## 258. Math Puzzle $11 / 86$

Hans is supposed to nail a $90 \mathrm{~cm}$ wide gap in a wooden fence. He has eight $10 \mathrm{~cm}$ wide and sixteen $8 \mathrm{~cm}$ wide boards of suitable length. A saw, with which he could split the wood lengthwise, is not available to him.

How many and which options does Hans have to close the opening without any gaps?
Besides empirical - i.e., found by trial and error - solutions, a system of equations also presented itself. The number of required boards is denoted by $m(10 \mathrm{~cm}$ wide) and $n(8 \mathrm{~cm}$ wide).

$$
10 m+8 n=90 \quad, \quad 5 m+4 n=45 \quad, \quad 4 n=5(9-m)
$$

From this, it follows that $n$ must be divisible by 5, i.e., $0 ; 5 ; 10 ; 15 \ldots$ Only two solutions arise:

1) 5 boards à $10 \mathrm{~cm}$ and 5 boards à $8 \mathrm{~cm}$
2) 1 board à $10 \mathrm{~cm}$ and 10 boards à $8 \mathrm{~cm}$