1985 AHSME Problems/Problem 12
Problem
Let , and be distinct prime numbers, where is not considered a prime. Which of the following is the smallest positive perfect cube having as a divisor?
Solution
A number of the form will be a perfect cube precisely when , , and are multiples of 3. (Clearly, since we are looking for the smallest possible perfect cube, we can assume that it has no other prime factors.) Furthermore, for it to be a multiple of , we must have , , and . The smallest multiple of that is at least is , the smallest that is at least is again , and the smallest that is at least is . Hence the smallest possible perfect cube with as a divisor is .
See Also
1985 AHSME (Problems • Answer Key • Resources) | ||
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Followed by Problem 13 | |
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