2000 AIME I Problems/Problem 11: Difference between revisions
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Using the formula for a [[geometric series]], this reduces to <math>S = \frac{2^{-3}(2^7 - 1)}{2-1} \cdot \frac{5^{-3}(5^{7} - 1)}{5-1} = \frac{127 \cdot 78124}{4000} = 2480 + \frac{437}{1000}</math>, and <math>\left\lfloor \frac{S}{10} \right\rfloor = \boxed{248}</math>. | Using the formula for a [[geometric series]], this reduces to <math>S = \frac{2^{-3}(2^7 - 1)}{2-1} \cdot \frac{5^{-3}(5^{7} - 1)}{5-1} = \frac{127 \cdot 78124}{4000} = 2480 + \frac{437}{1000}</math>, and <math>\left\lfloor \frac{S}{10} \right\rfloor = \boxed{248}</math>. | ||
== See also == | == See also == | ||
Revision as of 14:49, 21 January 2015
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
![\[(2^{-3} + 2^{-2} + 2^{-1} + 2^{0} + 2^{1} + 2^2 + 2^3)(5^{-3} + 5^{-2} +5^{-1} + 5^{0} + 5^{1} + 5^2 + 5^3)\]](http://latex.artofproblemsolving.com/4/5/8/45856060696df39fc49e29ee269d2a07c63d853d.png)
Using the formula for a geometric series, this reduces to 
, and 
.
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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