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

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==Solution 1==
 
==Solution 1==
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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) don't have over <math>3</math> factors, so the remaining <math>12-5 = \boxed{\textbf{(B) }7}</math> factors have more than <math>3</math> factors.
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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> factors; only <math>1, 2, 4, 5, 101</math> don't have over <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==
 
==Solution 2==
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==Solution 3==
 
==Solution 3==
  
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Let <math>d(n)</math> be the number of factors of n. We know by prime factorisation that <math>d(2022) = 12</math>. These <math>12</math> numbers can be divided into unordered pairs <math>{a,b}</math> where <math>ab = 2022</math>. Since <math>d(2022) = d(a)d(b)</math>, one of <math>d(a), d(b)</math> has <math>3</math> or less factors and the other has <math>4</math> or more - in to total <math>6</math> factors of <math>2022</math> with more than <math>3</math> factors. However, this argument has exceptions where <math>a</math> and <math>b</math> share a nontrivial common factor, which in this case can only be two. There are two cases - One in which <math>5</math> and <math>101</math> divide the same factor, WLOG assumed to be <math>a</math>, so that <math>d(a) = 2^3 > 3</math> and <math>d(b) = 2</math>, as otherwise. In the other case, <math>a = 5\cdot2</math> and <math>b = 101\cdot2</math>, so that <math>d(a) = d(b) = 4</math>. This adds one factor with more than <math>3</math> factors, so the answer is <math>\boxed{\textbf{(B) }7}</math>.
+
Let <math>d(n)</math> be the number of factors of n. We know by prime factorization that <math>d(2020) = 12</math>. These <math>12</math> numbers can be divided into unordered pairs <math>{a,b}</math> where <math>ab = 2020</math>. Since <math>d(2020) = d(a)d(b)</math>, one of <math>d(a), d(b)</math> has <math>3</math> or less factors and the other has <math>4</math> or more - in to total <math>6</math> factors of <math>2020</math> with more than <math>3</math> factors. However, this argument has exceptions where <math>a</math> and <math>b</math> share a nontrivial common factor, which in this case can only be two. There are two cases - One in which <math>5</math> and <math>101</math> divide the same factor, WLOG assumed to be <math>a</math>, so that <math>d(a) = 2^3 > 3</math> and <math>d(b) = 2</math>, as otherwise. In the other case, <math>a = 5\cdot2</math> and <math>b = 101\cdot2</math>, so that <math>d(a) = d(b) = 4</math>. This adds one factor with more than <math>3</math> factors, so the answer is <math>\boxed{\textbf{(B) }7}</math>.
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 +
==Video Solution by NiuniuMaths (Easy to understand!)==
 +
https://www.youtube.com/watch?v=bHNrBwwUCMI
 +
 
 +
~NiuniuMaths
 +
 
 +
==Video Solution by Math-X (First understand the problem!!!)==
 +
https://youtu.be/UnVo6jZ3Wnk?si=YsWfaht72aFTG3eZ&t=3020
 +
 
 +
~Math-X
 +
 
 +
==Video Solution (🚀Just 3 min🚀)==
 +
https://youtu.be/jJnxvFJuhQw
 +
 
 +
~Education, the Study of Everything
  
 
== Video Solution by OmegaLearn ==
 
== Video Solution by OmegaLearn ==

Latest revision as of 16:50, 24 May 2024

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 $12$ factors; only $1, 2, 4, 5, 101$ 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 factorization that $d(2020) = 12$. These $12$ numbers can be divided into unordered pairs ${a,b}$ where $ab = 2020$. Since $d(2020) = d(a)d(b)$, one of $d(a), d(b)$ has $3$ or less factors and the other has $4$ or more - in to total $6$ factors of $2020$ 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, $a = 5\cdot2$ and $b = 101\cdot2$, 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 NiuniuMaths (Easy to understand!)

https://www.youtube.com/watch?v=bHNrBwwUCMI

~NiuniuMaths

Video Solution by Math-X (First understand the problem!!!)

https://youtu.be/UnVo6jZ3Wnk?si=YsWfaht72aFTG3eZ&t=3020

~Math-X

Video Solution (🚀Just 3 min🚀)

https://youtu.be/jJnxvFJuhQw

~Education, the Study of Everything

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

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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