# 2020 AMC 8 Problems/Problem 17

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## 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 $12-5=\boxed{\textbf{(B) }7}$.

~savannahsolver

## Video Solution by Interstigation

~Interstigation

 2020 AMC 8 (Problems • Answer Key • Resources) Preceded byProblem 16 Followed byProblem 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