Which are those natural numbers $n$ for which

$$
5^{n}+12 n^{2}+12 n+3
$$

is divisible by $100$?
Solution. If $n=1$, then $N=5^{n}+12 n^{2}+12 n+3=32$, which is not divisible by 100. Therefore, we can assume that $n \geq 2$.

$N$ will be divisible by 100 if and only if it is divisible by 4 and 25, since $100=4 \cdot 25$ and $(4 ; 25)=1$. Given that $n \geq 2$, $5^{2}=25 \mid 5^{n}$, so it must be a divisor of $12 n^{2}+12 n+3=3(2 n+1)^{2}$. Since $(3 ; 25)=1$, 25 must be a divisor of $(2 n+1)^{2}$. This is true if and only if $2 n+1$ is divisible by 5, i.e., $n=5 k+2$. In this case, $2 n+1=2(5 k+2)+1=10 k+5$.

4 will be a divisor of $N$ if and only if it is a divisor of $\left(5^{n}+3\right)$. But $5^{n}=(4+1)^{n}$ leaves a remainder of 1 when divided by 4, so $5^{n}+3$ leaves a remainder of $1+3=4$, which is 0, when divided by 4. Therefore, it is divisible by 4 for all $n \geq 1$.

In summary: 100 is a divisor of $5^{n}+12 n^{2}+12 n+3$ if and only if $n=5 k+2$, where $k=0,1,2,3, \ldots$.