2000 AIME I Problems/Problem 11
Contents
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
![\[(2^{-3} + 2^{-2} + 2^{-1} + 2^{0} + 2^{1} + 2^2 + 2^3)(5^{-3} + 5^{-2} +5^{-1} + 5^{0} + 5^{1} + 5^2 + 5^3)\]](http://latex.artofproblemsolving.com/4/5/8/45856060696df39fc49e29ee269d2a07c63d853d.png)
Using the formula for a geometric series, this reduces to 
, and 
.
Solution 2
Essentially, the problem asks us to compute ![\[\sum_{a=-3}^3 \sum_{b=-3}^3 \frac{2^a}{5^b}\]](http://latex.artofproblemsolving.com/a/b/9/ab96b03ec3fab6b3b9afad62ccaa804264f99561.png)
which is pretty easy: ![\[\sum_{a=-3}^3 \sum_{b=-3}^3 \frac{2^a}{5^b} = \sum_{a=-3}^3 2^a \sum_{b=-3}^3 \frac{1}{5^b} = \sum_{a=-3}^3 2^a 5^{3}\bigg( \frac{1-5^{-7}}{1-\frac{1}{5}} \bigg) = 5^{3}\bigg( \frac{1-5^{-7}}{1-\frac{1}{5}} \bigg) \sum_{a=-3}^3 2^a = 5^{3}\bigg( \frac{1-5^{-7}}{1-\frac{1}{5}} \bigg)2^{-3} \bigg( \frac{1-2^7}{1-2} \bigg) = 2480 + \frac{437}{1000}\]](http://latex.artofproblemsolving.com/7/b/a/7ba344eb0b1f83da9307067a05bd49bdc53651b8.png)
so our answer is 
.
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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