Difference between revisions of "1984 AHSME Problems/Problem 21"
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Latest revision as of 11:51, 5 July 2013
Problem
The number of triples of positive integers which satisfy the simultaneous equations
is
Solution
We can factor the second equation to get , so we see that must be a factor of , and since this is prime, or . However, if , then , which is impossible for the field of positive integers. Therefore, for all possible solutions. Substituting this into the original equations gives
and
.
From the second equation, , and substituting this into the first equation yields , or . Factoring this gives , so or . Both of these yield integer solutions for , giving or , respectively. Therefore, the only solutions are and , yielding solutions, .
See Also
1984 AHSME (Problems • Answer Key • Resources) | ||
Preceded by Problem 20 |
Followed by Problem 22 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 • 26 • 27 • 28 • 29 • 30 | ||
All AHSME Problems and Solutions |
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