
Solution. The product of all the numbers written on the blackboard is

$$
S=2^{31} \cdot 3^{15} \cdot 5^{7} \cdot 7^{4} \cdot 11^{3} \cdot 13^{2} \cdot 17 \cdot 19 \cdot 23 \cdot 29 \cdot 31 .
$$

Apparently, the numbers $17,19,23,29$, and 31 can never be erased and can never be a part of any change. In any step, there is always left at least one other number, which gives in total $5+1=6$ numbers. Six numbers on the blackboard are achievable indeed: because of the odd exponents of primes 2, 3, 5, and 11 in $S$ we allocate the set $A=\{2,9,11,22,25\}$ and we put all the other numbers with exception of 17,19 , 23, 29 a 31 into set

$$
B=\{3,4,5,6,7,8,10,12,13,14,15,16,18,20,21,24,26,27,28,30,32,33\} .
$$

In the first step we choose the group $A$ and replace it by the number

$$
n=\sqrt{2 \cdot 9 \cdot 11 \cdot 22 \cdot 25}=\sqrt{2^{2} \cdot 3^{2} \cdot 5^{2} \cdot 11^{2}}=2 \cdot 3 \cdot 5 \cdot 11 .
$$

Because the product of all the numbers from $B$ is $2^{31-2} \cdot 3^{15-2} \cdot 5^{7-2} \cdot 7^{4} \cdot 11^{3-2}$. $13^{2}=2^{29} \cdot 3^{13} \cdot 5^{5} \cdot 7^{4} \cdot 11 \cdot 13^{2}$, in the second step we choose $n$ together with all the numbers from $B$ and replace it by

$$
\sqrt{(2 \cdot 3 \cdot 5 \cdot 11) \cdot\left(2^{29} \cdot 3^{13} \cdot 5^{5} \cdot 7^{4} \cdot 11 \cdot 13^{2}\right)}=2^{15} \cdot 3^{7} \cdot 5^{3} \cdot 7^{2} \cdot 11 \cdot 13,
$$

which leaves just six numbers on the blackboard.

Conclusion. The sought least amount is six.

## Second Round of the 59th Czech and Slovak Mathematical Olympiad (January 20th, 2010)

![](https://cdn.mathpix.com/cropped/2024_04_17_10b455106c778bb0ba10g-11.jpg?height=161&width=390&top_left_y=323&top_left_x=515)


3. There are numbers $1,2, \ldots, 33$ written on the blackboard. In one step we choose a group of numbers on the blackboard (at least two) such that their product is a square, we erase them and write the square root of their product instead. We proceed until no group can be chosen. What is the least amount of numbers left on the blackboard?
