1995 AIME Problems/Problem 6
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, using complementary counting, there are factors of that do not divide .
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 .
Consider divisors of such that . WLOG, let and
Then, it is easy to see that will always be less than as we go down the divisor list of until we hit .
Therefore, the median divisor of is .
Then, there are divisors of . Exactly of these divisors are
There are divisors of that are .
Therefore, the answer is .
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