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

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==Solution==
 
==Solution==
  
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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 cases, and therefore the answer is <math>121-2 = \boxed{D}</math>
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The prime factorization of 100,000 is <math>2^5 \cdot 5^5</math>. Thus, we choose two numbers <math>2^a5^b</math> and <math>2^c5^d</math> where <math>0 \le a,b,c,d \le 5</math> and <math>(a,b) \neq (c,d)</math>, whose product is <math>2^{a+c}5^{b+d}</math>, where <math>0 \le a+c \le 10</math> and <math>0 \le b+d \le 10</math>.
  
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- Robin's solution
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Consider <math>100000^2 = 2^{10}5^{10}</math>. The number of divisors is <math>(10+1)(10+1) = 121</math>. However, some of the divisors of <math>2^{10}5^{10}</math> cannot be written as a product of two distinct divisors of <math>2^5 \cdot 5^5</math>, namely: <math>1 = 2^05^0</math>, <math>2^{10}5^{10}</math>, <math>2^{10}</math>, and <math>5^{10}</math>. This gives <math>121-4 = 117</math> candidate numbers. It is not too hard to show that every number of the form <math>2^p5^q</math> where <math>0 \le p, q \le 10</math>, and <math>p,q</math> are not both 0 or 10, can be written as a product of two distinct elements in <math>S</math>. Hence the answer is <math>\boxed{\textbf{(C) } 117}</math>.
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-scrabbler94
  
 
==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 14:40, 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

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 $0 \le b+d \le 10$.

Consider $100000^2 = 2^{10}5^{10}$. The number of divisors is $(10+1)(10+1) = 121$. 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 $5^{10}$. 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 $\boxed{\textbf{(C) } 117}$.

-scrabbler94

See Also

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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