119. For what values of $n$ is the sum $n^{2}+(n+1)^{2}+(n+2)^{2}+$ $(n+3)^{2}$ divisible by 10 ( $n$ is a natural number)?
Solution. I with assistance. We have:

$$
n^{2}+(n+1)^{2}+(n+2)^{2}+(n+3)^{2}=2\left(2 n^{2}+6 n+7\right)
$$

Therefore, $2 n^{2}+6 n+7=5 k$. Solving this quadratic equation, we find:

$$
n=\frac{-3 \pm \sqrt{ } 5(2 k-1)}{2}
$$

The number $n$ will be a natural number if and only if $2 k-1=5(2 t+1)^{2}$. Thus, $n=\frac{-3+5(2 t+1)}{2}$, or $n=5 t+1 \quad(t-$ an integer non-negative number).

II with assistance. Since $n^{2}+(n+1)^{2}+(n+2)^{2}+(n+3)^{2}=$ $=(2 n+3)^{2}+5$, the last digit of the number $2 n+3$ must be 5 (for the given sum to be divisible by 10). Therefore, $2 n+3=$ $=10 t+5$, from which $n=5 t+1$.

III with assistance. Let's list the last digits of the squares of consecutive natural numbers:

$$
0,1,4,9,6,5,6,9,4,1
$$

In this sequence, there are only two consecutive sets of four numbers whose sum is divisible by 10:

$$
1+4+9+6=6+9+4+1=20
$$

Therefore, the last digit of the number $n$ must be 1 or 6. Thus, $n=5 t+1$.

Answer: $n=5 t+1$.