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> | ||
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== See also == | == See also == |
Revision as of 20:32, 21 October 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:
![$(1,2), (1,3), (2,3), (1,4), (2,4), (3,4), \dots , (8,9)$](http://latex.artofproblemsolving.com/1/4/4/14453015377b67cb285015b2c9d7cd8597a5cb43.png)
That makes 36. But , so we subtract all the extraneous pairs:
and
.
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.