Difference between revisions of "2020 AMC 8 Problems/Problem 17"
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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>. | ||
+ | |||
+ | ==Video Solution== | ||
+ | https://youtu.be/2dazhQ31I14 | ||
+ | |||
+ | ~savannahsolver | ||
==See also== | ==See also== |
Revision as of 17:40, 20 November 2020
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
of these; only
(i.e.
of them) have at most
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
.
Video Solution
~savannahsolver
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.