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
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
.
- NOTE: Hi, I'm just someone wondering: isn't the question saying that S is the sum of fractions where a and b are relatively prime divisors of 1000? So for example 5/8 would be okay but 10/25 wouldn't. If that's the case, then a and b would only be 5, 25, 125, 2, 4, and 8. So S would be:
(5+25+125)/8 + (5+25+125)/4 + (5+25+125)/2 + (2+4+8)/5 + (2+4+8)/25 + (2+4+8)/125 = 434/125 + 1085/8 then our solution would be the largest integer less than (3+135)/10, which would be 13.
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 | ||
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