Problem: Jane is years old. Dick is older than Jane. In years, where is a positive integer, Dick's age and Jane's age will both be two-digit numbers and will have the property that Jane's age is obtained by interchanging the digits of Dick's age. Let be Dick's present age. How many ordered pairs of positive integers are possible?
Solution:
Let Dick's age and Jane's age in years be and , respectively. At that time, Dick will be years older than Jane, and the sum of their ages will be . Dick's age must always exceed Jane's by a multiple of ; thus Dick's current age must be , or . Suppose that Dick is , so that the sum of their ages is . Their age-sum is therefore always odd, and it is not a multiple of until it reaches . This takes years. Every years thereafter, as long as Dick has a two-digit age, their ages will be reversals of each other; Dick's ages at those times are , and . Similar reasoning applies if Dick's current age is , and their age-sum is : the next age-sum that has the same parity and is divisible by is , when Dick is and . Every years thereafter, until Dick is , their ages are reversals of each other - five examples in all. Similarly, there are four examples if Dick's current age is , four examples if his current age is , three examples if his current age is , two examples if his current age is , one example if his current age is , and none if his current age is . The total number of ordered pairs is thus .
Since Dick must always be older than Jane, in years Jane may be or . Dick's age will be the result of reversing the digits of Jane's age. The total number of ordered pairs is thus .
The problems on this page are the property of the MAA's American Mathematics Competitions