2002 AIME I Problem 3

Question

Jane is

25

years old. Dick is older than Jane. In

n

years, where

n

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

d

be Dick’s present age. How many ordered pairs of positive integers

(d, n)

are possible?

Answer

Answer: 25

Concepts:digits ages systematic listing

Difficulty rating: 2300

Small hint:

Focus on the future ages: Jane’s is a two-digit number whose digit reversal, Dick’s age, is larger

Big hint:

Count the possible values of Jane’s future age: two-digit numbers old enough for Jane whose tens digit is smaller than the units digit. Each determines

(d, n).

Solution:

In

n

years Jane’s age is

25 + n,

and Dick’s age is its digit reversal. If Jane’s future age is

10a + b,

Dick’s is

10b + a,

which is larger exactly when

b \gt a.

Conversely, every two-digit value of

25 + n

with tens digit less than units digit yields exactly one valid pair:

n = 10a + b - 25

and

\begin{aligned} d &= 10b + a - n \\ &= 25 + 9(b - a) \gt 25, \end{aligned}

so Dick is indeed older than Jane now.

So we count two-digit numbers that are at least

26

and have tens digit less than units digit:

4

starting with

2

(namely

26

through

29

), then

6,

5,

4,

3,

2,

1