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

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== Solution ==
 
== Solution ==
{{solution}}
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Let Jane's age <math>n</math> years from now be <math>10a+b</math>, and let Dick's age be <math>10b+a</math>. If <math>10b+a>10a+b</math>, then <math>b>a</math>. The possible pairs of <math>a,b</math> are:
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<cmath>(1,2), (1,3), (2,3), (1,4), (2,4), (3,4), \dots , (8,9)</cmath>
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That makes 36. But <math>10a+b>25</math>, so we subtract all the extraneous pairs: <math>(1,2), (1,3), (2,3), (1,4), (2,4), (1,5), (2,5), (1,6), (1,7), (1,8),</math> and <math>(1,9)</math>. <math>36-11=\boxed{025}</math>
  
 
== See also ==
 
== See also ==
 
{{AIME box|year=2002|n=I|num-b=2|num-a=4}}
 
{{AIME box|year=2002|n=I|num-b=2|num-a=4}}
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[[Category:Intermediate Combinatorics Problems]]

Revision as of 09:49, 1 July 2008

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 $(1,9)$. $36-11=\boxed{025}$

See also

2002 AIME I (ProblemsAnswer KeyResources)
Preceded by
Problem 2
Followed by
Problem 4
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15
All AIME Problems and Solutions