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2019 AMC 10B Problems/Problem 19

Revision as of 08:36, 15 February 2019 by Mshell214 (talk | contribs) (→‎Solution 2)
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 1

To find the number of numbers that are the product of two distinct elements of $S$, we first square $100,000$ and factor it. Factoring, we find $100,000^2 = 2^{10} \cdot 5^{10}$. Therefore, $100,000^2$ has $(10 + 1)(10 + 1) = 121$ distinct factors. Each of these can be achieved by multiplying two factors of $S$. However, the factors must be distinct, so we eliminate $1$ and $100,000^2$, as well as $2^{10}$ and $5^{10}$, so the answer is $121 - 4 = 117$.

Solution by greersc. (Edited by AZAZ12345 and then by greersc once again)

Solution 2

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 $5^{10}$. The last two can not be created because the maximum factor of 100,000 involving only 2s or 5s is only $2^5 or 5^5$. Since the factors chosen must be distinct, the last two numbers cannot be created because those 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 $\boxed{\textbf{(C) } 117}$.

-scrabbler94 (edited by mshell214)

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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