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Difference between revisions of "2019 AMC 10B Problems/Problem 19"

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==Problem==
 
==Problem==
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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>
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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> @
  
 
<math>\textbf{(A) }98\qquad\textbf{(B) }100\qquad\textbf{(C) }117\qquad\textbf{(D) }119\qquad\textbf{(E) }121</math>
 
<math>\textbf{(A) }98\qquad\textbf{(B) }100\qquad\textbf{(C) }117\qquad\textbf{(D) }119\qquad\textbf{(E) }121</math>

Revision as of 18:33, 22 February 2019

The following problem is from both the 2019 AMC 10B #19 and 2019 AMC 12B #14, so both problems redirect to this page.

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

The prime factorization of $100,000$ is $2^5 \cdot 5^5$. Thus, we choose two numbers $2^a5^b$ and $2^c5^d$ where $0 \le a,b,c,d \le 5$ and $(a,b) \neq (c,d)$, whose product is $2^{a+c}5^{b+d}$, where $0 \le a+c \le 10$ and $0 \le b+d \le 10$.

Notice that this is analogous to choosing a divisor of $100,000^2 = 2^{10}5^{10}$, which has $(10+1)(10+1) = 121$ divisors. However, some of the divisors of $2^{10}5^{10}$ cannot be written as a product of two distinct divisors of $2^5 \cdot 5^5$, namely: $1 = 2^05^0$, $2^{10}5^{10}$, $2^{10}$, and $5^{10}$. The last two cannot be so written because the maximum factor of $100,000$ containing only $2$s or $5$s (and not both) is only $2^5$ or $5^5$. Since the factors chosen must be distinct, the last two numbers cannot be so written because they would require $2^5 \cdot 2^5$ or $5^5 \cdot 5^5$. This gives $121-4 = 117$ candidate numbers. It is not too hard to show that every number of the form $2^p5^q$, where $0 \le p, q \le 10$, and $p,q$ are not both $0$ or $10$, can be written as a product of two distinct elements in $S$. Hence the answer is $\boxed{\textbf{(C) } 117}$.

See Also

2019 AMC 10B (Problems • Answer Key • Resources)
Preceded by
Problem 18 		Followed by
Problem 20
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 10 Problems and Solutions
2019 AMC 12B (Problems • Answer Key • Resources)
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

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