Difference between revisions of "2020 AMC 8 Problems/Problem 17"
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<math>\textbf{(A) }6 \qquad \textbf{(B) }7 \qquad \textbf{(C) }8 \qquad \textbf{(D) }9 \qquad \textbf{(E) }10</math> | <math>\textbf{(A) }6 \qquad \textbf{(B) }7 \qquad \textbf{(C) }8 \qquad \textbf{(D) }9 \qquad \textbf{(E) }10</math> | ||
− | ==Solution== | + | ==Solution 1== |
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>. | 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>. | ||
Revision as of 21:09, 18 November 2020
How many positive integer factors of have more than factors?
Contents
Solution 1
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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