1996 AHSME Problems/Problem 29
Contents
Problem
If is a positive integer such that has positive divisors and has positive divisors, then how many positive divisors does have?
Solution 1
Working with the second part of the problem first, we know that has divisors. We try to find the various possible prime factorizations of by splitting into various products of or integers.
The variables are different prime factors, and one of them must be . We now try to count the factors of , to see which prime factorization is correct and has factors.
In the first case, is the only possibility. This gives , which has factors, which is way too many.
In the second case, gives . If , then there are factors, while if , there are factors.
In the second case, gives . If , then there are factors, while if , there are factors.
In the third case, gives . If , then there are factors, while if , there are factors.
In the third case, gives . If , then there are factors, while if , there are factors.
In the fourth case, gives . If , then there are factors. This is the factorization we want.
Thus, , which has factors, and , which has factors.
In this case, , which has factors, and the answer is
Solution 2
Because has factors and has factors, we should rewrite the number As the formula for the number of divisors for such a number gives: We plug in the variations we need to make for the cases and . has has
If we take the top and divide by the bottom, we get the following equation: . Letting and for convenience and expanding this out gives us:
We can use Simon's Favorite Factoring Trick (SFFT) to turn this back into:
As we want to be dealing with rather reasonable numbers for and , we try to make the term the slightly larger term and the term the slightly smaller term. This effect is achieved when and . Therefore, . We get that this already satisfies the requirements for the number we are looking for, and we take
See also
1996 AHSME (Problems • Answer Key • Resources) | ||
Preceded by Problem 28 |
Followed by Problem 30 | |
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