2000 AIME I Problems/Problem 11
Problem
Let be the sum of all numbers of the form
where
and
are relatively prime positive divisors of
What is the greatest integer that does not exceed
?
Solution 1
Since all divisors of can be written in the form of
, it follows that
can also be expressed in the form of
, where
. Thus every number in the form of
will be expressed one time in the product
Using the formula for a geometric series, this reduces to , and
.
Solution 2
Essentially, the problem asks us to compute which is pretty easy:
so our answer is
.
Solution 3
The sum is equivalent to
Therefore, it's the sum of the factors of
divided by
. The sum is
by the sum of factors formula. The answer is therefore
after some computation.
- whatRthose
See also
2000 AIME I (Problems • Answer Key • Resources) | ||
Preceded by Problem 10 |
Followed by Problem 12 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |
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