Difference between revisions of "2020 AMC 8 Problems/Problem 17"
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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> | 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> | ||
~ junaidmansuri | ~ 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>. | ||
+ | |||
+ | -franzliszt | ||
==See also== | ==See also== |
Revision as of 14:00, 18 November 2020
How many positive integer factors of have more than
factors?
Contents
[hide]Solution
We list out the factors of :
Of these, only
(
of them) do not have more than
factors. Therefore the answer is
.
Solution 2
The prime factorization of is
. The total number of factors of
is given by the product of one more than each of the prime powers which comes out to
. Instead of finding how many factors of
have more than three factors, we will instead find how many have one, two, or three factors and subtract this number from
to find the answer. The only number which has one factor is
. For a number to have exactly two factors, it must be prime. From the prime factorization of
, we know that these can only be
and
. 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
is
. Our list of factors is
and
which is a total five factors. Thus, the number of factors of
that have more than three factors is
.
~ junaidmansuri
Solution 3
The prime factorization of is
so it has
factors. Then we can count that
all have
or fewer divisors so by complementary counting our answer is
.
-franzliszt
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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