2002 AMC 10P Problems/Problem 8
Problem
How many ordered triples of positive integers satisfy
Solution 1
The given expression, , is equivalent to . Next, notice how must be a power of less than and the exponent of its prime factorization must be a factor of otherwise, won't be an integer. We can use a bit of case work to solve this problem.
Case 1:
Clearly, our only four solutions are and along with and
Case 2:
Clearly, our only two solutions are and
Case 3:
Clearly, our only two solutions are and
Case 4:
Clearly, our only solution is
Adding up all our cases gives
Thus, our answer is .
See also
2002 AMC 10P (Problems • Answer Key • Resources) | ||
Preceded by Problem 7 |
Followed by Problem 9 | |
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All AMC 10 Problems and Solutions |
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