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Find the last digit of the number $7^{7}$.

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$7^{2} \equiv-1(\bmod 10)$, so, $7^{4} \equiv 1(\bmod 10). 7^{7} \equiv(-1)^{7}=-1(\bmod 4)$, therefore $7^{7^{7}} \equiv 7^{3}=343(\bmod 10)$.

## Answer

3. 

Prove that the sum of the squares of five consecutive natural numbers is not a perfect square.

## Solution

First method. The remainder of the division by 4 of the square of an odd number is 1, and the remainder of the square of an even number is 0. Therefore, the specified sum when divided by 4 gives a remainder of 2 or 3, that is, it is not a square.

Second method. $(n-2)^{2}+(n-1)^{2}+n^{2}+(n+1)^{2}+(n+2)^{2}=5 n^{2}+10=5\left(n^{2}+2\right)$, and $n^{2}+2$ is not divisible by 5.