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

(I wrote the solution using LaTeX.)
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==Solution==
 
==Solution==
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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>, so the answer is <math>121 - 1 - 1 = 119</math>.
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Solution by greersc.
  
 
==See Also==
 
==See Also==
 
{{AMC10 box|year=2019|ab=B|num-b=18|num-a=20}}
 
{{AMC10 box|year=2019|ab=B|num-b=18|num-a=20}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 16:50, 14 February 2019

Problem

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 $S^2$, so the answer is $121 - 1 - 1 = 119$.

Solution by greersc.

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

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