The sum of two numbers is 581; the quotient of their least common multiple and greatest common divisor is 240. Which are these numbers?
I. Solution. Let $d$ be the greatest common divisor of the sought numbers $A$ and $B$, and $m$ their least common multiple, so

$$
A=d a \text { and } B=d b \text {, where } a \text { and } b \text { are coprime and thus } m=a b d \text {. }
$$

According to our data: $A+B=(a+b) d=581$ and $\frac{m}{d}=a b=240$.

Already now $581=7 \cdot 83$, so it is possible

$$
a+b=7, \quad a+b=83, \quad a+b=581
$$

The case $a+b=7$ can be ruled out, because $a b=240$. (Indeed, if $a+b=7$, then the value of the product $a b$ can be at most $\left(\frac{7}{2}\right)^{2}=\frac{49}{4}$.)

Accordingly, we need to examine two cases.
I. $a+b=83$ and $a b=240$. Then $a$ and $b$ are the roots of the equation $u^{2}-83 u+240=0$. These are: 80 and 3, indeed integers. In this case:

$$
d=\frac{581}{83}=7 \quad \text { and } \quad A=7 \cdot 80=560, \quad B=7 \cdot 3=21
$$

II. $a+b=581$ and $a b=240$. Now $a$ and $b$ are the roots of the equation $v^{2}-581 v+240=0$. However, these are not integers. Therefore, our problem has only one solution:

$$
A=560, \quad B=21 \text {. }
$$

György Margulis (Bolyai r. VI. o. Bp. V.)

II. Solution. We could have also started from the fact that $a b=240$, that is, 240 must be factored into two factors that are coprime. Since $240=2^{4} \cdot 3 \cdot 5$, the factorization into coprime factors is:

1 and $240, 2^{4}=16$ and 15, 3 and 80, 5 and 48.

Accordingly,

$$
a+b \text { value: } 240,31,83,53 .
$$

However, $a+b$ must be a divisor of 581; out of the four values, only 83 is a divisor of 581, so $d=\frac{581}{83}=7$ and thus the sought numbers are: $7 \cdot 3$, and $7 \cdot 80$.