Difference between revisions of "2019 AMC 12B Problems/Problem 14"

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==Solution==
 
==Solution==
  
First, find the prime factorization of <math>100,000</math>. It is <math>2^5 \cdot 5^5</math>. Thus, any factor will have the pattern <math>2^m \cdot 5^n</math>, where <math>m, n < 5</math>. Multiplying this by another factor with the same pattern <math>2^k \cdot 5^l</math> gets us <math>2^m+k \cdot 5^n+l</math>. It initially seems like we have <math>11</math> options for the power of <math>2</math> and <math>11</math> options for the power of <math>5</math>, giving us a total of <math>121</math> choices. However, note that the factors must be distinct. If they are distinct, we cannot have <math>1</math> (as it is only formed by <math>1 \cdot 1</math>), or <math>2^10 \cdot 5^10</math> (as it is only formed by <math>100,000 \cdot 100,000</math>). These are the only two cases where the distinction rule forces us to eliminate case, and therefore the answer is <math>121-2</math> D
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First, find the prime factorization of <math>100,000</math>. It is <math>2^5 \cdot 5^5</math>. Thus, any factor will have the pattern <math>2^m \cdot 5^n</math>, where <math>m, n < 5</math>. Multiplying this by another factor with the same pattern <math>2^k \cdot 5^l</math> gets us <math>2^m+k \cdot 5^n+l</math>. It initially seems like we have <math>11</math> options for the power of <math>2</math> and <math>11</math> options for the power of <math>5</math>, giving us a total of <math>121</math> choices. However, note that the factors must be distinct. If they are distinct, we cannot have <math>1</math> (as it is only formed by <math>1 \cdot 1</math>), or <math>2^{10} \cdot 5^{10}</math> (as it is only formed by <math>100,000 \cdot 100,000</math>). These are the only two cases where the distinction rule forces us to eliminate case, and therefore the answer is <math>121-2</math> D
  
 
==See Also==
 
==See Also==
 
{{AMC12 box|year=2019|ab=B|num-b=13|num-a=15}}
 
{{AMC12 box|year=2019|ab=B|num-b=13|num-a=15}}

Revision as of 13:07, 14 February 2019

Problem

Let $S$ be the set of all positive integer divisors of $100,000.$ How many numbers are the product of two distinct elements of $S?$

Solution

First, find the prime factorization of $100,000$. It is $2^5 \cdot 5^5$. Thus, any factor will have the pattern $2^m \cdot 5^n$, where $m, n < 5$. Multiplying this by another factor with the same pattern $2^k \cdot 5^l$ gets us $2^m+k \cdot 5^n+l$. It initially seems like we have $11$ options for the power of $2$ and $11$ options for the power of $5$, giving us a total of $121$ choices. However, note that the factors must be distinct. If they are distinct, we cannot have $1$ (as it is only formed by $1 \cdot 1$), or $2^{10} \cdot 5^{10}$ (as it is only formed by $100,000 \cdot 100,000$). These are the only two cases where the distinction rule forces us to eliminate case, and therefore the answer is $121-2$ D

See Also

2019 AMC 12B (ProblemsAnswer KeyResources)
Preceded by
Problem 13
Followed by
Problem 15
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All AMC 12 Problems and Solutions
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