Difference between revisions of "2002 AIME I Problems/Problem 3"

Revision as of 18:54, 4 July 2013

Problem

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 number 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?

Solution

Let Jane's age $n$ years from now be $10a+b$ , and let Dick's age be $10b+a$ . If $10b+a>10a+b$ , then $b>a$ . The possible pairs of $a,b$ are:

$(1,2), (1,3), (2,3), (1,4), (2,4), (3,4), \dots , (8,9)$

That makes 36. But $10a+b>25$ , so we subtract all the extraneous pairs: $(1,2), (1,3), (2,3), (1,4), (2,4), (1,5), (2,5), (1,6), (1,7), (1,8),$ and $(1,9)$ . $36-11=\boxed{025}$

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Difference between revisions of "2002 AIME I Problems/Problem 3"

Revision as of 18:54, 4 July 2013

Problem

Solution

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