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

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== Problem ==
 
== Problem ==
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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 number 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?
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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 ==
 
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:
 
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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<center><math>(1,2), (1,3), (2,3), (1,4), (2,4), (3,4), \dots , (8,9)</math></center>
  
 
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]]
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{{MAA Notice}}

Latest revision as of 16:02, 21 May 2023

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 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?

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

Solution 2

Start by assuming that $n < 5$ (essentially, Jane is in the 20s when their ages are 'reverses' of each other). Then we get the pairs \[(61,1),(70,2),(79,3),(88,4).\] Repeating this for the 30s gives \[(34,9),(43,10),(52,11),(61,12),(70,13),(79,14).\] From here, it's pretty clear that every decade we go up we get $(d,n+11)$ 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 $4+6+5+\dots+2+1=4+21=\boxed{025}.$

~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

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