Difference between revisions of "2020 AMC 8 Problems/Problem 17"
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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>. | 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!!!)== | ==Video Solution by Math-X (First understand the problem!!!)== |
Revision as of 15:35, 26 January 2024
Contents
- 1 Problem
- 2 Solution 1
- 3 Solution 2
- 4 Solution 3
- 5 Video Solution by NiuniuMaths (Easy to understand!)
- 6 Video Solution by Math-X (First understand the problem!!!)
- 7 Video Solution (🚀Just 3 min🚀)
- 8 Video Solution by OmegaLearn
- 9 Video Solution by North America Math Contest Go Go
- 10 Video Solution by WhyMath
- 11 Video Solution
- 12 Video Solution by Interstigation
- 13 See also
Problem
How many positive integer factors of have more than factors? (As an example, has factors, namely and )
Solution 1
Since , we can simply list its factors: There are factors; only don't have over factors, so the remaining factors have more than factors.
Solution 2
As in Solution 1, we prime factorize as , and we recall the standard formula that the number of positive factors of an integer is found by adding to each exponent in its prime factorization, and then multiplying these. Thus has factors. The only number which has one factor is . For a number to have exactly two factors, it must be prime, and the only prime factors of are , , and . 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 is . Thus, there are factors of which themselves have , , or factors (namely , , , , and ), so the number of factors of that have more than factors is .
Solution 3
Let be the number of factors of n. We know by prime factorization that . These numbers can be divided into unordered pairs where . Since , one of has or less factors and the other has or more - in to total factors of with more than factors. However, this argument has exceptions where and share a nontrivial common factor, which in this case can only be two. There are two cases - One in which and divide the same factor, WLOG assumed to be , so that and , as otherwise. In the other case, and , so that . This adds one factor with more than factors, so the answer is .
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🚀)
~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
~savannahsolver
Video Solution
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.