Difference between revisions of "2020 AMC 8 Problems/Problem 17"

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==Solution==
 
==Solution==
We list out the factors of <math>2020</math>: <cmath>1, 2, 4, 5, 10, 20, 101, 202, 404, 505, 1010, 2020.</cmath> Of these, only <math>1, 2, 4, 5, 101</math> (<math>5</math> of them) do not have more than <math>3</math> factors. Therefore the answer is <math>\tau{2020}-5=\boxed{\textbf{(B) }7}</math>.
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We list out the factors of <math>2020</math>: <cmath>1, 2, 4, 5, 10, 20, 101, 202, 404, 505, 1010, 2020.</cmath> Of these, only <math>1, 2, 4, 5, 101</math> (<math>5</math> of them) do not have more than <math>3</math> factors. Therefore the answer is <math>\tau\left({2020}\right)-5=\boxed{\textbf{(B) }7}</math>.
  
 
==See also==
 
==See also==

Revision as of 01:08, 18 November 2020

How many positive integer factors of $2020$ have more than $3$ factors?

$\textbf{(A) }6 \qquad \textbf{(B) }7 \qquad \textbf{(C) }8 \qquad \textbf{(D) }9 \qquad \textbf{(E) }10$

Solution

We list out the factors of $2020$: \[1, 2, 4, 5, 10, 20, 101, 202, 404, 505, 1010, 2020.\] Of these, only $1, 2, 4, 5, 101$ ($5$ of them) do not have more than $3$ factors. Therefore the answer is $\tau\left({2020}\right)-5=\boxed{\textbf{(B) }7}$.

See also

2020 AMC 8 (ProblemsAnswer KeyResources)
Preceded by
Problem 16
Followed by
Problem 18
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 AJHSME/AMC 8 Problems and Solutions

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