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

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{{duplicate|[[2019 AMC 10B Problems|2019 AMC 10B #19]] and [[2019 AMC 12B Problems|2019 AMC 12B #14]]}}
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==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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<math>\textbf{(A) }98\qquad\textbf{(B) }100\qquad\textbf{(C) }117\qquad\textbf{(D) }119\qquad\textbf{(E) }121</math>
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==Solution==
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The prime factorization of <math>100,000</math> 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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Notice that this is analogous to choosing a divisor of <math>100,000^2 = 2^{10}5^{10}</math>, which has <math>(10+1)(10+1) = 121</math> divisors. 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>. The last two cannot be so written because the maximum factor of <math>100,000</math> containing only <math>2</math>s or <math>5</math>s (and not both) is only <math>2^5</math> or <math>5^5</math>. Since the factors chosen must be distinct, the last two numbers cannot be so written because they would require <math>2^5 \cdot 2^5</math> or <math>5^5 \cdot 5^5</math>. The first two would require <math>1 \cdot 1</math> and <math>2^{5}5^{5} \cdot 2^{5}5^{5}</math>, respectively. 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 <math>0</math> or <math>10</math>, 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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==See Also==
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{{AMC10 box|year=2019|ab=B|num-b=18|num-a=20}}
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{{AMC12 box|year=2019|ab=B|num-b=13|num-a=15}}
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{{MAA Notice}}

Latest revision as of 18:04, 14 July 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 $5^5 \cdot 5^5$. The first two would require $1 \cdot 1$ and $2^{5}5^{5} \cdot 2^{5}5^{5}$, respectively. 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 (ProblemsAnswer KeyResources)
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 (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

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