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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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We list out the factors of <math>2020</math>: <cmath>1, 2, 4, 5, 10, 20, 101, 202, 404, 505, 1010, 2020.</cmath> Of these, only <math>1, 2, 4, 5, 101</math> (<math>5</math> of them) do not have more than <math>3</math> factors. Therefore the answer is <math>\tau\left({2020}\right)-5=\boxed{\textbf{(B) }7}</math>.
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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> (<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.
  
 
==Solution 2==
 
==Solution 2==
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The prime factorization of <math>2020</math> is <math>2^2\cdot5\cdot101</math>. The total number of factors of <math>2020</math> is given by the product of one more than each of the prime powers which comes out to <math>3\cdot2\cdot2=12</math>. Instead of finding how many factors of <math>2020</math> have more than three factors, we will instead find how many have one, two, or three factors and subtract this number from <math>12</math> to find the answer. The only number which has one factor is <math>1</math>. For a number to have exactly two factors, it must be prime. From the prime factorization of <math>2020</math>, we know that these can only be <math>2,5,</math> and <math>101</math>. For a number to have three factors, it must be a square of a prime. The only square of a prime that is a factor of <math>2020</math> is <math>4</math>. Our list of factors is <math>1,2,4,5,</math> and <math>101</math> which is a total five factors. Thus, the number of factors of <math>2020</math> that have more than three factors is <math>12-5=7 \implies\boxed{\textbf{(B) }7}</math>.<br>
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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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~[http://artofproblemsolving.com/community/user/jmansuri junaidmansuri]
 
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==Solution 3==
 
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The prime factorization of <math>2020</math> is <math>2^2\cdot5\cdot101</math> so it has <math>(2+1)(1+1)(1+1)=12</math> factors. Then we can count that <math>1,2,4,5,101</math> all have <math>3</math> or fewer divisors so by complementary counting our answer is <math>12-5=\textbf{(B)}\ 7</math>.
 
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-franzliszt
 
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==Solution 4==
 
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The prime factorization of <math>2020</math> is <math>2^2\cdot5\cdot101</math> so it has <math>(2+1)(1+1)(1+1)=12</math> factors. Then we can count that <math>1,2,4,5,101</math> which gets <math>12-5=\textbf{(B)}\ 7</math>.
 
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~oceanxia~
 
  
 
==See also==
 
==See also==

Revision as of 08:21, 20 November 2020

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 $1, 2, 4, 5, 101$ ($5$ of them) have at most $3$ 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 $12-5=\boxed{\textbf{(B) }7}$.

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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