2020 AMC 8 Problems/Problem 17

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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 $\tau\left({2020}\right)-5=\boxed{\textbf{(B) }7}$.

Solution 2

The prime factorization of 2020 is $2^2\cdot5\cdot101$. The total number of factors of $2020$ is given by the product of one more than each of the prime powers which comes out to $3\cdot2\cdot2=12$. Instead of finding how many factors of $2020$ have more than three factors, we will instead find how many have one, two, or three factors and subtract this number from $12$ to find the answer. The only number which has one factor is $1$. For a number to have exactly two factors, it must be prime. From the prime factorization of $2020$, we know that these can only be $2,5,$ and $101$. For a number to have three factors, it must be a square of a prime. The only square of a prime that is a factor of $2020$ is 4. Thus our list of factors is $1,2,4,5,$ and $101$ which is a total five factors. Thus, the number of factors of $2020$ that have more than three factors is $12-5=7 \implies\boxed{\textbf{(B) }7}$.
~ junaidmansuri


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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