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

(→‎Video Solution by WhyMath)
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==Solution 2==
 
==Solution 2==
 
As in Solution 1, we prime factorize <math>2020</math> as <math>2^2\cdot 5\cdot 101</math>, and we recall the standard formula that the number of positive factors of an integer is found by adding <math>1</math> to each exponent in its prime factorization, and then multiplying these. Thus <math>2020</math> has <math>(2+1)(1+1)(1+1) = 12</math> factors. The only number which has one factor is <math>1</math>. For a number to have exactly two factors, it must be prime, and the only prime factors of <math>2020</math> are <math>2</math>, <math>5</math>, and <math>101</math>. 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 <math>2020</math> is <math>4</math>. Thus, there are <math>5</math> factors of <math>2020</math> which themselves have <math>1</math>, <math>2</math>, or <math>3</math> factors (namely <math>1</math>, <math>2</math>, <math>4</math>, <math>5</math>, and <math>101</math>), so the number of factors of <math>2020</math> that have more than <math>3</math> factors is <math>12-5=\boxed{\textbf{(B) }7}</math>.
 
As in Solution 1, we prime factorize <math>2020</math> as <math>2^2\cdot 5\cdot 101</math>, and we recall the standard formula that the number of positive factors of an integer is found by adding <math>1</math> to each exponent in its prime factorization, and then multiplying these. Thus <math>2020</math> has <math>(2+1)(1+1)(1+1) = 12</math> factors. The only number which has one factor is <math>1</math>. For a number to have exactly two factors, it must be prime, and the only prime factors of <math>2020</math> are <math>2</math>, <math>5</math>, and <math>101</math>. 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 <math>2020</math> is <math>4</math>. Thus, there are <math>5</math> factors of <math>2020</math> which themselves have <math>1</math>, <math>2</math>, or <math>3</math> factors (namely <math>1</math>, <math>2</math>, <math>4</math>, <math>5</math>, and <math>101</math>), so the number of factors of <math>2020</math> that have more than <math>3</math> factors is <math>12-5=\boxed{\textbf{(B) }7}</math>.
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 +
==Solution 3==
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 +
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 = <math>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 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 <math>d(a) = d(b) = 4</math>. This adds one factor with more than 3 factors, so the answer is \boxed{\textbf{(B) }7}$
  
 
==Video Solution by North America Math Contest Go Go==
 
==Video Solution by North America Math Contest Go Go==

Revision as of 04:17, 25 December 2022

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 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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