1984 AHSME Problems/Problem 21: Difference between revisions
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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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