# 2014 UNM-PNM Statewide High School Mathematics Contest II Problems/Problem 3

## Problem

Two people, call them $A$ and $B$, are having a discussion about the ages of $B$’s children.

A: “What are the ages, in years only, of your four children?”

B: “The product of their ages is $72$.”

A: “Not enough information.”

B: “The sum of their ages equals your eldest daughter’s age.”

A: “Still not enough information.”

B: “My oldest child who is at least a year older than her siblings took the AMC 8 for the first time this year.”

A: “Still not enough information.”

B: “My youngest child is my only son.”

A: “Now I know their ages..”

What are their ages?

## Solution

The product ages is 72. The largest age is unique. The smallest age is unique.

Assume that the youngest child is 1. Then the remaining 3 ages multiply to 72. The next smallest age possible is 2. The product of the final 2 ages is then 36 and this results in several solutions:

1. (1,2,2,18) (can be disregarded)

2. (1,2,3,12)

3. (1,2,4,9)

If the 2nd youngest is 3:

1. (1,3,3,8)

2. (1,3,4,6)

If the 2nd youngest is 4, the final 2 ages multiply to 18. This leads to an impossibility.

Assume the youngest age is 2. Then the remaining ages multiply to 36. The next smallest age possible is 3: leads to a final solution of (2,3,3,4).

So we have 6 solutions. The first solution can be disregarded with knowledge of the age limit of AMC 8.