Problem 1. Determine for how many triples $(m, n, p)$ of non-zero natural numbers less than or equal to 5 the number

$$
A=2^{m}+3^{n}+5^{p}
$$

is divisible by 10.
Solution. The last digit of $A$ is 0 and the last digit of $5^{p}$ is 5 (regardless of the value of $p$), so the last digit of the sum $2^{m}+3^{n}$ is equal to 5 $2 p$

The last digits of the numbers $2^{1}, 2^{2}, 2^{3}, 2^{4}$, and $2^{5}$ are $2,4,8,6$, and $2 \mathbf{1 p}$

The last digits of the numbers $3^{1}, 3^{2}, 3^{3}, 3^{4}$, and $3^{5}$ are $3,9,7,1$, and $3 \mathbf{1 p}$ As a result, the pair $(m, n)$ can be chosen in 7 ways, namely: $(1,1)$; $(1,5)$; $(2,4)$; $(3,3)$; $(4,2)$; $(5,1)$; $(5,5)$ $2 p$

Since $p$ can be any of the numbers $1,2,3,4$ or 5, there are $7 \cdot 5=35$ triples $(m, n, p)$ with the property from the statement $1 p$