Problem 9.8. How many pairs of natural numbers $a$ and $b$ exist such that $a \geqslant b$ and

$$
\frac{1}{a}+\frac{1}{b}=\frac{1}{6} ?
$$
Answer: 5.

Solution. From the condition, it follows that $\frac{1}{6}=\frac{1}{a}+\frac{1}{b} \leqslant \frac{2}{b}$, from which $b \leqslant 12$. Also, $\frac{1}{6}=\frac{1}{a}+\frac{1}{b}>\frac{1}{b}$, so $b>6$. Therefore, $b$ can take values from 7 to 12 inclusive.

Using $\frac{1}{a}=\frac{1}{6}-\frac{1}{b}=\frac{b-6}{6 b}$, we get

$$
a=\frac{6 b}{b-6}=6+\frac{36}{b-6}
$$

Substitute the possible values of $b$ and check if the number $a$ is natural.

If $b=7$, then $a=6+\frac{36}{7-6}=42$ - fits.

If $b=8$, then $a=6+\frac{36}{8-6}=24$ - fits.

If $b=9$, then $a=6+\frac{36}{9-6}=18$ - fits.

If $b=10$, then $a=6+\frac{36}{10-6}=15$ - fits.

If $b=11$, then $a=6+\frac{36}{11-6}=6+\frac{36}{5}$ - not an integer, does not fit.

If $b=12$, then $a=6+\frac{36}{12-6}=12$ - fits.

It is easy to see that all 5 pairs $(42,7),(24,8),(18,9),(15,10),(12,12)$ are solutions to the original equation.

Another solution. Multiplying the equation by the denominators (since $a$ and $b$ are natural, they can be multiplied and divided), we get

$$
6 a+6 b=a b
$$

This equation can be easily transformed into the following:

$$
(a-6)(b-6)=36
$$

From this, it is clear that the numbers $a-6$ and $b-6$ must be divisors of 36, and natural, because if they were negative integers, the smaller of them would be no more than -6, which is impossible.

The first factor is no less than the second by condition. By considering the possible factorizations

$$
(a-6)(b-6)=36 \cdot 1=18 \cdot 2=12 \cdot 3=9 \cdot 4=6 \cdot 6
$$

we get the same 5 solutions.

## 10th grade