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**Soroban** Hello, Obsidantion!

An interesting problem ... a bit tricky for a magazine, though.

We assume that their ages are positive integers.

Let the ages be: .$\displaystyle a \:<\: b \:<\: c$

On the first occasion: .$\displaystyle c \:=\:a+b\;\;{\color{blue}[1]}$

Their ages were: .$\displaystyle \begin{array}{|c|c|c|} a & b & a +b \end{array}$

A few years later: .$\displaystyle b \:=\:\frac{a+c}{2}$ . . . $\displaystyle b$ is the *average* of $\displaystyle a$ and $\displaystyle c$

Substitute [1]: .$\displaystyle b \:=\:\frac{a + (a+b)}{2} \quad\Rightarrow\quad b \:=\:2a$

Their ages were: .$\displaystyle \begin{array}{|c|c|c|} a & 2a & 3a \end{array} $

$\displaystyle \frac{a+b+c}{2}$ years after the first occasion .$\displaystyle = \;\frac{a+2a+3a}{2} \:=\:3a$ years later

They are all $\displaystyle 3a$ years older. .Their ages are: .$\displaystyle \begin{array}{|c|c|c|} 4a & 5a & 6a \end{array}$

At this time, one of is 18 years old . . . It must have been the oldest, $\displaystyle 6a$

. . $\displaystyle 6a = 18 \quad\Rightarrow\quad a = 3 \quad\Longrightarrow\quad 4a = 12,\;5a = 15$

Therefore, the other two were 12 and 15 years old.