Blinkov A.d:

The square of the sum of the digits of the number $A$ is equal to the sum of the digits of the number $A^{2}$. Find all such two-digit numbers $A$.
Answer: $10,11,12,13,20,21,22,30,31$.

Solution 1. Note that $A^{2} \leq 99^{2}=9801<9999$. Therefore, the sum of the digits of $A^{2}$ is less than $9 * 4=36$. Since it is equal to the square of the sum of the digits of $A$, the sum of the digits of $A$ is less than $36^{1 / 2}=6$, that is, less than or equal to 5. There remain 15 options: $10,11,12,13,14,20,21,22,23,30,31,32,40,41,50$, of which only 9 options satisfy the condition, as indicated in the answer.

Solution 2. Let $S(n)$ denote the sum of the digits of the number $n$. Note that when adding two numbers "vertically," the only possible carry is a one to the higher place, and each such carry reduces the sum of the digits by 9. Therefore, the sum of the digits of the sum of any number of addends does not exceed the sum of the digits of the addends. As a consequence, $S(n) \leq n$, since $n=1+1+\ldots+1$ ($n$ ones), and equality is achieved only for single-digit numbers, because when adding another one to the sum of nine ones, a carry will already occur. In addition, appending a zero to the right does not change the sum of the digits, that is, $S(10 n)=S(n)$.

Let the desired two-digit number $A=\underline{a b=10 a+b}$. Then

$S\left(A^{2}\right)=S\left((10 a+b)^{2}\right)=S\left(100 a^{2}+10^{*} 2 a b+b^{2}\right) \leq$

$\leq S\left(100 a^{2}\right)+S\left(10^{*} 2 a b\right)+S\left(b^{2}\right)=S\left(a^{2}\right)+S(2 a b)+S\left(b^{2}\right) \leq$

$\leq a^{2}+2 a b+b^{2}=(a+b)^{2}=(S(A))^{2}$, and equality is achieved if and only if $a^{2}, 2 a b$, and $b^{2}$ are single-digit numbers (if so, then when adding $100 a^{2}, 10 * 2 a b$, and $b^{2}$, no carries occur, since they contain all their non-zero digits in different places). Therefore, it is necessary and sufficient that the numbers $a^{2}, b^{2}$, and $2 a b$ be less than 10, that is, $1 \leq a \leq 3,0 \leq b \leq 3$ and $a b \leq 4$.

From this, we find all the options listed in the answer.

Author: Kalinin D.A.

The product of five numbers is not zero. Each of these numbers was decreased by one, and their product did not change. Provide an example of such numbers.

## Solution

Answer: for example, $5,6,7,8,-1$.

Another example can be constructed: let the first four numbers be twos. We get an equation for the fifth number $x$: $16 x=x-1$, from which $x=-1 / 15$.