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Difference between revisions of "2000 AIME I Problems/Problem 11"

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== Problem ==
 
== Problem ==
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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 divisors of <math>1000.</math> What is the greatest integer that does not exceed <math>S/10</math>?
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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>?
  
 
== Solution ==
 
== Solution ==
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{{solution}}
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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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<cmath>(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)</cmath>
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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>.
  
 
== See also ==
 
== See also ==
 
{{AIME box|year=2000|n=I|num-b=10|num-a=12}}
 
{{AIME box|year=2000|n=I|num-b=10|num-a=12}}
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[[Category:Intermediate Algebra Problems]]

Revision as of 11:56, 1 January 2008

Problem

Let $S$ be the sum of all numbers of the form $a/b,$ where $a$ and $b$ are relatively prime positive divisors of $1000.$ What is the greatest integer that does not exceed $S/10$?

Solution

Since all divisors of $1000 = 2^35^3$ can be written in the form of $2^{m}5^{n}$, it follows that $\frac{a}{b}$ can also be expressed in the form of $2^{x}5^{y}$, where $-3 \le x,y \le 3$. Thus every number in the form of $a/b$ will be expressed one time in the product

\[(2^{-3} + 2^{-2} + 2^{-1} + 2^{0} + 2^{1} + 2^2 + 2^3)(5^{-3} + 5^{-2} +5^{-1} + 5^{0} + 5^{1} + 5^2 + 5^3)\]

Using the formula for a geometric series, this reduces to $S = \frac{2^{-3}(2^7 - 1)}{2-1} \cdot \frac{5^{-3}(5^{7} - 1)}{5-1} = \frac{127 \cdot 78124}{4000} = 2480 + \frac{437}{1000}$, and $\left\lfloor \frac{S}{10} \right\rfloor = \boxed{248}$.

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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