## Task A-1.2.

Determine all ordered triples $(a, b, c)$ of natural numbers for which $a \leqslant b \leqslant c$ and

$$
\frac{3}{7}=\frac{1}{a}+\frac{1}{a b}+\frac{1}{a b c}
$$
## First solution.

By multiplying the entire expression by $7 a b c$ we have

$$
3 a b c=7(b c+c+1)=(7 b+7) c+7
$$

Since the left side of the equation is divisible by $c$, the right side must also be, meaning 7 must be divisible by $c$, i.e., $c=1$ or $c=7$.

If $c=1$, from $a \leqslant b \leqslant c$ we conclude $a=b=c=1$, which is not a solution upon substitution. Therefore, we conclude $c=7$.

Substituting into the above equation, we get $3 a b \cdot 7=(7 b+7) \cdot 7+7$, or $3 a b=7 b+8$.

As before, it is necessary that 8 is divisible by $b$, i.e., $b=1,2,4$ or 8.

1 point

For $b=1$, we get $3 a=15$, or $a=5$, which is not possible due to $a \leqslant b$.

1 point

For $b=2$, we get $6 a=22$, which is not possible.

1 point

For $b=4$, we get $12 a=36$, or $a=3$.

1 point

The case $b=8$ is not possible due to $b \leqslant c$.

1 point

Therefore, the only possible solution is $(a, b, c)=(3,4,7)$.