## PROBLEM 2

Knowing that $3^{6 n+12}+9^{3 n+6}+27^{2 n+4}=3^{4(n+3)+255}, n \in \mathbb{N}$.

Find the remainder of the division of A by 5, where $A=2^{n}+3^{n}+4^{n}+7^{n}$.
## Solution:

Applying the power calculation rule we obtain $\mathrm{n}=127$

$.2 \mathrm{p}$

$A=2^{n}+3^{n}+4^{n}+7^{n}$

$U\left(2^{127}\right)=8$ $1 \mathrm{p}$

$U\left(3^{127}\right)=7$ $1 \mathrm{p}$

$U\left(4^{127}\right)=4$ $1 \mathrm{p}$

$U\left(7^{127}\right)=3$ $1 \mathrm{p}$

the last digit of A, i.e., $U(A)=2$, so the remainder of the division of A by 5 is 2 . $1 \mathrm{p}$.