Difference between revisions of "2002 AIME I Problems/Problem 3"
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== Problem == | == Problem == | ||
| − | Jane is 25 years old. Dick is older than Jane. In <math>n</math> years, where <math>n</math> is a positive integer, Dick's age and Jane's age will both be two-digit | + | Jane is 25 years old. Dick is older than Jane. In <math>n</math> years, where <math>n</math> 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 <math>d</math> be Dick's present age. How many ordered pairs of positive integers <math>(d,n)</math> are possible? |
== Solution == | == Solution == | ||
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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> | 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> | ||
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| + | == Solution 2 == | ||
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| + | Start by assuming that <math>n < 5</math> (essentially, Jane is in the 20s when their ages are 'reverses' of each other). Then we get the pairs <cmath>(61,1),(70,2),(79,3),(88,4).</cmath> Repeating this for the 30s gives <cmath>(34,9),(43,10),(52,11),(61,12),(70,13),(79,14).</cmath> From here, it's pretty clear that every decade we go up we get <math>(d,n+11)</math> as a pair. Since both ages must always be two-digit numbers, we can show that each decade after the 30s, we get 1 fewer option. Therefore, our answer is <math>4+6+5+\dots+2+1=4+21=\boxed{025}.</math> | ||
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| + | ~Dhillonr25 | ||
== See also == | == See also == | ||
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[[Category:Intermediate Combinatorics Problems]] | [[Category:Intermediate Combinatorics Problems]] | ||
| + | {{MAA Notice}} | ||
Latest revision as of 16:02, 21 May 2023
Contents
Problem
Jane is 25 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 Jane's age years from now be 
, and let Dick's age be 
. If 
, then 
. The possible pairs of 
are:
That makes 36. But 
, so we subtract all the extraneous pairs: 
and 
. 
Solution 2
Start by assuming that 
(essentially, Jane is in the 20s when their ages are 'reverses' of each other). Then we get the pairs ![\[(61,1),(70,2),(79,3),(88,4).\]](http://latex.artofproblemsolving.com/e/1/f/e1f55dfbef298d9189565fe7fd2265d08e7277ce.png)
Repeating this for the 30s gives ![\[(34,9),(43,10),(52,11),(61,12),(70,13),(79,14).\]](http://latex.artofproblemsolving.com/a/b/e/abe202b40f3ef28bddc6a62b026f656e84577455.png)
From here, it's pretty clear that every decade we go up we get 
as a pair. Since both ages must always be two-digit numbers, we can show that each decade after the 30s, we get 1 fewer option. Therefore, our answer is 
~Dhillonr25
See also
| 2002 AIME I (Problems • Answer Key • Resources) | ||
| 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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
