The sum of two natural numbers and their greatest common divisor is equal to their least common multiple. Determine the ratio of the two numbers.





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Let the two numbers be $A$ and $B$ where $A \geq B$, their greatest common divisor $D$, and their least common multiple $M$. Then $A = aD$, $B = bD$, and $A / B = a / b$, where $a$ and $b$ are natural numbers that are relatively prime to each other, $a \geq b$, and the least common multiple, as known, can be expressed as $M = abD$.

According to the requirement,

$$
A + B + D = M
$$

Substituting the above and rearranging,

$$
a + b + 1 = ab, \quad ab - a - b + 1 = (a - 1)(b - 1) = 2.
$$

Here, $a - 1$ and $b - 1$ are integers, and $a - 1 \geq b - 1$, so only $a - 1 = 2, b - 1 = 1$ is possible, which gives $A : B = a : b = 3 : 2$.

Herneczki István (Sopron, Széchenyi I. Gimn., II. o. t.)