Difference between revisions of "2002 AIME I Problems/Problem 3"
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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> | ||
| + | Notice how the answer is equal to Jane's age | ||
== See also == | == See also == | ||
Revision as of 13:12, 28 May 2017
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 number 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 
. 
Notice how the answer is equal to Jane's age
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. 
