1995 AIME Problems/Problem 6
Contents
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, using complementary counting, 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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