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Difference between revisions of "1990 AIME Problems/Problem 5"

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== Problem ==
 
== Problem ==
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Let <math>n^{}_{}</math> be the smallest positive integer that is a multiple of <math>75_{}^{}</math> and has exactly <math>75_{}^{}</math> positive integral divisors, including <math>1_{}^{}</math> and itself. Find <math>n/75^{}_{}</math>.
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Let <math>n^{}_{}</math> be the smallest positive [[integer]] that is a multiple of <math>75_{}^{}</math> and has exactly <math>75_{}^{}</math> positive integral divisors, including <math>1_{}^{}</math> and itself. Find <math>n/75^{}_{}</math>.
  
 
== Solution ==
 
== Solution ==
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The [[prime factorization]] of <math>75 = 3^15^2</math>. Thus, for <math>n</math> to have exactly <math>75</math> integral divisors, we need to have <math>\displaystyle n = a^{x-1}b^{y-1}\ldots</math> such that <math>x \cdot y \cdot \ldots = 75</math>. Since we know that <math>n</math> is [[divisible]] by <math>75</math>, two of the [[prime]] [[factor]]s must be <math>3</math> and <math>5</math>. To minimize <math>n</math>, a third factor which is less than <math>5</math> can be used; the only possible [[prime]] number is <math>2</math>. Also to minimize <math>n</math>, we want <math>5</math>, the greatest of all the factors, to be raised to the least power. Therefore, <math>n = 2^43^45^2</math> and <math>\frac{n}{75} = \frac{2^43^45^2}{3 \cdot 5^2} = 16 \cdot 27 = 432</math>.
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The [[prime factorization]] of <math>75 = 3^15^2</math>. Thus, for <math>n</math> to have exactly <math>75</math> integral divisors, we need to have <math>n = a^{x-1}b^{y-1}\ldots</math> such that <math>x \cdot y \cdot \ldots = 75</math>. Since we know that <math>n</math> is [[divisible]] by <math>75</math>, two of the [[prime]] [[factor]]s must be <math>3</math> and <math>5</math>. To minimize <math>n</math>, a third factor which is less than <math>5</math> can be used; the only possible [[prime]] number is <math>2</math>. Also to minimize <math>n</math>, we want <math>5</math>, the greatest of all the factors, to be raised to the least power. Therefore, <math>n = 2^43^45^2</math> and <math>\frac{n}{75} = \frac{2^43^45^2}{3 \cdot 5^2} = 16 \cdot 27 = 432</math>.
  
 
== See also ==
 
== See also ==
 
{{AIME box|year=1990|num-b=4|num-a=6}}
 
{{AIME box|year=1990|num-b=4|num-a=6}}
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[[Category:Intermediate Number Theory Problems]]

Revision as of 22:00, 24 November 2007

Problem

Let $n^{}_{}$ be the smallest positive integer that is a multiple of $75_{}^{}$ and has exactly $75_{}^{}$ positive integral divisors, including $1_{}^{}$ and itself. Find $n/75^{}_{}$.

Solution

The prime factorization of $75 = 3^15^2$. Thus, for $n$ to have exactly $75$ integral divisors, we need to have $n = a^{x-1}b^{y-1}\ldots$ such that $x \cdot y \cdot \ldots = 75$. Since we know that $n$ is divisible by $75$, two of the prime factors must be $3$ and $5$. To minimize $n$, a third factor which is less than $5$ can be used; the only possible prime number is $2$. Also to minimize $n$, we want $5$, the greatest of all the factors, to be raised to the least power. Therefore, $n = 2^43^45^2$ and $\frac{n}{75} = \frac{2^43^45^2}{3 \cdot 5^2} = 16 \cdot 27 = 432$.

See also

1990 AIME (Problems • Answer Key • Resources)
Preceded by
Problem 4 		Followed by
Problem 6
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15
All AIME Problems and Solutions
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