Difference between revisions of "2000 AIME I Problems/Problem 11"
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Let <math>S</math> be the sum of all numbers of the form <math>a/b,</math> where <math>a</math> and <math>b</math> are [[relatively prime]] positive [[divisor]]s of <math>1000.</math> What is the [[floor function|greatest integer]] that does not exceed <math>S/10</math>? | Let <math>S</math> be the sum of all numbers of the form <math>a/b,</math> where <math>a</math> and <math>b</math> are [[relatively prime]] positive [[divisor]]s of <math>1000.</math> What is the [[floor function|greatest integer]] that does not exceed <math>S/10</math>? | ||
− | == Solution == | + | == Solution 1 == |
Since all divisors of <math>1000 = 2^35^3</math> can be written in the form of <math>2^{m}5^{n}</math>, it follows that <math>\frac{a}{b}</math> can also be expressed in the form of <math>2^{x}5^{y}</math>, where <math>-3 \le x,y \le 3</math>. Thus every number in the form of <math>a/b</math> will be expressed one time in the product | Since all divisors of <math>1000 = 2^35^3</math> can be written in the form of <math>2^{m}5^{n}</math>, it follows that <math>\frac{a}{b}</math> can also be expressed in the form of <math>2^{x}5^{y}</math>, where <math>-3 \le x,y \le 3</math>. Thus every number in the form of <math>a/b</math> will be expressed one time in the product | ||
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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>. | ||
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+ | == 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>. | ||
== See also == | == See also == |
Revision as of 20:46, 3 December 2016
Contents
[hide]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 .
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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