Difference between revisions of "2020 AMC 8 Problems/Problem 17"
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==Solution 1== | ==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> | + | 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== | ||
− | Let <math>d(n)</math> be the number of factors of n. We know by prime | + | 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>. |
+ | |||
+ | ==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== | ==Video Solution by North America Math Contest Go Go== |
Latest revision as of 17:50, 24 May 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.