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

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Let <math>S</math> be the set of all positive integer divisors of <math>100,000.</math> How many numbers are the product of two distinct elements of <math>S?</math>
 
Let <math>S</math> be the set of all positive integer divisors of <math>100,000.</math> How many numbers are the product of two distinct elements of <math>S?</math>
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<math>\textbf{(A) }98\qquad\textbf{(B) }100\qquad\textbf{(C) }117\qquad\textbf{(D) }119\qquad\textbf{(E) }121</math>
  
 
==Solution==
 
==Solution==

Revision as of 15:31, 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 $S?$

$\textbf{(A) }98\qquad\textbf{(B) }100\qquad\textbf{(C) }117\qquad\textbf{(D) }119\qquad\textbf{(E) }121$

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 cases, and therefore the answer is $121-2 = \boxed{D}$

- Robin's solution

See Also

2019 AMC 12B (ProblemsAnswer KeyResources)
Preceded by
Problem 13
Followed by
Problem 15
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25
All AMC 12 Problems and Solutions