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

m (Solution)
(Solution)
Line 5: Line 5:
  
 
==Solution==
 
==Solution==
To find the number of numbers that are the product of two distinct elements of <math>S</math>, we first square <math>S</math> and factor it. Factoring, we find <math>S^2 = 2^{10} \cdot 5^{10}</math>. Therefore, <math>S^2</math> has <math>(10 + 1)(10 + 1) = 121</math> distinct factors. Each of these can be achieved by multiplying two factors of <math>S</math>. However, the factors must be distinct, so we eliminate <math>1</math> and <math>S^2</math>, which is <math>2^10 /cdot 5^10</math> so the answer is <math>121 - 4 = 117</math>.  
+
To find the number of numbers that are the product of two distinct elements of <math>S</math>, we first square <math>S</math> and factor it. Factoring, we find <math>S^2 = 2^{10} \cdot 5^{10}</math>. Therefore, <math>S^2</math> has <math>(10 + 1)(10 + 1) = 121</math> distinct factors. Each of these can be achieved by multiplying two factors of <math>S</math>. However, the factors must be distinct, so we eliminate <math>1</math> and <math>S^2</math>, which is <math>2^{10}\cdot 5^{10}</math>, so the answer is <math>121 - 4 = 117</math>.  
  
 
Solution by greersc. (Edited by AZAZ12345 and then by greersc once again)
 
Solution by greersc. (Edited by AZAZ12345 and then by greersc once again)

Revision as of 17:49, 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

To find the number of numbers that are the product of two distinct elements of $S$, we first square $S$ and factor it. Factoring, we find $S^2 = 2^{10} \cdot 5^{10}$. Therefore, $S^2$ 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 $S^2$, which is $2^{10}\cdot 5^{10}$, so the answer is $121 - 4 = 117$.

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

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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png