height is 8, so we multiply 4 and 8, then divide by 2 to get 24. We add 24 and 12 to get 36. Then, we calculate the area of rectangle ABCD, which is 48. We subtract 36 from 48, resulting in $\boxed{12}$ | 12 |
331 | https://artofproblemsolving.com/wiki/index.php/2016_AMC_8_Problems/Problem_3 | 1 | Four students take an exam. Three of their scores are $70, 80,$ and $90$ . If the average of their four scores is $70$ , then what is the remaining score?
$\textbf{(A) }40\qquad\textbf{(B) }50\qquad\textbf{(C) }55\qquad\textbf{(D) }60\qquad \textbf{(E) }70$ | Let $r$ be the remaining student's score. We know that the average, 70, is equal to $\frac{70 + 80 + 90 + r}{4}$ . We can use basic algebra to solve for $r$ \[\frac{70 + 80 + 90 + r}{4} = 70\] \[\frac{240 + r}{4} = 70\] \[240 + r = 280\] \[r = 40\] giving us the answer of $\boxed{40}$ | 40 |
332 | https://artofproblemsolving.com/wiki/index.php/2016_AMC_8_Problems/Problem_3 | 2 | Four students take an exam. Three of their scores are $70, 80,$ and $90$ . If the average of their four scores is $70$ , then what is the remaining score?
$\textbf{(A) }40\qquad\textbf{(B) }50\qquad\textbf{(C) }55\qquad\textbf{(D) }60\qquad \textbf{(E) }70$ | Since $90$ is $20$ more than $70$ , and $80$ is $10$ more than $70$ , for $70$ to be the average, the other number must be $30$ less than $70$ , or $\boxed{40}$ | 40 |
333 | https://artofproblemsolving.com/wiki/index.php/2016_AMC_8_Problems/Problem_4 | 1 | When Cheenu was a boy, he could run $15$ miles in $3$ hours and $30$ minutes. As an old man, he can now walk $10$ miles in $4$ hours. How many minutes longer does it take for him to walk a mile now compared to when he was a boy?
$\textbf{(A) }6\qquad\textbf{(B) }10\qquad\textbf{(C) }15\qquad\textbf{(D) }18\qquad \textbf{(E) }30$ | When Cheenu was a boy, he could run $15$ miles in $3$ hours and $30$ minutes $= 3\times60 + 30$ minutes $= 210$ minutes, thus running $\frac{210}{15} = 14$ minutes per mile. Now that he is an old man, he can walk $10$ miles in $4$ hours $= 4 \times 60$ minutes $= 240$ minutes, thus walking $\frac{240}{10} = 24$ minutes per mile. Therefore, it takes him $\boxed{10}$ minutes longer to walk a mile now compared to when he was a boy. | 10 |
334 | https://artofproblemsolving.com/wiki/index.php/2016_AMC_8_Problems/Problem_5 | 1 | The number $N$ is a two-digit number.
• When $N$ is divided by $9$ , the remainder is $1$
• When $N$ is divided by $10$ , the remainder is $3$
What is the remainder when $N$ is divided by $11$
$\textbf{(A) }0\qquad\textbf{(B) }2\qquad\textbf{(C) }4\qquad\textbf{(D) }5\qquad \textbf{(E) }7$ | From the second bullet point, we know that the second digit must be $3$ , for a number divisible by $10$ ends in zero. Since there is a remainder of $1$ when $N$ is divided by $9$ , the multiple of $9$ must end in a $2$ for it to have the desired remainder $\pmod {10}.$ We now look for this one:
$9(1)=9\\ 9(2)=18\\ 9(3)=27\\ 9(4)=36\\ 9(5)=45\\ 9(6)=54\\ 9(7)=63\\ 9(8)=72$
The number $72+1=73$ satisfies both conditions. We subtract the biggest multiple of $11$ less than $73$ to get the remainder. Thus, $73-11(6)=73-66=\boxed{7}$ | 7 |
335 | https://artofproblemsolving.com/wiki/index.php/2016_AMC_8_Problems/Problem_5 | 2 | The number $N$ is a two-digit number.
• When $N$ is divided by $9$ , the remainder is $1$
• When $N$ is divided by $10$ , the remainder is $3$
What is the remainder when $N$ is divided by $11$
$\textbf{(A) }0\qquad\textbf{(B) }2\qquad\textbf{(C) }4\qquad\textbf{(D) }5\qquad \textbf{(E) }7$ | This two digit number must take the form of $10x+y,$ where $x$ and $y$ are integers $0$ to $9.$ However, if x is an integer, we must have $y=3.$ So, the number's new form is $10x+3.$ This needs to have a remainder of $1$ when divided by $9.$ Because of the $9$ divisibility rule, we have \[10x+3 \equiv 1 \pmod 9.\] We subtract the three, getting \[10x \equiv -2 \pmod 9.\] which simplifies to \[10x \equiv 7 \pmod 9.\] However, $9x \equiv 0 \pmod 9,$ so \[10x - 9x \equiv 7 - 0 \pmod 9\] and \[x \equiv 7 \pmod 9.\]
Let the quotient of $9$ in our modular equation be $c,$ and let our desired number be $z,$ so $x=9c+7$ and $z = 10x+3.$ We substitute these values into $z = 10x+3,$ and get \[z = 10(9c+7) + 3\] so \[z = 90c+73.\] As a result, $z \equiv 73 \pmod {90}.$
To prove generalization vigorously, we can let $a$ be the remainder when $z$ is divided by $11.$ Setting up a modular equation, we have \[90c + 73 \equiv a \pmod {11}.\] Simplifying, \[90c+7 \equiv a \pmod {11}\] If $c = 1,$ then we don't have a 2 digit number! Thus, $c=0$ and $a=\boxed{7}$ | 7 |
336 | https://artofproblemsolving.com/wiki/index.php/2016_AMC_8_Problems/Problem_6 | 1 | The following bar graph represents the length (in letters) of the names of 19 people. What is the median length of these names? [asy] unitsize(0.9cm); draw((-0.5,0)--(10,0), linewidth(1.5)); draw((-0.5,1)--(10,1)); draw((-0.5,2)--(10,2)); draw((-0.5,3)--(10,3)); draw((-0.5,4)--(10,4)); draw((-0.5,5)--(10,5)); draw((-0.5,6)--(10,6)); draw((-0.5,7)--(10,7)); label("frequency",(-0.5,8)); label("0", (-1, 0)); label("1", (-1, 1)); label("2", (-1, 2)); label("3", (-1, 3)); label("4", (-1, 4)); label("5", (-1, 5)); label("6", (-1, 6)); label("7", (-1, 7)); filldraw((0,0)--(0,7)--(1,7)--(1,0)--cycle, black); filldraw((2,0)--(2,3)--(3,3)--(3,0)--cycle, black); filldraw((4,0)--(4,1)--(5,1)--(5,0)--cycle, black); filldraw((6,0)--(6,4)--(7,4)--(7,0)--cycle, black); filldraw((8,0)--(8,4)--(9,4)--(9,0)--cycle, black); label("3", (0.5, -0.5)); label("4", (2.5, -0.5)); label("5", (4.5, -0.5)); label("6", (6.5, -0.5)); label("7", (8.5, -0.5)); label("name length", (4.5, -1)); [/asy]
$\textbf{(A) }3\qquad\textbf{(B) }4\qquad\textbf{(C) }5\qquad\textbf{(D) }6\qquad \textbf{(E) }7$ | We first notice that the median name will be the $(19+1)/2=10^{\mbox{th}}$ name. The $10^{\mbox{th}}$ name is $\boxed{4}$ | 4 |
337 | https://artofproblemsolving.com/wiki/index.php/2016_AMC_8_Problems/Problem_6 | 2 | The following bar graph represents the length (in letters) of the names of 19 people. What is the median length of these names? [asy] unitsize(0.9cm); draw((-0.5,0)--(10,0), linewidth(1.5)); draw((-0.5,1)--(10,1)); draw((-0.5,2)--(10,2)); draw((-0.5,3)--(10,3)); draw((-0.5,4)--(10,4)); draw((-0.5,5)--(10,5)); draw((-0.5,6)--(10,6)); draw((-0.5,7)--(10,7)); label("frequency",(-0.5,8)); label("0", (-1, 0)); label("1", (-1, 1)); label("2", (-1, 2)); label("3", (-1, 3)); label("4", (-1, 4)); label("5", (-1, 5)); label("6", (-1, 6)); label("7", (-1, 7)); filldraw((0,0)--(0,7)--(1,7)--(1,0)--cycle, black); filldraw((2,0)--(2,3)--(3,3)--(3,0)--cycle, black); filldraw((4,0)--(4,1)--(5,1)--(5,0)--cycle, black); filldraw((6,0)--(6,4)--(7,4)--(7,0)--cycle, black); filldraw((8,0)--(8,4)--(9,4)--(9,0)--cycle, black); label("3", (0.5, -0.5)); label("4", (2.5, -0.5)); label("5", (4.5, -0.5)); label("6", (6.5, -0.5)); label("7", (8.5, -0.5)); label("name length", (4.5, -1)); [/asy]
$\textbf{(A) }3\qquad\textbf{(B) }4\qquad\textbf{(C) }5\qquad\textbf{(D) }6\qquad \textbf{(E) }7$ | To find the median length of a name from a bar graph, we must add up the number of names. Doing so gives us $7 + 3 + 1 + 4 + 4 = 19$ . Thus the index of the median length would be the 10th name. Since there are $7$ names with length $3$ , and $3$ names with length $4$ , the $10$ th name would have $4$ letters. Thus our answer is $\boxed{4}$ | 4 |
338 | https://artofproblemsolving.com/wiki/index.php/2016_AMC_8_Problems/Problem_8 | 1 | Find the value of the expression \[100-98+96-94+92-90+\cdots+8-6+4-2.\] $\textbf{(A) }20\qquad\textbf{(B) }40\qquad\textbf{(C) }50\qquad\textbf{(D) }80\qquad \textbf{(E) }100$ | We can group each subtracting pair together: \[(100-98)+(96-94)+(92-90)+ \ldots +(8-6)+(4-2).\] After subtracting, we have: \[2+2+2+\ldots+2+2=2(1+1+1+\ldots+1+1).\] There are $50$ even numbers, therefore there are $\dfrac{50}{2}=25$ even pairs. Therefore the sum is $2 \cdot 25=\boxed{50}$ | 50 |
339 | https://artofproblemsolving.com/wiki/index.php/2016_AMC_8_Problems/Problem_8 | 2 | Find the value of the expression \[100-98+96-94+92-90+\cdots+8-6+4-2.\] $\textbf{(A) }20\qquad\textbf{(B) }40\qquad\textbf{(C) }50\qquad\textbf{(D) }80\qquad \textbf{(E) }100$ | Since our list does not end with one, we divide every number by 2 and we end up with \[50-49+48-47+ \ldots +4-3+2-1\] We can group each subtracting pair together: \[(50-49)+(48-47)+(46-45)+ \ldots +(4-3)+(2-1).\] There are now $25$ pairs of numbers, and the value of each pair is $1$ . This sum is $25$ . However, we divided by $2$ originally so we will multiply $2*25$ to get the final answer of $\boxed{50}$ | 50 |
340 | https://artofproblemsolving.com/wiki/index.php/2016_AMC_8_Problems/Problem_9 | 1 | What is the sum of the distinct prime integer divisors of $2016$
$\textbf{(A) }9\qquad\textbf{(B) }12\qquad\textbf{(C) }16\qquad\textbf{(D) }49\qquad \textbf{(E) }63$ | The prime factorization is $2016=2^5\times3^2\times7$ . Since the problem is only asking us for the distinct prime factors, we have $2,3,7$ . Their desired sum is then $\boxed{12}$ | 12 |
341 | https://artofproblemsolving.com/wiki/index.php/2016_AMC_8_Problems/Problem_9 | 2 | What is the sum of the distinct prime integer divisors of $2016$
$\textbf{(A) }9\qquad\textbf{(B) }12\qquad\textbf{(C) }16\qquad\textbf{(D) }49\qquad \textbf{(E) }63$ | We notice that $9 \mid 2016$ , since $2+0+1+6 = 9$ , and $9 \mid 9$ . We can divide $2016$ by $9$ to get $224$ . This is divisible by $4$ , as $4 \mid 24$ . Dividing $224$ by $4$ , we have $56$ . This is clearly divisible by $7$ , leaving $8$ . We have $2016 = 9\cdot 4\cdot 7\cdot 8$ . We know that $4$ and $8$ are both multiples of $2$ $9$ is $3^2$ , and $7$ is prime. This means that the distinct prime factors are $2,3,$ and $7$ . Their sum is $\boxed{12}$ | 12 |
342 | https://artofproblemsolving.com/wiki/index.php/2016_AMC_8_Problems/Problem_10 | 1 | Suppose that $a * b$ means $3a-b.$ What is the value of $x$ if \[2 * (5 * x)=1\] $\textbf{(A) }\frac{1}{10} \qquad\textbf{(B) }2\qquad\textbf{(C) }\frac{10}{3} \qquad\textbf{(D) }10\qquad \textbf{(E) }14$ | Let us plug in $(5 * x)=1$ into $3a-b$ . Thus it would be $3(5)-x$ . Now we have $2*(15-x)=1$ . Plugging $2*(15-x)$ into $3a-b$ , we have $6-15+x=1$ . Solving for $x$ we have \[-9+x=1\] \[x=\boxed{10}\] | 10 |
343 | https://artofproblemsolving.com/wiki/index.php/2016_AMC_8_Problems/Problem_10 | 2 | Suppose that $a * b$ means $3a-b.$ What is the value of $x$ if \[2 * (5 * x)=1\] $\textbf{(A) }\frac{1}{10} \qquad\textbf{(B) }2\qquad\textbf{(C) }\frac{10}{3} \qquad\textbf{(D) }10\qquad \textbf{(E) }14$ | Let us set a variable $y$ equal to $5 * x$ . Solving for y in the equation $3(2)-y=1$ , we see that y is equal to five. By substitution, we see that $5 * x$ = 5. Solving for x in the equation $5(3)-x = 5$ we get \[x=\boxed{10}\] | 10 |
344 | https://artofproblemsolving.com/wiki/index.php/2016_AMC_8_Problems/Problem_11 | 1 | Determine how many two-digit numbers satisfy the following property: when the number is added to the number obtained by reversing its digits, the sum is $132.$
$\textbf{(A) }5\qquad\textbf{(B) }7\qquad\textbf{(C) }9\qquad\textbf{(D) }11\qquad \textbf{(E) }12$ | We can write the two digit number in the form of $10a+b$ ; reverse of $10a+b$ is $10b+a$ . The sum of those numbers is: \[(10a+b)+(10b+a)=132\] \[11a+11b=132\] \[a+b=12\] We can use brute force to find order pairs $(a,b)$ such that $a+b=12$ . Since $a$ and $b$ are both digits, both $a$ and $b$ have to be integers less than $10$ . Thus, our ordered pairs are $(3,9); (4,8); (5,7); (6,6); (7,5); (8,4); (9,3)$ ; or $\boxed{7}$ ordered pairs. | 7 |
345 | https://artofproblemsolving.com/wiki/index.php/2016_AMC_8_Problems/Problem_11 | 2 | Determine how many two-digit numbers satisfy the following property: when the number is added to the number obtained by reversing its digits, the sum is $132.$
$\textbf{(A) }5\qquad\textbf{(B) }7\qquad\textbf{(C) }9\qquad\textbf{(D) }11\qquad \textbf{(E) }12$ | Since the numbers are “mirror images,” their average has to be $\frac{132}{2}=66$ . The highest possible value for the tens digit is $9$ because it is a two-digit number. $9-6=3$ and $6-3=3$ , so our lowest tens digit is $3$ . The numbers between $9$ and $3$ inclusive is $9-3+1=\boxed{7}$ total possibilities. | 7 |
346 | https://artofproblemsolving.com/wiki/index.php/2016_AMC_8_Problems/Problem_14 | 1 | Karl's car uses a gallon of gas every $35$ miles, and his gas tank holds $14$ gallons when it is full. One day, Karl started with a full tank of gas, drove $350$ miles, bought $8$ gallons of gas, and continued driving to his destination. When he arrived, his gas tank was half full. How many miles did Karl drive that day?
$\textbf{(A) }525\qquad\textbf{(B) }560\qquad\textbf{(C) }595\qquad\textbf{(D) }665\qquad \textbf{(E) }735$ | Since he uses a gallon of gas every $35$ miles, he had used $\frac{350}{35} = 10$ gallons after $350$ miles. Therefore, after the first leg of his trip he had $14 - 10 = 4$ gallons of gas left. Then, he bought $8$ gallons of gas, which brought him up to $12$ gallons of gas in his gas tank. When he arrived, he had $\frac{1}{2} \cdot 14 = 7$ gallons of gas. So he used $5$ gallons of gas on the second leg of his trip. Therefore, the second part of his trip covered $5 \cdot 35 = 175$ miles. Adding this to the $350$ miles, we see that he drove $350 + 175 = \boxed{525}$ miles. | 525 |
347 | https://artofproblemsolving.com/wiki/index.php/2016_AMC_8_Problems/Problem_15 | 1 | What is the largest power of $2$ that is a divisor of $13^4 - 11^4$
$\textbf{(A)}\mbox{ }8\qquad \textbf{(B)}\mbox{ }16\qquad \textbf{(C)}\mbox{ }32\qquad \textbf{(D)}\mbox{ }64\qquad \textbf{(E)}\mbox{ }128$ | First, we use difference of squares on $13^4 - 11^4 = (13^2)^2 - (11^2)^2$ to get $13^4 - 11^4 = (13^2 + 11^2)(13^2 - 11^2)$ . Using difference of squares again and simplifying, we get $(169 + 121)(13+11)(13-11) = 290 \cdot 24 \cdot 2 = (2\cdot 8 \cdot 2) \cdot (3 \cdot 145)$ . Realizing that we don't need the right-hand side because it doesn't contain any factor of 2, we see that the greatest power of $2$ that is a divisor $13^4 - 11^4$ is $\boxed{32}$ | 32 |
348 | https://artofproblemsolving.com/wiki/index.php/2016_AMC_8_Problems/Problem_15 | 2 | What is the largest power of $2$ that is a divisor of $13^4 - 11^4$
$\textbf{(A)}\mbox{ }8\qquad \textbf{(B)}\mbox{ }16\qquad \textbf{(C)}\mbox{ }32\qquad \textbf{(D)}\mbox{ }64\qquad \textbf{(E)}\mbox{ }128$ | Just like in the above solution, we use the difference-of-squares factorization, but only once to get $13^4-11^4=(13^2-11^2)(13^2+11^2).$ We can then compute that this is equal to $48\cdot290.$ Note that $290=2\cdot145$ (we don't need to factorize any further as $145$ is already odd) thus the largest power of $2$ that divides $290$ is only $2^1=2,$ while $48=2^4\cdot3,$ so the largest power of $2$ that divides $48$ is $2^4=16.$ Hence, the largest power of $2$ that is a divisor of $13^4-11^4$ is $2\cdot16=\boxed{32}.$ | 32 |
349 | https://artofproblemsolving.com/wiki/index.php/2016_AMC_8_Problems/Problem_15 | 3 | What is the largest power of $2$ that is a divisor of $13^4 - 11^4$
$\textbf{(A)}\mbox{ }8\qquad \textbf{(B)}\mbox{ }16\qquad \textbf{(C)}\mbox{ }32\qquad \textbf{(D)}\mbox{ }64\qquad \textbf{(E)}\mbox{ }128$ | Let $n=13^4-11^4.$ We wish to find the largest power of $2$ that divides $n$
Denote $v_p(k)$ as the largest exponent of $p$ in the prime factorization of $n$ . In this problem, we have $p=2$
By the Lifting the Exponent Lemma on $n$
\[v_2(13^4-11^4)=v_2(13-11)+v_2(4)+v_2(13+11)-1\] \[=v_2(2)+v_2(4)+v_2(24)-1\] \[=1+2+3-1=5.\]
Therefore, exponent of the largest power of $2$ that divids $13^4-11^4$ is $5,$ so the largest power of $2$ that divides this number is $2^5=\boxed{32}$ | 32 |
350 | https://artofproblemsolving.com/wiki/index.php/2016_AMC_8_Problems/Problem_15 | 4 | What is the largest power of $2$ that is a divisor of $13^4 - 11^4$
$\textbf{(A)}\mbox{ }8\qquad \textbf{(B)}\mbox{ }16\qquad \textbf{(C)}\mbox{ }32\qquad \textbf{(D)}\mbox{ }64\qquad \textbf{(E)}\mbox{ }128$ | We can simply take 13 to the 4th power, which is 28561. We subtract that by 11 to the 4th power, which is 14641 (You can use Pascal's Triangle to find this). Finally, subtract the numbers to get 13920.
To test the options, since we need the largest one, we can go from top down. Testing, we see that both D and E are decimals,
and 32 works. So, our answer is $\boxed{32}.$ | 32 |
351 | https://artofproblemsolving.com/wiki/index.php/2016_AMC_8_Problems/Problem_16 | 1 | Annie and Bonnie are running laps around a $400$ -meter oval track. They started together, but Annie has pulled ahead, because she runs $25\%$ faster than Bonnie. How many laps will Annie have run when she first passes Bonnie?
$\textbf{(A) }1\dfrac{1}{4}\qquad\textbf{(B) }3\dfrac{1}{3}\qquad\textbf{(C) }4\qquad\textbf{(D) }5\qquad \textbf{(E) }25$ | Each lap Bonnie runs, Annie runs another quarter lap, so Bonnie will run four laps before she is overtaken. This means that Annie and Bonnie are equal so that Annie needs to run another lap to overtake Bonnie. That means Annie will have run $\boxed{5}$ laps. | 5 |
352 | https://artofproblemsolving.com/wiki/index.php/2016_AMC_8_Problems/Problem_16 | 2 | Annie and Bonnie are running laps around a $400$ -meter oval track. They started together, but Annie has pulled ahead, because she runs $25\%$ faster than Bonnie. How many laps will Annie have run when she first passes Bonnie?
$\textbf{(A) }1\dfrac{1}{4}\qquad\textbf{(B) }3\dfrac{1}{3}\qquad\textbf{(C) }4\qquad\textbf{(D) }5\qquad \textbf{(E) }25$ | Call $x$ the distance Annie runs. If Annie is $25\%$ faster than Bonnie, then Bonnie will run a distance of $\frac{4}{5}x$ . For Annie to meet Bonnie, she must run an extra $400$ meters, the length of the track. So $x-\left(\frac{4}{5}\right)x=400 \implies x=2000$ , which is $\boxed{5}$ laps. | 5 |
353 | https://artofproblemsolving.com/wiki/index.php/2016_AMC_8_Problems/Problem_17 | 1 | An ATM password at Fred's Bank is composed of four digits from $0$ to $9$ , with repeated digits allowable. If no password may begin with the sequence $9,1,1,$ then how many passwords are possible?
$\textbf{(A)}\mbox{ }30\qquad\textbf{(B)}\mbox{ }7290\qquad\textbf{(C)}\mbox{ }9000\qquad\textbf{(D)}\mbox{ }9990\qquad\textbf{(E)}\mbox{ }9999$ | For the first three digits, there are $10^3-1=999$ combinations since $911$ is not allowed. For the final digit, any of the $10$ numbers are allowed. $999 \cdot 10 = 9990 \rightarrow \boxed{9990}$ | 990 |
354 | https://artofproblemsolving.com/wiki/index.php/2016_AMC_8_Problems/Problem_17 | 2 | An ATM password at Fred's Bank is composed of four digits from $0$ to $9$ , with repeated digits allowable. If no password may begin with the sequence $9,1,1,$ then how many passwords are possible?
$\textbf{(A)}\mbox{ }30\qquad\textbf{(B)}\mbox{ }7290\qquad\textbf{(C)}\mbox{ }9000\qquad\textbf{(D)}\mbox{ }9990\qquad\textbf{(E)}\mbox{ }9999$ | Counting the prohibited cases, we find that there are 10 of them. This is because, when we start with 9,1, and 1, we can have any of the 10 digits for the last digit. So, our answer is $10^4-10=\boxed{9990}.$ | 990 |
355 | https://artofproblemsolving.com/wiki/index.php/2016_AMC_8_Problems/Problem_18 | 1 | In an All-Area track meet, $216$ sprinters enter a $100-$ meter dash competition. The track has $6$ lanes, so only $6$ sprinters can compete at a time. At the end of each race, the five non-winners are eliminated, and the winner will compete again in a later race. How many races are needed to determine the champion sprinter?
$\textbf{(A)}\mbox{ }36\qquad\textbf{(B)}\mbox{ }42\qquad\textbf{(C)}\mbox{ }43\qquad\textbf{(D)}\mbox{ }60\qquad\textbf{(E)}\mbox{ }72$ | From any $n-$ th race, only $\frac{1}{6}$ will continue on. Since we wish to find the total number of races, a column representing the races over time is ideal.
Starting with the first race: \[\frac{216}{6}=36\] \[\frac{36}{6}=6\] \[\frac{6}{6}=1\] Adding all of the numbers in the second column yields $\boxed{43}$ | 43 |
356 | https://artofproblemsolving.com/wiki/index.php/2016_AMC_8_Problems/Problem_18 | 2 | In an All-Area track meet, $216$ sprinters enter a $100-$ meter dash competition. The track has $6$ lanes, so only $6$ sprinters can compete at a time. At the end of each race, the five non-winners are eliminated, and the winner will compete again in a later race. How many races are needed to determine the champion sprinter?
$\textbf{(A)}\mbox{ }36\qquad\textbf{(B)}\mbox{ }42\qquad\textbf{(C)}\mbox{ }43\qquad\textbf{(D)}\mbox{ }60\qquad\textbf{(E)}\mbox{ }72$ | Every race eliminates $5$ players. The winner is decided when there is only $1$ runner left. You can construct the equation: $216$ $5x$ $1$ . Thus, $215$ players have to be eliminated. Therefore, we need $\frac{215}{5}$ games to decide the winner, or $\boxed{43}$ | 43 |
357 | https://artofproblemsolving.com/wiki/index.php/2016_AMC_8_Problems/Problem_19 | 1 | The sum of $25$ consecutive even integers is $10,000$ . What is the largest of these $25$ consecutive integers?
$\textbf{(A)}\mbox{ }360\qquad\textbf{(B)}\mbox{ }388\qquad\textbf{(C)}\mbox{ }412\qquad\textbf{(D)}\mbox{ }416\qquad\textbf{(E)}\mbox{ }424$ | Let $n$ be the 13th consecutive even integer that's being added up. By doing this, we can see that the sum of all 25 even numbers will simplify to $25n$ since $(n-2k)+\dots+(n-4)+(n-2)+(n)+(n+2)+(n+4)+ \dots +(n+2k)=25n$ . Now, $25n=10000 \rightarrow n=400$ . Remembering that this is the 13th integer, we wish to find the 25th, which is $400+2(25-13)=\boxed{424}$ | 424 |
358 | https://artofproblemsolving.com/wiki/index.php/2016_AMC_8_Problems/Problem_19 | 2 | The sum of $25$ consecutive even integers is $10,000$ . What is the largest of these $25$ consecutive integers?
$\textbf{(A)}\mbox{ }360\qquad\textbf{(B)}\mbox{ }388\qquad\textbf{(C)}\mbox{ }412\qquad\textbf{(D)}\mbox{ }416\qquad\textbf{(E)}\mbox{ }424$ | Let $x$ be the largest number. Then, $x+(x-2)+(x-4)+\cdots +(x-48)=10000$ . Factoring this gives $2\left(\frac{x}{2} + \left(\frac{x}{2} - 1\right) + \left(\frac{x}{2} - 2\right) +\cdots + \left(\frac{x}{2} - 24\right)\right)=10,000$ . Grouping like terms gives $25\left(\frac{x}{2}\right) - 300=5000$ , and continuing down the line, we find $x=\boxed{424}$ | 424 |
359 | https://artofproblemsolving.com/wiki/index.php/2016_AMC_8_Problems/Problem_19 | 3 | The sum of $25$ consecutive even integers is $10,000$ . What is the largest of these $25$ consecutive integers?
$\textbf{(A)}\mbox{ }360\qquad\textbf{(B)}\mbox{ }388\qquad\textbf{(C)}\mbox{ }412\qquad\textbf{(D)}\mbox{ }416\qquad\textbf{(E)}\mbox{ }424$ | Let $x$ be the smallest number. The equation will become, $x+(x+2)+(x+4)+\cdots +(x+48)=10,000$ . After you combine like terms, you get $25x+(50*12)=10,000$ which turns into $10,000-600=25x$ $25x=9400$ , so $x=376$ . Then, you add $376+48 = \boxed{424}$ | 424 |
360 | https://artofproblemsolving.com/wiki/index.php/2016_AMC_8_Problems/Problem_19 | 4 | The sum of $25$ consecutive even integers is $10,000$ . What is the largest of these $25$ consecutive integers?
$\textbf{(A)}\mbox{ }360\qquad\textbf{(B)}\mbox{ }388\qquad\textbf{(C)}\mbox{ }412\qquad\textbf{(D)}\mbox{ }416\qquad\textbf{(E)}\mbox{ }424$ | Dividing the series by $2$ , we get that the sum of $25$ consecutive integers is $5000$ . Let the middle number be $k$ we know that the sum is $25k$ , so $25k=5000$ . Solving, $k=200$ $2k=400$ is the middle term of the original sequence, so the original last term is $400+\frac{25-1}{2}\cdot 2=424$ . So the answer is $\boxed{424}$ | 424 |
361 | https://artofproblemsolving.com/wiki/index.php/2016_AMC_8_Problems/Problem_20 | 1 | The least common multiple of $a$ and $b$ is $12$ , and the least common multiple of $b$ and $c$ is $15$ . What is the least possible value of the least common multiple of $a$ and $c$
$\textbf{(A) }20\qquad\textbf{(B) }30\qquad\textbf{(C) }60\qquad\textbf{(D) }120\qquad \textbf{(E) }180$ | We wish to find possible values of $a$ $b$ , and $c$ . By finding the greatest common factor of $12$ and $15$ , we can find that $b$ is 3. Moving on to $a$ and $c$ , in order to minimize them, we wish to find the least such that the least common multiple of $a$ and $3$ is $12$ $\rightarrow 4$ . Similarly, with $3$ and $c$ , we obtain $5$ . The least common multiple of $4$ and $5$ is $20 \rightarrow \boxed{20}$ | 20 |
362 | https://artofproblemsolving.com/wiki/index.php/2016_AMC_8_Problems/Problem_22 | 1 | Rectangle $DEFA$ below is a $3 \times 4$ rectangle with $DC=CB=BA=1$ . The area of the "bat wings" (shaded area) is
[asy] draw((0,0)--(3,0)--(3,4)--(0,4)--(0,0)--(2,4)--(3,0)); draw((3,0)--(1,4)--(0,0)); fill((0,0)--(1,4)--(1.5,3)--cycle, black); fill((3,0)--(2,4)--(1.5,3)--cycle, black); label("$A$",(3.05,4.2)); label("$B$",(2,4.2)); label("$C$",(1,4.2)); label("$D$",(0,4.2)); label("$E$", (0,-0.2)); label("$F$", (3,-0.2)); label("$1$", (0.5, 4), N); label("$1$", (1.5, 4), N); label("$1$", (2.5, 4), N); label("$4$", (3.2, 2), E); [/asy]
$\textbf{(A) }2\qquad\textbf{(B) }2 \frac{1}{2}\qquad\textbf{(C) }3\qquad\textbf{(D) }3 \frac{1}{2}\qquad \textbf{(E) }5$ | Let G be the midpoint B and C
Draw H, J, K beneath C, G, B, respectively.
[asy] draw((0,0)--(3,0)--(3,4)--(0,4)--(0,0)--(2,4)--(3,0)); draw((3,0)--(1,4)--(0,0)); fill((0,0)--(1,4)--(1.5,3)--cycle, grey); fill((3,0)--(2,4)--(1.5,3)--cycle, grey); draw((1,0)--(1,4)); draw((1.5,0)--(1.5,4)); draw((2,0)--(2,4)); label("$A$",(3.05,4.2)); label("$B$",(2,4.2)); label("$C$",(1,4.2)); label("$D$",(0,4.2)); label("$E$", (0,-0.2)); label("$F$", (3,-0.2)); label("$G$", (1.5, 4.2)); label("$H$", (1, -0.2)); label("$J$", (1.5, -0.2)); label("$K$", (2, -0.2)); label("$1$", (0.5, 4), N); label("$1$", (2.5, 4), N); label("$4$", (3.2, 2), E); [/asy]
Let us take a look at rectangle CDEH. I have labeled E' for convenience. First of all, we can see that EE'H and CE'B are similar triangles because all their three angles are the same. Furthermore, since EH=CB, we can confirm that EE'H and CE'B are identical triangles. Thus, CE'=E'H, which is half of CH.
[asy] fill((0,0)--(1,4)--(1,2)--cycle, grey); draw((0,0)--(1,0)--(1,4)--(0,4)--(0,0)); draw((0,0)--(1,4)--(1,2)--(0,0)); label("$C$",(1,4.2)); label("$D$",(0,4.2)); label("$E$", (0,-0.2)); label("$H$", (1, -0.2)); label("$E'$", (1.2, 2)); [/asy]
Then we can see that CEE' has $\frac{1}{4}$ the area of CDEH because it has half the base and the same height. Similarly, the shaded regions in CGJH, BGJK, and ABKF all have $\frac{1}{4}$ the area of their rectangle. So, the total shaded region is just $\frac{1}{4}$ the area of the total region, or $\frac{1}{4} \times 3 \times 4$ , or $\boxed{3}$ | 3 |
363 | https://artofproblemsolving.com/wiki/index.php/2016_AMC_8_Problems/Problem_22 | 2 | Rectangle $DEFA$ below is a $3 \times 4$ rectangle with $DC=CB=BA=1$ . The area of the "bat wings" (shaded area) is
[asy] draw((0,0)--(3,0)--(3,4)--(0,4)--(0,0)--(2,4)--(3,0)); draw((3,0)--(1,4)--(0,0)); fill((0,0)--(1,4)--(1.5,3)--cycle, black); fill((3,0)--(2,4)--(1.5,3)--cycle, black); label("$A$",(3.05,4.2)); label("$B$",(2,4.2)); label("$C$",(1,4.2)); label("$D$",(0,4.2)); label("$E$", (0,-0.2)); label("$F$", (3,-0.2)); label("$1$", (0.5, 4), N); label("$1$", (1.5, 4), N); label("$1$", (2.5, 4), N); label("$4$", (3.2, 2), E); [/asy]
$\textbf{(A) }2\qquad\textbf{(B) }2 \frac{1}{2}\qquad\textbf{(C) }3\qquad\textbf{(D) }3 \frac{1}{2}\qquad \textbf{(E) }5$ | The area of trapezoid $CBFE$ is $\frac{1+3}2\cdot 4=8$ . Next, we find the height of each triangle to calculate their area. The two non-colored isosceles triangles are similar, and are in a $3:1$ ratio by AA similarity (alternate interior and vertical angles) so the height of the larger is $3,$ while the height of the smaller one is $1.$ Thus, their areas are $\frac12$ and $\frac92$ . Subtracting these areas from the trapezoid, we get $8-\frac12-\frac92 =\boxed3$ . Therefore, the answer to this problem is $\boxed{3}$ | 3 |
364 | https://artofproblemsolving.com/wiki/index.php/2016_AMC_8_Problems/Problem_22 | 3 | Rectangle $DEFA$ below is a $3 \times 4$ rectangle with $DC=CB=BA=1$ . The area of the "bat wings" (shaded area) is
[asy] draw((0,0)--(3,0)--(3,4)--(0,4)--(0,0)--(2,4)--(3,0)); draw((3,0)--(1,4)--(0,0)); fill((0,0)--(1,4)--(1.5,3)--cycle, black); fill((3,0)--(2,4)--(1.5,3)--cycle, black); label("$A$",(3.05,4.2)); label("$B$",(2,4.2)); label("$C$",(1,4.2)); label("$D$",(0,4.2)); label("$E$", (0,-0.2)); label("$F$", (3,-0.2)); label("$1$", (0.5, 4), N); label("$1$", (1.5, 4), N); label("$1$", (2.5, 4), N); label("$4$", (3.2, 2), E); [/asy]
$\textbf{(A) }2\qquad\textbf{(B) }2 \frac{1}{2}\qquad\textbf{(C) }3\qquad\textbf{(D) }3 \frac{1}{2}\qquad \textbf{(E) }5$ | Set coordinates to the points:
Let $E=(0,0)$ $F=(3,0)$
[asy] draw((0,0)--(3,0)--(3,4)--(0,4)--(0,0)--(2,4)--(3,0)); draw((3,0)--(1,4)--(0,0)); fill((0,0)--(1,4)--(1.5,3)--cycle, black); fill((3,0)--(2,4)--(1.5,3)--cycle, black); label(scale(0.7)*"$A(3,4)$",(3.25,4.2)); label(scale(0.7)*"$B(2,4)$",(2.1,4.2)); label(scale(0.7)*"$C(1,4)$",(0.9,4.2)); label(scale(0.7)*"$D(0,4)$",(-0.3,4.2)); label(scale(0.7)*"$E(0,0)$", (0,-0.2)); label(scale(0.7)*"$Z(\frac{3}{2},3)$", (1.5,1.8)); label(scale(0.7)*"$F(3,0)$", (3,-0.2)); label(scale(0.7)*"$1$", (0.3, 4), N); label(scale(0.7)*"$1$", (1.5, 4), N); label(scale(0.7)*"$1$", (2.7, 4), N); label(scale(0.7)*"$4$", (3.2, 2), E); [/asy]
Now, we easily discover that line $CF$ has lattice coordinates at $(1,4)$ and $(3,0)$ . Hence, the slope of line $CF=-2$
Plugging in the rest of the coordinate points, we find that line $CF=-2x+6$
Doing the same process to line $BE$ , we find that line $BE=2x$
Hence, setting them equal to find the intersection point...
$y=2x=-2x+6\implies 4x=6\implies x=\frac{3}{2}\implies y=3$
Hence, we find that the intersection point is $(\frac{3}{2},3)$ . Call it Z.
Now, we can see that
$E=(0,0)$
$Z=(\dfrac{3}{2},3)$
$C=(1,4)$
Now use the Shoelace Theorem
$\frac{(0\cdot 3 + \dfrac{3}{2}\cdot 4 + 1\cdot 0)-(\dfrac{3}{2}\cdot 0 + 1\cdot 3 + 4\cdot 0)}{2} = \frac{6-3}{2} = \frac{3}{2}$
Using the Shoelace Theorem , we find that the area of one of those small shaded triangles is $\frac{3}{2}$
Now because there are two of them, we multiple that area by $2$ to get $\boxed{3}$ | 3 |
365 | https://artofproblemsolving.com/wiki/index.php/2016_AMC_8_Problems/Problem_22 | 4 | Rectangle $DEFA$ below is a $3 \times 4$ rectangle with $DC=CB=BA=1$ . The area of the "bat wings" (shaded area) is
[asy] draw((0,0)--(3,0)--(3,4)--(0,4)--(0,0)--(2,4)--(3,0)); draw((3,0)--(1,4)--(0,0)); fill((0,0)--(1,4)--(1.5,3)--cycle, black); fill((3,0)--(2,4)--(1.5,3)--cycle, black); label("$A$",(3.05,4.2)); label("$B$",(2,4.2)); label("$C$",(1,4.2)); label("$D$",(0,4.2)); label("$E$", (0,-0.2)); label("$F$", (3,-0.2)); label("$1$", (0.5, 4), N); label("$1$", (1.5, 4), N); label("$1$", (2.5, 4), N); label("$4$", (3.2, 2), E); [/asy]
$\textbf{(A) }2\qquad\textbf{(B) }2 \frac{1}{2}\qquad\textbf{(C) }3\qquad\textbf{(D) }3 \frac{1}{2}\qquad \textbf{(E) }5$ | [asy] draw((0,0)--(3,0)--(3,4)--(0,4)--(0,0)--(2,4)--(3,0)); draw((3,0)--(1,4)--(0,0)); fill((0,0)--(1,4)--(1.5,3)--cycle, black); fill((3,0)--(2,4)--(1.5,3)--cycle, black); label("$A$",(3.05,4.2)); label("$B$",(2,4.2)); label("$C$",(1,4.2)); label("$D$",(0,4.2)); label("$E$", (0,-0.2)); label("$F$", (3,-0.2)); label("$G$", (1.5, 3.2), N); label("$1$", (0.5, 4), N); label("$1$", (1.5, 4), N); label("$1$", (2.5, 4), N); label("$4$", (3.2, 2), E); [/asy]
First, it is easy to see that $\triangle CGB \sim \triangle EGF$ . Therefore, the ratio of the height of $\triangle CBG$ to the height of $\triangle EFG$ is $\frac{1}{3}$ . Thus, the area of $\triangle CBG$ is $\frac{1\cdot1}{2} = \frac{1}{2}$ , and the area of $\triangle CBE$ is $\frac{1\cdot4}{2} = 2$ . So, the area of $\triangle CGE$ is $2-\frac{1}{2}$ . Besides, since trapezoid $CBEF$ is isosceles, $\triangle CGE \cong \triangle BGF$ . Hence, the area of the "bat wings" is $2\cdot(2-\frac{1}{2})= \boxed{3}$ | 3 |
366 | https://artofproblemsolving.com/wiki/index.php/2016_AMC_8_Problems/Problem_22 | 5 | Rectangle $DEFA$ below is a $3 \times 4$ rectangle with $DC=CB=BA=1$ . The area of the "bat wings" (shaded area) is
[asy] draw((0,0)--(3,0)--(3,4)--(0,4)--(0,0)--(2,4)--(3,0)); draw((3,0)--(1,4)--(0,0)); fill((0,0)--(1,4)--(1.5,3)--cycle, black); fill((3,0)--(2,4)--(1.5,3)--cycle, black); label("$A$",(3.05,4.2)); label("$B$",(2,4.2)); label("$C$",(1,4.2)); label("$D$",(0,4.2)); label("$E$", (0,-0.2)); label("$F$", (3,-0.2)); label("$1$", (0.5, 4), N); label("$1$", (1.5, 4), N); label("$1$", (2.5, 4), N); label("$4$", (3.2, 2), E); [/asy]
$\textbf{(A) }2\qquad\textbf{(B) }2 \frac{1}{2}\qquad\textbf{(C) }3\qquad\textbf{(D) }3 \frac{1}{2}\qquad \textbf{(E) }5$ | Solution 2/4 are easily better, but if you really wanted to you could use Pick's Theorem for each half of the "bat wings". Unfortunately it isn't immediately applicable since the point common to each bat wing does not lie on a lattice point. We can remedy this by pretending the figure is twice as big and at the end divide the area by 4 (since the area of similar shapes scales quadratically with the scaling factor).
[asy] // Original drawing code draw((0,0)--(3,0)--(3,4)--(0,4)--(0,0)--(2,4)--(3,0)); draw((3,0)--(1,4)--(0,0)); fill((0,0)--(1,4)--(1.5,3)--cycle, black); fill((3,0)--(2,4)--(1.5,3)--cycle, black); label("$A$",(3.05,4.2)); label("$B$",(2,4.2)); label("$C$",(1,4.2)); label("$D$",(0,4.2)); label("$E$", (0,-0.2)); label("$F$", (3,-0.2)); label("$2$", (0.5, 4), N); label("$2$", (1.5, 4), N); label("$2$", (2.5, 4), N); label("$8$", (3.2, 2), E); // Draw the grid lines for (real i=0.5; i<3; i+=0.5) { draw((i,0)--(i,4), gray+linewidth(0.5)); // Vertical grid lines } for (real j=0.5; j<4; j+=0.5) { draw((0,j)--(3,j), gray+linewidth(0.5)); // Horizontal grid lines } // Boundary points with green dots and black border filldraw(circle((0,0), 0.05), green, black+linewidth(0.5)); filldraw(circle((.5,1), 0.05), green, black+linewidth(0.5)); filldraw(circle((1,2), 0.05), green, black+linewidth(0.5)); filldraw(circle((1.5,3), 0.05), green, black+linewidth(0.5)); filldraw(circle((1,4), 0.05), green, black+linewidth(0.5)); filldraw(circle((.5,2), 0.05), green, black+linewidth(0.5)); // Interior points with red dots and black border filldraw(circle((.5,1.5), 0.05), red, black+linewidth(0.5)); filldraw(circle((1,2.5), 0.05), red, black+linewidth(0.5)); filldraw(circle((1,3), 0.05), red, black+linewidth(0.5)); filldraw(circle((1,3.5), 0.05), red, black+linewidth(0.5)); [/asy]
Now we can safely use Pick's Theorem on the scaled-up wings:
\[A'=2\left(\frac{\textcolor{green}{b}}{2}+\textcolor{red}{i}-1\right)=2\left(\frac{6}{2}+4-1\right)=12\]
And finally we scale this down to get the original area:
\[A=\frac14A'=\frac14 12=\boxed{3}\] | 3 |
367 | https://artofproblemsolving.com/wiki/index.php/2016_AMC_8_Problems/Problem_23 | 1 | Two congruent circles centered at points $A$ and $B$ each pass through the other circle's center. The line containing both $A$ and $B$ is extended to intersect the circles at points $C$ and $D$ . The circles intersect at two points, one of which is $E$ . What is the degree measure of $\angle CED$
$\textbf{(A) }90\qquad\textbf{(B) }105\qquad\textbf{(C) }120\qquad\textbf{(D) }135\qquad \textbf{(E) }150$ | Observe that $\triangle{EAB}$ is equilateral. Therefore, $m\angle{AEB}=m\angle{EAB}=m\angle{EBA} = 60^{\circ}$ . Since $CD$ is a straight line, we conclude that $m\angle{EBD} = 180^{\circ}-60^{\circ}=120^{\circ}$ . Since $BE=BD$ (both are radii of the same circle), $\triangle{BED}$ is isosceles, meaning that $m\angle{BED}=m\angle{BDE}=30^{\circ}$ . Similarly, $m\angle{AEC}=m\angle{ACE}=30^{\circ}$
Now, $\angle{CED}=m\angle{AEC}+m\angle{AEB}+m\angle{BED} = 30^{\circ}+60^{\circ}+30^{\circ} = 120^{\circ}$ . Therefore, the answer is $\boxed{120}$ | 120 |
368 | https://artofproblemsolving.com/wiki/index.php/2016_AMC_8_Problems/Problem_24 | 1 | The digits $1$ $2$ $3$ $4$ , and $5$ are each used once to write a five-digit number $PQRST$ . The three-digit number $PQR$ is divisible by $4$ , the three-digit number $QRS$ is divisible by $5$ , and the three-digit number $RST$ is divisible by $3$ . What is $P$
$\textbf{(A) }1\qquad\textbf{(B) }2\qquad\textbf{(C) }3\qquad\textbf{(D) }4\qquad \textbf{(E) }5$ | We see that since $QRS$ is divisible by $5$ $S$ must equal either $0$ or $5$ , but it cannot equal $0$ , so $S=5$ . We notice that since $PQR$ must be even, $R$ must be either $2$ or $4$ . However, when $R=2$ , we see that $T \equiv 2 \pmod{3}$ , which cannot happen because $2$ and $5$ are already used up; so $R=4$ . This gives $T \equiv 3 \pmod{4}$ , meaning $T=3$ . Now, we see that $Q$ could be either $1$ or $2$ , but $14$ is not divisible by $4$ , but $24$ is. This means that $Q=2$ and $P=\boxed{1}$ | 1 |
369 | https://artofproblemsolving.com/wiki/index.php/2016_AMC_8_Problems/Problem_24 | 2 | The digits $1$ $2$ $3$ $4$ , and $5$ are each used once to write a five-digit number $PQRST$ . The three-digit number $PQR$ is divisible by $4$ , the three-digit number $QRS$ is divisible by $5$ , and the three-digit number $RST$ is divisible by $3$ . What is $P$
$\textbf{(A) }1\qquad\textbf{(B) }2\qquad\textbf{(C) }3\qquad\textbf{(D) }4\qquad \textbf{(E) }5$ | We know that out of $PQRST,$ $QRS$ is divisible by $5$ . Therefore $S$ is obviously 5 because $QRS$ is divisible by 5. So we now have $PQR5T$ as our number. Next, let's move on to the second piece of information that was given to us. $RST$ is divisible by 3. So, according to the divisibility by 3 rule, the sum of $RST$ has to be a multiple of 3. The only 2 big enough are 9 and 12 and since 5 is already given. The possible sums of $RT$ are 4 and 7. So, the possible values for $R$ are 1,3,4,3 and the possible values of $T$ are 3,1,3,4. So, using this we can move on to the fact that $PQR$ is divisible by 4. So, using that we know that $R$ has to be even so 4 is the only possible value for $R$ . Using that we also know that 3 is the only possible value for 3. So, we have $PQRST$ $PQ453$ so the possible values are 1 and 2 for $P$ and $Q$ . Using the divisibility rule of 4 we know that $QR$ has to be divisible by 4. So, either 14 or 24 are the possibilities, and 24 is divisible by 4. So the only value left for $P$ is 1. $P=\boxed{1}$ | 1 |
370 | https://artofproblemsolving.com/wiki/index.php/2016_AMC_8_Problems/Problem_24 | 3 | The digits $1$ $2$ $3$ $4$ , and $5$ are each used once to write a five-digit number $PQRST$ . The three-digit number $PQR$ is divisible by $4$ , the three-digit number $QRS$ is divisible by $5$ , and the three-digit number $RST$ is divisible by $3$ . What is $P$
$\textbf{(A) }1\qquad\textbf{(B) }2\qquad\textbf{(C) }3\qquad\textbf{(D) }4\qquad \textbf{(E) }5$ | We know that $QRS$ is divisible by $5$ , so $S$ would be either $5$ or $0$ . However, $0$ is not a choice, so $S=5$ . Also, $PQR$ is divisible by $4$ , so this means that $QR$ is $12$ $32$ $24$ , or $52$ . If $R=2$ , then $T$ has to be $2$ or $5$ $RST$ is divisible by $3$ ), but both are taken. So, $R=4 \Rightarrow QR=24$ $R+S+T$ must equal $9$ or $12$ , but because $4+5=9$ $R+S+T=12 \Rightarrow T=3$ . This leaves $P=\boxed{1}$ | 1 |
371 | https://artofproblemsolving.com/wiki/index.php/2016_AMC_8_Problems/Problem_24 | 4 | The digits $1$ $2$ $3$ $4$ , and $5$ are each used once to write a five-digit number $PQRST$ . The three-digit number $PQR$ is divisible by $4$ , the three-digit number $QRS$ is divisible by $5$ , and the three-digit number $RST$ is divisible by $3$ . What is $P$
$\textbf{(A) }1\qquad\textbf{(B) }2\qquad\textbf{(C) }3\qquad\textbf{(D) }4\qquad \textbf{(E) }5$ | We can simply try each of the answer choice, and we will see which one works. Trying $P=\boxed{1}$ | 1 |
372 | https://artofproblemsolving.com/wiki/index.php/2015_AMC_8_Problems/Problem_1 | 1 | Onkon wants to cover his room's floor with his favourite red carpet. How many square yards of red carpet are required to cover a rectangular floor that is $12$ feet long and $9$ feet wide? (There are 3 feet in a yard.)
$\textbf{(A) }12\qquad\textbf{(B) }36\qquad\textbf{(C) }108\qquad\textbf{(D) }324\qquad \textbf{(E) }972$ | First, we multiply $12\cdot9$ . To get that, we need $108$ square feet of carpet to cover the room's floor. Since there are $9$ square feet in a square yard, you divide $108$ by $9$ to get $12$ square yards, so our answer is $\bold{\boxed{12}$ | 12 |
373 | https://artofproblemsolving.com/wiki/index.php/2015_AMC_8_Problems/Problem_1 | 2 | Onkon wants to cover his room's floor with his favourite red carpet. How many square yards of red carpet are required to cover a rectangular floor that is $12$ feet long and $9$ feet wide? (There are 3 feet in a yard.)
$\textbf{(A) }12\qquad\textbf{(B) }36\qquad\textbf{(C) }108\qquad\textbf{(D) }324\qquad \textbf{(E) }972$ | Since there are $3$ feet in a yard, we divide $9$ by $3$ to get $3$ , and $12$ by $3$ to get $4$ . To find the area of the carpet, we then multiply these two values together to get $\boxed{12}$ | 12 |
374 | https://artofproblemsolving.com/wiki/index.php/2015_AMC_8_Problems/Problem_3 | 1 | Jack and Jill are going swimming at a pool that is one mile from their house. They leave home simultaneously. Jill rides her bicycle to the pool at a constant speed of $10$ miles per hour. Jack walks to the pool at a constant speed of $4$ miles per hour. How many minutes before Jack does Jill arrive?
$\textbf{(A) }5\qquad\textbf{(B) }6\qquad\textbf{(C) }8\qquad\textbf{(D) }9\qquad \textbf{(E) }10$ | Using $d=rt$ , we can set up an equation for when Jill arrives at the swimming pool:
$1=10t$
Solving for $t$ , we get that Jill gets to the pool in $\frac{1}{10}$ of an hour, which is $6$ minutes. Doing the same for Jack, we get that
Jack arrives at the pool in $\frac{1}{4}$ of an hour, which in turn is $15$ minutes. Thus, Jill has to wait $15-6=\boxed{9}$ | 9 |
375 | https://artofproblemsolving.com/wiki/index.php/2015_AMC_8_Problems/Problem_4 | 1 | The Blue Bird High School chess team consists of two boys and three girls. A photographer wants to take a picture of the team to appear in the local newspaper. She decides to have them sit in a row with a boy at each end and the three girls in the middle. How many such arrangements are possible?
$\textbf{(A) }2\qquad\textbf{(B) }4\qquad\textbf{(C) }5\qquad\textbf{(D) }6\qquad \textbf{(E) }12$ | There are $2! = 2$ ways to order the boys on the ends, and there are $3!=6$ ways to order the girls in the middle. We get the answer to be $2 \cdot 6 = \boxed{12}$ | 12 |
376 | https://artofproblemsolving.com/wiki/index.php/2015_AMC_8_Problems/Problem_6 | 1 | In $\bigtriangleup ABC$ $AB=BC=29$ , and $AC=42$ . What is the area of $\bigtriangleup ABC$
$\textbf{(A) }100\qquad\textbf{(B) }420\qquad\textbf{(C) }500\qquad\textbf{(D) }609\qquad \textbf{(E) }701$ | We know the semi-perimeter of $\triangle ABC$ is $\frac{29+29+42}{2}=50$ . Next, we use Heron's Formula to find that the area of the triangle is just $\sqrt{50(50-29)^2(50-42)}=\sqrt{50 \cdot 21^2 \cdot 8}=\boxed{420}$ | 420 |
377 | https://artofproblemsolving.com/wiki/index.php/2015_AMC_8_Problems/Problem_6 | 2 | In $\bigtriangleup ABC$ $AB=BC=29$ , and $AC=42$ . What is the area of $\bigtriangleup ABC$
$\textbf{(A) }100\qquad\textbf{(B) }420\qquad\textbf{(C) }500\qquad\textbf{(D) }609\qquad \textbf{(E) }701$ | Splitting the isosceles triangle in half, we get a right triangle with hypotenuse $29$ and leg $21$ . Using the Pythagorean Theorem , we know the height is $\sqrt{29^2-21^2}=20$ . Now that we know the height, the area is $\dfrac{(20)(42)}{2} = \boxed{420}$ | 420 |
378 | https://artofproblemsolving.com/wiki/index.php/2015_AMC_8_Problems/Problem_8 | 1 | What is the smallest whole number larger than the perimeter of any triangle with a side of length $5$ and a side of length $19$
$\textbf{(A) }24\qquad\textbf{(B) }29\qquad\textbf{(C) }43\qquad\textbf{(D) }48\qquad \textbf{(E) }57$ | We know from the Triangle Inequality that the last side, $s$ , fulfills $s<5+19=24$ . Adding $5+19$ to both sides of the inequality, we get $s+5+19<48$ , and because $s+5+19$ is the perimeter of our triangle, $\boxed{48}$ is our answer. | 48 |
379 | https://artofproblemsolving.com/wiki/index.php/2015_AMC_8_Problems/Problem_9 | 1 | On her first day of work, Janabel sold one widget. On day two, she sold three widgets. On day three, she sold five widgets, and on each succeeding day, she sold two more widgets than she had sold on the previous day. How many widgets in total had Janabel sold after working $20$ days?
$\textbf{(A) }39\qquad\textbf{(B) }40\qquad\textbf{(C) }210\qquad\textbf{(D) }400\qquad \textbf{(E) }401$ | First, we have to find how many widgets she makes on Day $20$ . We can write the linear equation $y=-1+2x$ to represent this situation. Then, we can plug in $20$ for $x$ $y=-1+2(20)$ -- $y=-1+40$ -- $y=39$ . The sum of $1,3,5, ... 39$ is $\dfrac{(1 + 39)(20)}{2}= \boxed{400}$ | 400 |
380 | https://artofproblemsolving.com/wiki/index.php/2015_AMC_8_Problems/Problem_9 | 2 | On her first day of work, Janabel sold one widget. On day two, she sold three widgets. On day three, she sold five widgets, and on each succeeding day, she sold two more widgets than she had sold on the previous day. How many widgets in total had Janabel sold after working $20$ days?
$\textbf{(A) }39\qquad\textbf{(B) }40\qquad\textbf{(C) }210\qquad\textbf{(D) }400\qquad \textbf{(E) }401$ | The sum is just the first $20$ odd counting/natural numbers, which is $20^2=\boxed{400}$ | 400 |
381 | https://artofproblemsolving.com/wiki/index.php/2015_AMC_8_Problems/Problem_9 | 3 | On her first day of work, Janabel sold one widget. On day two, she sold three widgets. On day three, she sold five widgets, and on each succeeding day, she sold two more widgets than she had sold on the previous day. How many widgets in total had Janabel sold after working $20$ days?
$\textbf{(A) }39\qquad\textbf{(B) }40\qquad\textbf{(C) }210\qquad\textbf{(D) }400\qquad \textbf{(E) }401$ | We can easily find out she makes $2\cdot20-1 = 39$ widgets on Day $20$ . Then, we make the sum of $1,3, 5, ... ,35,37,39$ by adding in this way: $(1+39)+(3+37)+(5+35)+...+(19+21)$ , which include $10$ pairs of $40$ . So, the sum of $1,3,5, ...~39$ is $(40\cdot10)=\boxed{400}$ | 400 |
382 | https://artofproblemsolving.com/wiki/index.php/2015_AMC_8_Problems/Problem_12 | 1 | How many pairs of parallel edges, such as $\overline{AB}$ and $\overline{GH}$ or $\overline{EH}$ and $\overline{FG}$ , does a cube have?
[asy] import three; currentprojection=orthographic(1/2,-1,1/2); /* three - currentprojection, orthographic */ draw((0,0,0)--(1,0,0)--(1,1,0)--(0,1,0)--cycle); draw((0,0,0)--(0,0,1)); draw((0,1,0)--(0,1,1)); draw((1,1,0)--(1,1,1)); draw((1,0,0)--(1,0,1)); draw((0,0,1)--(1,0,1)--(1,1,1)--(0,1,1)--cycle); label("$D$",(0,0,0),S); label("$A$",(0,0,1),N); label("$H$",(0,1,0),S); label("$E$",(0,1,1),N); label("$C$",(1,0,0),S); label("$B$",(1,0,1),N); label("$G$",(1,1,0),S); label("$F$",(1,1,1),N); [/asy]
$\textbf{(A) }6\qquad\textbf{(B) }12\qquad\textbf{(C) }18\qquad\textbf{(D) }24\qquad \textbf{(E) }36$ | We first count the number of pairs of parallel lines that are in the same direction as $\overline{AB}$ . The pairs of parallel lines are $\overline{AB}\text{ and }\overline{EF}$ $\overline{CD}\text{ and }\overline{GH}$ $\overline{AB}\text{ and }\overline{CD}$ $\overline{EF}\text{ and }\overline{GH}$ $\overline{AB}\text{ and }\overline{GH}$ , and $\overline{CD}\text{ and }\overline{EF}$ . These are $6$ pairs total. We can do the same for the lines in the same direction as $\overline{AE}$ and $\overline{AD}$ . This means there are $6\cdot 3=\boxed{18}$ total pairs of parallel lines. | 18 |
383 | https://artofproblemsolving.com/wiki/index.php/2015_AMC_8_Problems/Problem_12 | 2 | How many pairs of parallel edges, such as $\overline{AB}$ and $\overline{GH}$ or $\overline{EH}$ and $\overline{FG}$ , does a cube have?
[asy] import three; currentprojection=orthographic(1/2,-1,1/2); /* three - currentprojection, orthographic */ draw((0,0,0)--(1,0,0)--(1,1,0)--(0,1,0)--cycle); draw((0,0,0)--(0,0,1)); draw((0,1,0)--(0,1,1)); draw((1,1,0)--(1,1,1)); draw((1,0,0)--(1,0,1)); draw((0,0,1)--(1,0,1)--(1,1,1)--(0,1,1)--cycle); label("$D$",(0,0,0),S); label("$A$",(0,0,1),N); label("$H$",(0,1,0),S); label("$E$",(0,1,1),N); label("$C$",(1,0,0),S); label("$B$",(1,0,1),N); label("$G$",(1,1,0),S); label("$F$",(1,1,1),N); [/asy]
$\textbf{(A) }6\qquad\textbf{(B) }12\qquad\textbf{(C) }18\qquad\textbf{(D) }24\qquad \textbf{(E) }36$ | Look at any edge, let's say $\overline{AB}$ . There are three ways we can pair $\overline{AB}$ with another edge. $\overline{AB}\text{ and }\overline{EF}$ $\overline{AB}\text{ and }\overline{HG}$ , and $\overline{AB}\text{ and }\overline{DC}$ . There are 12 edges on a cube. 3 times 12 is 36. We have to divide by 2 because every pair is counted twice, so $\frac{36}{2}$ is $\boxed{18}$ total pairs of parallel lines. | 18 |
384 | https://artofproblemsolving.com/wiki/index.php/2015_AMC_8_Problems/Problem_12 | 3 | How many pairs of parallel edges, such as $\overline{AB}$ and $\overline{GH}$ or $\overline{EH}$ and $\overline{FG}$ , does a cube have?
[asy] import three; currentprojection=orthographic(1/2,-1,1/2); /* three - currentprojection, orthographic */ draw((0,0,0)--(1,0,0)--(1,1,0)--(0,1,0)--cycle); draw((0,0,0)--(0,0,1)); draw((0,1,0)--(0,1,1)); draw((1,1,0)--(1,1,1)); draw((1,0,0)--(1,0,1)); draw((0,0,1)--(1,0,1)--(1,1,1)--(0,1,1)--cycle); label("$D$",(0,0,0),S); label("$A$",(0,0,1),N); label("$H$",(0,1,0),S); label("$E$",(0,1,1),N); label("$C$",(1,0,0),S); label("$B$",(1,0,1),N); label("$G$",(1,1,0),S); label("$F$",(1,1,1),N); [/asy]
$\textbf{(A) }6\qquad\textbf{(B) }12\qquad\textbf{(C) }18\qquad\textbf{(D) }24\qquad \textbf{(E) }36$ | We can use the feature of 3-Dimension in a cube to solve the problem systematically. For example, in the 3-D of the cube, $\overline{AB}$ $\overline{BC}$ , and $\overline{BF}$ have $4$ different parallel edges respectively. So it gives us the total pairs of parallel lines are $\binom{4}{2}\cdot3 =\boxed{18}$ | 18 |
385 | https://artofproblemsolving.com/wiki/index.php/2015_AMC_8_Problems/Problem_13 | 1 | How many subsets of two elements can be removed from the set $\{1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11\}$ so that the mean (average) of the remaining numbers is 6?
$\textbf{(A)}\text{ 1}\qquad\textbf{(B)}\text{ 2}\qquad\textbf{(C)}\text{ 3}\qquad\textbf{(D)}\text{ 5}\qquad\textbf{(E)}\text{ 6}$ | Since there will be $9$ elements after removal, and their mean is $6$ , we know their sum is $54$ . We also know that the sum of the set pre-removal is $66$ . Thus, the sum of the $2$ elements removed is $66-54=12$ . There are only $\boxed{5}$ | 5 |
386 | https://artofproblemsolving.com/wiki/index.php/2015_AMC_8_Problems/Problem_13 | 2 | How many subsets of two elements can be removed from the set $\{1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11\}$ so that the mean (average) of the remaining numbers is 6?
$\textbf{(A)}\text{ 1}\qquad\textbf{(B)}\text{ 2}\qquad\textbf{(C)}\text{ 3}\qquad\textbf{(D)}\text{ 5}\qquad\textbf{(E)}\text{ 6}$ | We can simply remove $5$ subsets of $2$ numbers while leaving only $6$ behind. The average of this one-number set is still $6$ , so the answer is $\boxed{5}$ | 5 |
387 | https://artofproblemsolving.com/wiki/index.php/2015_AMC_8_Problems/Problem_14 | 1 | Which of the following integers cannot be written as the sum of four consecutive odd integers?
$\textbf{(A)}\text{ 16}\quad\textbf{(B)}\text{ 40}\quad\textbf{(C)}\text{ 72}\quad\textbf{(D)}\text{ 100}\quad\textbf{(E)}\text{ 200}$ | Let our $4$ numbers be $n, n+2, n+4, n+6$ , where $n$ is odd. Then, our sum is $4n+12$ . The only answer choice that cannot be written as $4n+12$ , where $n$ is odd, is $\boxed{100}$ | 100 |
388 | https://artofproblemsolving.com/wiki/index.php/2015_AMC_8_Problems/Problem_14 | 2 | Which of the following integers cannot be written as the sum of four consecutive odd integers?
$\textbf{(A)}\text{ 16}\quad\textbf{(B)}\text{ 40}\quad\textbf{(C)}\text{ 72}\quad\textbf{(D)}\text{ 100}\quad\textbf{(E)}\text{ 200}$ | If the four consecutive odd integers are $2n-3,~ 2n-1, ~2n+1$ and $2n+3$ ; then, the sum is $8n$ . All the integers are divisible by $8$ except $\boxed{100}$ | 100 |
389 | https://artofproblemsolving.com/wiki/index.php/2015_AMC_8_Problems/Problem_14 | 3 | Which of the following integers cannot be written as the sum of four consecutive odd integers?
$\textbf{(A)}\text{ 16}\quad\textbf{(B)}\text{ 40}\quad\textbf{(C)}\text{ 72}\quad\textbf{(D)}\text{ 100}\quad\textbf{(E)}\text{ 200}$ | If the four consecutive odd integers are $a,~ a+2, ~a+4$ and $a+6$ , the sum is $4a+12$ , and $4a+12$ divided by $4$ gives $a+3$ . This means that $a+3$ must be even. The only integer that does not give an even integer when divided by $4$ is $100$ , so the answer is $\boxed{100}$ | 100 |
390 | https://artofproblemsolving.com/wiki/index.php/2015_AMC_8_Problems/Problem_14 | 4 | Which of the following integers cannot be written as the sum of four consecutive odd integers?
$\textbf{(A)}\text{ 16}\quad\textbf{(B)}\text{ 40}\quad\textbf{(C)}\text{ 72}\quad\textbf{(D)}\text{ 100}\quad\textbf{(E)}\text{ 200}$ | From Solution 1, we have the sum of the $4$ numbers to be equal to $4n + 12$ . Taking mod 8 gives us $4n + 4 \equiv b \pmod8$ for some residue $b$ and for some odd integer $n$ . Since $n \equiv 1 \pmod{2}$ , we can express it as the equation $n = 2a + 1$ for some integer $a$ . Multiplying 4 to each side of the equation yields $4n = 8a + 4$ , and taking mod 8 gets us $4n \equiv 4 \pmod{8}$ , so $b = 0$ . All the answer choices except choice D is a multiple of 8, and since 100 satisfies all the conditions the answer is $\boxed{100}$ | 100 |
391 | https://artofproblemsolving.com/wiki/index.php/2015_AMC_8_Problems/Problem_14 | 5 | Which of the following integers cannot be written as the sum of four consecutive odd integers?
$\textbf{(A)}\text{ 16}\quad\textbf{(B)}\text{ 40}\quad\textbf{(C)}\text{ 72}\quad\textbf{(D)}\text{ 100}\quad\textbf{(E)}\text{ 200}$ | Since they want CONSECUTIVE odd numbers, it won't be hard to just list the sums out: \[16=1+3+5+7\] \[40=7+9+11+13\] \[72=15+17+19+21\] \[200=47+49+51+53\] All of the answer choices can be a sum of consecutive odd numbers except $100$ , so the answer is $\boxed{100}$ | 100 |
392 | https://artofproblemsolving.com/wiki/index.php/2015_AMC_8_Problems/Problem_15 | 1 | At Euler Middle School, $198$ students voted on two issues in a school referendum with the following results: $149$ voted in favor of the first issue and $119$ voted in favor of the second issue. If there were exactly $29$ students who voted against both issues, how many students voted in favor of both issues?
$\textbf{(A) }49\qquad\textbf{(B) }70\qquad\textbf{(C) }79\qquad\textbf{(D) }99\qquad \textbf{(E) }149$ | We can see that this is a Venn Diagram Problem.
First, we analyze the information given. There are $198$ students. Let's use A as the first issue and B as the second issue.
$149$ students were for A, and $119$ students were for B. There were also $29$ students against both A and B.
Solving this without a Venn Diagram, we subtract $29$ away from the total, $198$ . Out of the remaining $169$ , we have $149$ people for A and
$119$ people for B. We add this up to get $268$ . Since that is more than what we need, we subtract $169$ from $268$ to get
$\boxed{99}$ | 99 |
393 | https://artofproblemsolving.com/wiki/index.php/2015_AMC_8_Problems/Problem_17 | 1 | Jeremy's father drives him to school in rush hour traffic in 20 minutes. One day there is no traffic, so his father can drive him 18 miles per hour faster and gets him to school in 12 minutes. How far in miles is it to school?
$\textbf{(A) } 4 \qquad \textbf{(B) } 6 \qquad \textbf{(C) } 8 \qquad \textbf{(D) } 9 \qquad \textbf{(E) } 12$ | For starters, we identify d as distance and v as velocity (speed)
Writing the equation gives us: $\frac{d}{v}=\frac{1}{3}$ and $\frac{d}{v+18}=\frac{1}{5}$
This gives $d=\frac{1}{5}v+3.6=\frac{1}{3}v$ , which gives $v=27$ , which then gives $d=\boxed{9}$ | 9 |
394 | https://artofproblemsolving.com/wiki/index.php/2015_AMC_8_Problems/Problem_17 | 2 | Jeremy's father drives him to school in rush hour traffic in 20 minutes. One day there is no traffic, so his father can drive him 18 miles per hour faster and gets him to school in 12 minutes. How far in miles is it to school?
$\textbf{(A) } 4 \qquad \textbf{(B) } 6 \qquad \textbf{(C) } 8 \qquad \textbf{(D) } 9 \qquad \textbf{(E) } 12$ | $d = rt$ $d=\frac{1}{3} \times r = \frac{1}{5} \times (r + 18)$
$\frac{r}{3} = \frac{r}{5} + \frac{18}{5}$
$10r = 270$ so $r = 27$ , plug into the first one and it's $\boxed{9}$ miles to school. | 9 |
395 | https://artofproblemsolving.com/wiki/index.php/2015_AMC_8_Problems/Problem_17 | 3 | Jeremy's father drives him to school in rush hour traffic in 20 minutes. One day there is no traffic, so his father can drive him 18 miles per hour faster and gets him to school in 12 minutes. How far in miles is it to school?
$\textbf{(A) } 4 \qquad \textbf{(B) } 6 \qquad \textbf{(C) } 8 \qquad \textbf{(D) } 9 \qquad \textbf{(E) } 12$ | We set up an equation in terms of $d$ the distance and $x$ the speed In miles per hour. We have $d=\left (\dfrac{1}{3}\right )(x)=\left (\dfrac{1}{5}\right) (x+18)$ , giving \[(5)(x)=(3)(x+18)\] \[5x=3x+54\] \[2x=54\] \[x=27\] Hence, $d=\dfrac{27}{3}=\boxed{9}$ | 9 |
396 | https://artofproblemsolving.com/wiki/index.php/2015_AMC_8_Problems/Problem_17 | 4 | Jeremy's father drives him to school in rush hour traffic in 20 minutes. One day there is no traffic, so his father can drive him 18 miles per hour faster and gets him to school in 12 minutes. How far in miles is it to school?
$\textbf{(A) } 4 \qquad \textbf{(B) } 6 \qquad \textbf{(C) } 8 \qquad \textbf{(D) } 9 \qquad \textbf{(E) } 12$ | Since it takes $\frac{3}{5}$ of the original time for him to get to school when there is no traffic, the speed must be $\frac{5}{3}$ of the speed in no traffic or $\frac{2}{3}$ more. Letting $x$ be the rate and we know that $\frac{5}{3}x = x + 18$ , so we have $\frac{2x}{3} = 18$ miles per hour. Solving for $x$ gives us $27$ miles per hour. Because $20$ minutes is a third of an hour, the distance would then be $9$ miles $\boxed{9}$ | 9 |
397 | https://artofproblemsolving.com/wiki/index.php/2015_AMC_8_Problems/Problem_17 | 5 | Jeremy's father drives him to school in rush hour traffic in 20 minutes. One day there is no traffic, so his father can drive him 18 miles per hour faster and gets him to school in 12 minutes. How far in miles is it to school?
$\textbf{(A) } 4 \qquad \textbf{(B) } 6 \qquad \textbf{(C) } 8 \qquad \textbf{(D) } 9 \qquad \textbf{(E) } 12$ | When driving in rush hour traffic, he drives $20$ minutes for one distance ( $1d$ ) to the school. It means he drives $60$ minutes for $3$ distances ( $3d$ ) to the school. When driving in no traffic hours, he drives $12$ minutes for one distance ( $1d$ ) to the school. It means he drives $60$ minutes for $5$ distances ( $5d$ ) to the school. Subtracting these two situations, it gives us $5d-3d = 18 = 2d$ , then $d=\frac{18}{2}=9$ . So the distance to the school would be $\boxed{9}$ miles. ----LarryFlora | 9 |
398 | https://artofproblemsolving.com/wiki/index.php/2015_AMC_8_Problems/Problem_17 | 6 | Jeremy's father drives him to school in rush hour traffic in 20 minutes. One day there is no traffic, so his father can drive him 18 miles per hour faster and gets him to school in 12 minutes. How far in miles is it to school?
$\textbf{(A) } 4 \qquad \textbf{(B) } 6 \qquad \textbf{(C) } 8 \qquad \textbf{(D) } 9 \qquad \textbf{(E) } 12$ | (Ratios) $\textbf{requires edits}$
In rush hour traffic, we can create this ratio for each distance we are going to try supposing the distance is $d$
$d$ $20$ minutes
In no traffic, we can do the same:
$d$ $12$ minutes
We want the ratio to be the distance to $60$ minutes:
$d$ $20$ minutes = $3$ d : $60$ minutes
$d$ $12$ minutes = $5$ d : $60$ minutes
This shows the speed in miles per hour. This means that the speed in the 2nd ratio has to be 18 faster than the first ratio. We get
$3d+18 = 5d \cdot d=9$ , or $\boxed{9}$ | 9 |
399 | https://artofproblemsolving.com/wiki/index.php/2015_AMC_8_Problems/Problem_18 | 1 | An arithmetic sequence is a sequence in which each term after the first is obtained by adding a constant to the previous term. For example, $2,5,8,11,14$ is an arithmetic sequence with five terms, in which the first term is $2$ and the constant added is $3$ . Each row and each column in this $5\times5$ array is an arithmetic sequence with five terms. The square in the center is labelled $X$ as shown. What is the value of $X$
$\textbf{(A) }21\qquad\textbf{(B) }31\qquad\textbf{(C) }36\qquad\textbf{(D) }40\qquad \textbf{(E) }42$
[asy] size(3.85cm); label("$X$",(2.5,2.1),N); for (int i=0; i<=5; ++i) draw((i,0)--(i,5), linewidth(.5)); for (int j=0; j<=5; ++j) draw((0,j)--(5,j), linewidth(.5)); void draw_num(pair ll_corner, int num) { label(string(num), ll_corner + (0.5, 0.5), p = fontsize(19pt)); } draw_num((0,0), 17); draw_num((4, 0), 81); draw_num((0, 4), 1); draw_num((4,4), 25); void foo(int x, int y, string n) { label(n, (x+0.5,y+0.5), p = fontsize(19pt)); } foo(2, 4, " "); foo(3, 4, " "); foo(0, 3, " "); foo(2, 3, " "); foo(1, 2, " "); foo(3, 2, " "); foo(1, 1, " "); foo(2, 1, " "); foo(3, 1, " "); foo(4, 1, " "); foo(2, 0, " "); foo(3, 0, " "); foo(0, 1, " "); foo(0, 2, " "); foo(1, 0, " "); foo(1, 3, " "); foo(1, 4, " "); foo(3, 3, " "); foo(4, 2, " "); foo(4, 3, " "); [/asy] | We begin filling in the table. The top row has a first term $1$ and a fifth term $25$ , so we have the common difference is $\frac{25-1}4=6$ . This means we can fill in the first row of the table: [asy] size(3.85cm); label("$X$",(2.5,2.1),N); for (int i=0; i<=5; ++i) draw((i,0)--(i,5), linewidth(.5)); for (int j=0; j<=5; ++j) draw((0,j)--(5,j), linewidth(.5)); void draw_num(pair ll_corner, int num) { label(string(num), ll_corner + (0.5, 0.5), p = fontsize(19pt)); } draw_num((0,0), 17); draw_num((4, 0), 81); draw_num((1,4), 7); draw_num((2,4), 13); draw_num((3,4), 19); draw_num((0, 4), 1); draw_num((4,4), 25); void foo(int x, int y, string n) { label(n, (x+0.5,y+0.5), p = fontsize(19pt)); } foo(2, 4, " "); foo(3, 4, " "); foo(0, 3, " "); foo(2, 3, " "); foo(1, 2, " "); foo(3, 2, " "); foo(1, 1, " "); foo(2, 1, " "); foo(3, 1, " "); foo(4, 1, " "); foo(2, 0, " "); foo(3, 0, " "); foo(0, 1, " "); foo(0, 2, " "); foo(1, 0, " "); foo(1, 3, " "); foo(1, 4, " "); foo(3, 3, " "); foo(4, 2, " "); foo(4, 3, " "); [/asy]
The fifth row has a first term of $17$ and a fifth term of $81$ , so the common difference is $\frac{81-17}4=16$ . We can fill in the fifth row of the table as shown: [asy] size(3.85cm); label("$X$",(2.5,2.1),N); for (int i=0; i<=5; ++i) draw((i,0)--(i,5), linewidth(.5)); for (int j=0; j<=5; ++j) draw((0,j)--(5,j), linewidth(.5)); void draw_num(pair ll_corner, int num) { label(string(num), ll_corner + (0.5, 0.5), p = fontsize(19pt)); } draw_num((0,0), 17); draw_num((4, 0), 81); draw_num((1,4), 7); draw_num((2,4), 13); draw_num((3,4), 19); draw_num((4, 4), 25); draw_num((0, 4), 1); draw_num((1, 0), 33); draw_num((2, 0), 49); draw_num((3, 0), 65); void foo(int x, int y, string n) { label(n, (x+0.5,y+0.5), p = fontsize(19pt)); } foo(2, 4, " "); foo(3, 4, " "); foo(0, 3, " "); foo(2, 3, " "); foo(1, 2, " "); foo(3, 2, " "); foo(1, 1, " "); foo(2, 1, " "); foo(3, 1, " "); foo(4, 1, " "); foo(2, 0, " "); foo(3, 0, " "); foo(0, 1, " "); foo(0, 2, " "); foo(1, 0, " "); foo(1, 3, " "); foo(1, 4, " "); foo(3, 3, " "); foo(4, 2, " "); foo(4, 3, " "); [/asy]
We must find the third term of the arithmetic sequence with a first term of $13$ and a fifth term of $49$ . The common difference of this sequence is $\frac{49-13}4=9$ , so the third term is $13+2\cdot 9=\boxed{31}$ | 31 |
400 | https://artofproblemsolving.com/wiki/index.php/2015_AMC_8_Problems/Problem_18 | 2 | An arithmetic sequence is a sequence in which each term after the first is obtained by adding a constant to the previous term. For example, $2,5,8,11,14$ is an arithmetic sequence with five terms, in which the first term is $2$ and the constant added is $3$ . Each row and each column in this $5\times5$ array is an arithmetic sequence with five terms. The square in the center is labelled $X$ as shown. What is the value of $X$
$\textbf{(A) }21\qquad\textbf{(B) }31\qquad\textbf{(C) }36\qquad\textbf{(D) }40\qquad \textbf{(E) }42$
[asy] size(3.85cm); label("$X$",(2.5,2.1),N); for (int i=0; i<=5; ++i) draw((i,0)--(i,5), linewidth(.5)); for (int j=0; j<=5; ++j) draw((0,j)--(5,j), linewidth(.5)); void draw_num(pair ll_corner, int num) { label(string(num), ll_corner + (0.5, 0.5), p = fontsize(19pt)); } draw_num((0,0), 17); draw_num((4, 0), 81); draw_num((0, 4), 1); draw_num((4,4), 25); void foo(int x, int y, string n) { label(n, (x+0.5,y+0.5), p = fontsize(19pt)); } foo(2, 4, " "); foo(3, 4, " "); foo(0, 3, " "); foo(2, 3, " "); foo(1, 2, " "); foo(3, 2, " "); foo(1, 1, " "); foo(2, 1, " "); foo(3, 1, " "); foo(4, 1, " "); foo(2, 0, " "); foo(3, 0, " "); foo(0, 1, " "); foo(0, 2, " "); foo(1, 0, " "); foo(1, 3, " "); foo(1, 4, " "); foo(3, 3, " "); foo(4, 2, " "); foo(4, 3, " "); [/asy] | The middle term of the first row is $\frac{25+1}{2}=13$ , since the middle number is just the average in an arithmetic sequence. Similarly, the middle of the bottom row is $\frac{17+81}{2}=49$ . Applying this again for the middle column, the answer is $\frac{49+13}{2}=\boxed{31}$ | 31 |
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