## Task 15/77

In what way can the number 92 be decomposed into two natural number addends, one of which is divisible by 5 and the other leaves a remainder of 3 when divided by 7?
Let $x$ and $y$ be the sought addends. Then

$$
x+y=92 \quad(1) \quad ; \quad x=5 k \quad ; \quad y=7 l+3
$$

with $k ; l \in \mathbb{N}$. From (1) and (1') it follows that $5 k+7 l=89$.

We consider equation (2) modulo 5 and $\bmod 7$ and obtain

$$
2 l \equiv 4(\bmod 5) ; \quad 5 k \equiv 5(\bmod 7) ; \quad l \equiv 2(\bmod 5) \quad k \equiv 1(\bmod 7)
$$

thus $l=2+5 t_{1}$ and $k=1+7 t_{2}$ with $t_{1} ; t_{2} \in \mathbb{N}$. Substituting this into (2), we get

$$
35\left(t_{1}+t_{2}\right)=70 \rightarrow t_{1}+t_{2}=2
$$

This yields the following three possibilities:

$$
\begin{array}{rrrrr}
t_{1}=0 ; & t_{2}=2 ; & k=2 ; \quad l=15 ; & x=75 ; & y=17 \\
t_{1}=1 ; & t_{2}=1 ; & k=7 ; \quad l=8 ; & x=40 ; & y=52 \\
t_{1}=2 ; & t_{2}=0 ; & k=12 ; \quad l=1 ; & x=5 ; & y=87
\end{array}
$$

As the verification shows, the pairs $(75 ; 17),(40 ; 52),(5 ; 87)$ are solutions.