158. It is known that the number $a$ is $n$ times greater than the number $b$, and the sum of the numbers $a$ and $b$ is $m$ times greater than their difference. Find the sum of the numbers $m$ and $n$, if $m$ and $n$ are natural numbers.

94
Given the problem, we set up the system of equations:

$$
\left\{\begin{array}{l}
\frac{a}{b}=n \\
\frac{a+b}{a-b}=m
\end{array}\right.
$$

From the second equation, express $\frac{a}{b}$. We have:

$$
\begin{gathered}
a+b=m a-m b, a(1-m)=-b(m+1) \\
\frac{a}{b}=\frac{m+1}{m-1}=\frac{m-1+2}{m-1}=1+\frac{2}{m-1} .
\end{gathered}
$$

Therefore,

$$
\frac{a}{b}=n=1+\frac{2}{m-1}
$$

Since $m$ and $n$ are natural numbers, then $n=3$ when $m=2$, $n=2$ when $m=3$, and $m+n=5$.

Answer: $m+n=5$.

$\triangle$ 159. Justify the rule for multiplying two-digit numbers by 11.

Solution.

$$
\overline{a b} \cdot 11=(10 a+b)(10+1)=100 a+10(a+b)+b
$$

where $a$ is the hundreds digit, $a+b$ is the tens digit, and $b$ is the units digit.