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2020 AMC 8 Problems/Problem 17

Revision as of 18:06, 19 November 2020 by Oceanxia (talk | contribs)

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 1

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 $4$. 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

Solution 3

The prime factorization of $2020$ is $2^2\cdot5\cdot101$ so it has $(2+1)(1+1)(1+1)=12$ factors. Then we can count that $1,2,4,5,101$ all have $3$ or fewer divisors so by complementary counting our answer is $12-5=\textbf{(B)}\ 7$.

-franzliszt

Solution 4

The prime factorization of $2020$ is $2^2\cdot5\cdot101$ so it has $(2+1)(1+1)(1+1)=12$ factors. Then we can count that $1,2,4,5,101$ which gets $12-5=\textbf{(B)}\ 7$. ~oceanxia~

See also

2020 AMC 8 (Problems • Answer Key • Resources)
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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