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 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 ==
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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
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== Remark ==
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The 25 ordered pairs <math>(d, n)</math> are:
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(34, 9)
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(34, 20)
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(34, 31)
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(34, 42)
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(34, 53)
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(34, 64)
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(43, 10)
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(43, 21)
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(43, 32)
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(43, 43)
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(43, 54)
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(52, 11)
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(52, 22)
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(52, 33)
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(52, 44)
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(61, 1)
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(61, 12)
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(61, 23)
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(61, 34)
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(70, 2)
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(70, 13)
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(70, 24)
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(79, 3)
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(79, 14)
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(88, 4)
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~Puck_0
 
== 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}}
  
 
[[Category:Intermediate Combinatorics Problems]]
 
[[Category:Intermediate Combinatorics Problems]]
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{{MAA Notice}}

Latest revision as of 10:57, 22 December 2024

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

Remark

The 25 ordered pairs $(d, n)$ are: (34, 9) (34, 20) (34, 31) (34, 42) (34, 53) (34, 64) (43, 10) (43, 21) (43, 32) (43, 43) (43, 54) (52, 11) (52, 22) (52, 33) (52, 44) (61, 1) (61, 12) (61, 23) (61, 34) (70, 2) (70, 13) (70, 24) (79, 3) (79, 14) (88, 4) ~Puck_0

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

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