Let the three positive integers be $x, y$ and $z$.

From the given information $(x+y) \times z=14$ and $(x \times y)+z=14$.

Let us look at the third number, $z$, first.

From the first equation $(x+y) \times z=14$, we see that $z$ must be a factor of 14 .

Therefore, the only possible values of $z$ are $1,2,7$ and 14 .

If $z=1$, then $x+y=14$ and $x y=13$. From this, $x=1$ and $y=13$ or $x=13$ and $y=1$. (We can find this by seeing that $x y=13$ so the only possible values of $x$ and $y$ are 1 and 13 , and then checking the first equation to see if they work.)

If $z=2$, then $x+y=7$ and $x y=12$. From this, $x=3$ and $y=4$ or $x=4$ and $y=3$.

(We can find these by looking at pairs of positive integers which add to 7 and checking if they multiply to 12 .)

If $z=7$, then $x+y=2$ and $x y=7$. There are no possibilities for $x$ and $y$ here. (Since $x$ and $y$ are positive integers and $x+y=2$ then we must have $x=1$ and $y=1$, but this doesn't work in the second equation.)

If $z=14$, then $x+y=1$ and $x y=0$. There are no possibilities for $x$ and $y$ here. (This is because one of them must be 0 , which is not a positive integer.)

Therefore, the four possible values for $x$ are 1, 13, 3 and 4 .

ANSWER: (B)
Pete is given three positive integers and is told to add the first two, and then multiply the result by the third. Instead, he multiplies the first two and adds the third to that result. Surprisingly, he still gets the correct answer of 14 . How many different values could the first number have been?
(A) 5
(B) 4
(C) 6
(D) 3
(E) 7