Permalink Submitted by Anonymous on April 12, 2013

There are seven distinct ways of factoring 36 into groups of three integers:

1 3 12 (sum is 16)
1 4 9 (sum is 14)
1 1 36 (sum is 38)
1 6 6 (sum is 13)
2 2 9 (sum is 13)
2 3 6 (sum is 11)
3 4 3 (sum is 10)

The 2nd clue is not about the house number but the fact that, even knowing the
house number is insufficient to work out their ages. This means that the sum
cannot be unique.

Of all the possibilities listed above, only (1 6 6) and (2 2 9) provide sums
that are non-unique.

The 3rd clue tells us that there is an eldest. Hence the ages are 2, 2 and 9.

## Here is my solution

There are seven distinct ways of factoring 36 into groups of three integers:

1 3 12 (sum is 16)

1 4 9 (sum is 14)

1 1 36 (sum is 38)

1 6 6 (sum is 13)

2 2 9 (sum is 13)

2 3 6 (sum is 11)

3 4 3 (sum is 10)

The 2nd clue is not about the house number but the fact that, even knowing the

house number is insufficient to work out their ages. This means that the sum

cannot be unique.

Of all the possibilities listed above, only (1 6 6) and (2 2 9) provide sums

that are non-unique.

The 3rd clue tells us that there is an eldest. Hence the ages are 2, 2 and 9.