2000 AIME I Problems/Problem 11: Difference between revisions
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== Solution 2 == | == Solution 2 == | ||
Essentially, the problem asks us to compute <cmath>\sum_{a=-3}^3 \sum_{b=-3}^3 \frac{2^a}{5^b}</cmath> which is pretty easy: <cmath>\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}</cmath> so our answer is <math>\left\lfloor \frac{2480 + \frac{437}{1000}}{10} \right\rfloor = \boxed{248}</math>. | Essentially, the problem asks us to compute <cmath>\sum_{a=-3}^3 \sum_{b=-3}^3 \frac{2^a}{5^b}</cmath> which is pretty easy: <cmath>\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}</cmath> so our answer is <math>\left\lfloor \frac{2480 + \frac{437}{1000}}{10} \right\rfloor = \boxed{248}</math>. | ||
== Solution 3 == | |||
The sum is equivalent to | |||
<math>\sum_{i | 10^6}^{} \frac{i}{1000}</math> | |||
Therefore, it's the sum of the factors of <math>10^6</math> | |||
divided by <math>1000</math>. The sum is <math>\frac{127 \times 19531}{1000}</math> by the sum of factors formula. The answer is therefore <math>\boxed{248}</math> after some computation. | |||
== See also == | == See also == | ||
Revision as of 12:39, 19 June 2019
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 
.
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.
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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