2020 AMC 8 Problems/Problem 17

Problem

How many positive integer factors of $2020$ have more than $3$ factors? (As an example, $12$ has $6$ factors, namely $1,2,3,4,6,$ and $12.$ )

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

Solution 1

Since $2020 = 2^2 \cdot 5 \cdot 101$ , we can simply list its factors: $1, 2, 4, 5, 10, 20, 101, 202, 404, 505, 1010, 2020.$ There are $12$ of these; only $1, 2, 4, 5, 101$ (i.e. $5$ of them) don't have over $3$ factors, so the remaining $12-5 = \boxed{\textbf{(B) }7}$ factors have more than $3$ factors.

Solution 2

As in Solution 1, we prime factorize $2020$ as $2^2\cdot 5\cdot 101$ , and we recall the standard formula that the number of positive factors of an integer is found by adding $1$ to each exponent in its prime factorization, and then multiplying these. Thus $2020$ has $(2+1)(1+1)(1+1) = 12$ factors. The only number which has one factor is $1$ . For a number to have exactly two factors, it must be prime, and the only prime factors of $2020$ are $2$ , $5$ , and $101$ . For a number to have three factors, it must be a square of a prime (this follows from the standard formula mentioned above), and from the prime factorization, the only square of a prime that is a factor of $2020$ is $4$ . Thus, there are $5$ factors of $2020$ which themselves have $1$ , $2$ , or $3$ factors (namely $1$ , $2$ , $4$ , $5$ , and $101$ ), so the number of factors of $2020$ that have more than $3$ factors is $12-5=\boxed{\textbf{(B) }7}$ .

Solution 3

Let $d(n)$ be the number of factors of n. We know by prime factorisation that $d(2022) = 12$ . These $12$ numbers can be divided into unordered pairs ${a,b}$ where $ab = 2022$ . Since $d(2022) = d(a)d(b)$ , one of $d(a), d(b)$ has $3$ or less factors and the other has $4$ or more - in to total $6$ factors of $2022$ with more than $3$ factors. However, this argument has exceptions where $a$ and $b$ share a nontrivial common factor, which in this case can only be two. There are two cases - One in which $5$ and $101$ divide the same factor, WLOG assumed to be $a$ , so that $d(a) = 2^3 > 3$ and $d(b) = 2$ , as otherwise. In the other case, $a = 5*2$ and $b = 101*2$ , so that $d(a) = d(b) = 4$ . This adds one factor with more than $3$ factors, so the answer is $\boxed{\textbf{(B) }7}$ .

Video Solution by OmegaLearn

https://youtu.be/Of-ZGiWgXyY?t=56

~ pi_is_3.14

Video Solution by North America Math Contest Go Go

https://www.youtube.com/watch?v=tUTFLUJ6a-4

~North America Math Contest Go Go Go

Video Solution by WhyMath

https://youtu.be/2dazhQ31I14

~savannahsolver

Video Solution

https://youtu.be/VnOecUiP-SA

Video Solution by Interstigation

https://youtu.be/YnwkBZTv5Fw?t=782

~Interstigation

2020 AMC 8 (Problems • Answer Key • Resources)
Preceded by Problem 16	Followed by Problem 18
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