1995 AIME Problems/Problem 6
Contents
[hide]Problem
Let How many positive integer divisors of are less than but do not divide ?
Solution 1
We know that must have factors by its prime factorization. If we group all of these factors (excluding ) into pairs that multiply to , then one factor per pair is less than , and so there are factors of that are less than . There are factors of , which clearly are less than , but are still factors of . Therefore, there are factors of that do not divide .
Solution 2
Let for some prime . Then has factors less than .
This simplifies to .
The number of factors of less than is equal to .
Thus, our general formula for is
Number of factors that satisfy the above
Incorporating this into our problem gives .
See also
1995 AIME (Problems • Answer Key • Resources) | ||
Preceded by Problem 5 |
Followed by Problem 7 | |
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