Solution

The ages of the three children are 3, 3, and 8 years. Here's how I arrived at the answer:

The product of the ages of the three children is 72. The sum of the ages is the same as the house number.

Now, 72 = 1 × 2 × 2 × 2 × 3 × 3. Three numbers whose product is 72 could be:

a)1, 1, 72; sum is 74

b)1, 2, 36; sum is 39

c)1, 3, 24; sum is 28

d)1, 4, 18; sum is 23

e)1, 6, 12; sum is 19

f)1, 8, 9; sum is 18

g)2, 2, 18; sum is 22

h)2, 3, 12; sum is 17

i)2, 4, 9; sum is 15

j)2, 6, 6; sum is 14

k)3, 3, 8; sum is 14

l)3, 4, 6; sum is 13

These are the twelve possibilities; there do not exist any other possibilities.

Ajit has prepared this list. He has only to look up the house number, and he can tell the ages. For example, if the house number is 19, then option (e) will apply, and the ages of the children will be 1, 6, and 12. Likewise, if the house number is 18, option (f) will apply, and the ages of the children will be 1, 8, and 9. Every sum is unique, so it would be easy to find the ages based on the house number.

But, still, Ajit cannot figure out the ages, although he is good at mathematics.

It means the house number is such that it occurs in more than one option. There are two options — (j) and (k), both of which have the same sum 14. The house number, therefore, is 14. Ajit cannot tell whether the children are aged 2, 6, and 6 or 3, 3, and 8. So he asks for a further hint.

The teacher's hint is: "The oldest child likes adventure sports."

It is not the sport that matters, but now Ajit knows there is one oldest child. In the case of option (j), the ages are 2, 6, and 6. Two children are of the same age, six; there is, thus, no single eldest child. It, therefore, cannot be the answer. So the only possibility is the option (k) where there are a single oldest child aged 8, and twins who are three years old.