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

Revision as of 14:19, 29 December 2020

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) have at least $3$ 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 $12-5=\boxed{\textbf{(B) }7}$ .

Video Solution

https://youtu.be/SPNobOd4t1c (Includes all the problems and has a free class update)

https://youtu.be/VnOecUiP-SA

@@ Line 5: / Line 5: @@
 ==Solution 1==
-Since <math>2020 = 2^2 \cdot 5 \cdot 101</math>, we can simply list its factors: <cmath>1, 2, 4, 5, 10, 20, 101, 202, 404, 505, 1010, 2020.</cmath> There are <math>12</math> of these; only <math>1, 2, 4, 5, 101</math> (i.e. <math>5</math> of them) have at most <math>3</math> factors, so the remaining <math>12-5 = \boxed{\textbf{(B) }7}</math> factors have more than <math>3</math> factors.
+Since <math>2020 = 2^2 \cdot 5 \cdot 101</math>, we can simply list its factors: <cmath>1, 2, 4, 5, 10, 20, 101, 202, 404, 505, 1010, 2020.</cmath> There are <math>12</math> of these; only <math>1, 2, 4, 5, 101</math> (i.e. <math>5</math> of them) have at least <math>3</math> factors, so the remaining <math>12-5 = \boxed{\textbf{(B) }7}</math> factors have more than <math>3</math> factors.
 ==Solution 2==

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