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Difference between revisions of "1995 AIME Problems/Problem 6"

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== Problem ==
 
== Problem ==
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Let <math>\displaystyle n=2^{31}3^{19}.</math>  How many positive [[integer]] [[divisor]]s of <math>\displaystyle n^2</math> are less than <math>\displaystyle n_{}</math> but do not divide <math>\displaystyle n_{}</math>?
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Let <math>n=2^{31}3^{19}.</math>  How many positive [[integer]] [[divisor]]s of <math>n^2</math> are less than <math>n_{}</math> but do not divide <math>n_{}</math>?
  
 
== Solution ==
 
== Solution ==
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We know that <math>n^2</math> must have <math>63\times 39</math> [[factor]]s by its [[prime factorization]]. There are <math>\frac{63\times 39-1}{2} = 1228</math> factors of <math>n^2</math> that are less than <math>n</math>, because if they form pairs <math>a</math>, then there is one factor per pair that is less than <math>n</math>. There are <math>32\times20-1 = 639</math> factors of <math>n</math> that are less than <math>n</math> itself. These are also factors of <math>n^2</math>. Therefore, there are <math>1228-639=539</math> factors of <math>n</math> that do not divide <math>n</math>.
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We know that <math>n^2</math> must have <math>63\times 39</math> [[factor]]s by its [[prime factorization]]. There are <math>\frac{63\times 39-1}{2} = 1228</math> factors of <math>n^2</math> that are less than <math>n</math>, because if they form pairs <math>a</math>, then there is one factor per pair that is less than <math>n</math>. There are <math>32\times20-1 = 639</math> factors of <math>n</math> that are less than <math>n</math> itself. These are also factors of <math>n^2</math>. Therefore, there are <math>1228-639=\boxed{589}</math> factors of <math>n</math> that do not divide <math>n</math>.
  
 
== See also ==
 
== See also ==
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* [[1995 AIME Problems]]
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{{AIME box|year=1995|num-b=5|num-a=7}}
  
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{{AIME box|year=1995|num-b=5|num-a=7}}
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[[Category:Intermediate Number Theory Problems]]

Revision as of 14:14, 15 March 2008

Problem

Let $n=2^{31}3^{19}.$ How many positive integer divisors of $n^2$ are less than $n_{}$ but do not divide $n_{}$?

Solution

We know that $n^2$ must have $63\times 39$ factors by its prime factorization. There are $\frac{63\times 39-1}{2} = 1228$ factors of $n^2$ that are less than $n$, because if they form pairs $a$, then there is one factor per pair that is less than $n$. There are $32\times20-1 = 639$ factors of $n$ that are less than $n$ itself. These are also factors of $n^2$. Therefore, there are $1228-639=\boxed{589}$ factors of $n$ that do not divide $n$.

See also

1995 AIME (Problems • Answer Key • Resources)
Preceded by
Problem 5 		Followed by
Problem 7
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15
All AIME Problems and Solutions
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