Henry Ernest Dudeney/Puzzles and Curious Problems/42 - Family Ages/Solution

Puzzles and Curious Problems by Henry Ernest Dudeney: $42$

A man and his wife had three children, John, Ben, and Mary,

and the difference between their parents' ages was the same as between John and Ben and between Ben and Mary.

The ages of John and Ben, multiplied together, equalled the age of the father,

and the ages of Ben and Mary multiplied together equalled the age of the mother.

The combined ages of the family amounted to ninety years.

What was the age of each person?

The children are triplets, all aged $6$.

The mother and father are both aged $36$.

Let $j$, $b$ and $m$ denote the ages of John, Ben and Mary respectively.

Let $F$ and $M$ denote the ages of the father and mother respectively.

We have that:

$\size {F - M} = \size {j - b} = \size {b - m}$

It is immediately apparent that this can be solved simply by:

$F = M$

and so:

$j - b - m$

where:

$j^2 = b^2 = m^2 = F = M$

such that:

$3 j + 2 F = 90$

The very $3$-ish nature of the final equation tempts us to try $j = 6$, and we see:

$3 \times 6 + 2 \times 36 = 90$

Note that we are asked to specify the age of each person.

But the only information we have is the difference between the ages, and not which is older than the other.

So if the children have different ages, we cannot tell which is the eldest.

Hence it is impossible in that case to state which child is of which age.

When they are all the same age, the age of each is obvious.

$\blacksquare$