[ Evenness and Oddness ]  [ Counting in Two Ways ]

In the cells of a $10 \times 10$ square table, numbers from 1 to 100 are arranged. Let $S_{1}, S_{2}, \ldots, S_{10}$ be the sums of the numbers in the columns of the table.

Could it be that among the numbers $S_{1}, S_{2}, \ldots, S_{10}$, every two consecutive numbers differ by 1?
If the condition were met, then in the sequence $S_{1}, S_{2}, \ldots, S_{10}$, even and odd numbers would strictly alternate.

## Solution

If $S_{i}$ and $S_{i+1}$ differ by 1, then these two numbers have different parity, meaning that in the sequence $S_{1}, S_{2}, \ldots, S_{10}$, even and odd numbers strictly alternate. Therefore, among the numbers $S_{1}, \ldots, S_{10}$, there are exactly five even and five odd numbers. Hence, the sum $S_{1}+\ldots+S_{10}$ is odd. On the other hand, $S_{1}+\ldots+S_{10}=1+2+\ldots+100$, and in this sum, there are 50 odd terms, so it is even. This is a contradiction.

## Answer

It could not.

Given polynomials $P_{1}, P_{2}, \ldots, P_{5}$, whose sums of coefficients are $1, 2, 3, 4, 5$ respectively.

Find the sum of the coefficients of the polynomial $Q=P_{1} P_{2} \ldots P_{5}$.

## Hint

The sum of the coefficients of a polynomial $P$ is equal to $P(1)$.

## Solution

The sum of the coefficients of the polynomial $Q$ is $Q(1)=P_{1}(1) P_{2}(1) \ldots P_{5}(1)=1 \cdot 2 \cdot 3 \cdot 4 \cdot 5=120$.

## Answer

120.