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 21:46, 3 December 2016
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
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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