2020 AMC 8

Complete problem set with solutions and individual problem pages

Problem 17 Hard

How many positive integer factors of 2020 have more than 3 factors? (As an example, 12 has 6 factors, namely 1,2,3,4,6, and 12.)

  • A.

    6

  • B.

    7

  • C.

    8

  • D.

    9

  • E.

    10

Answer:B

Solution 1

Since 2020 = 2^2 \cdot 5 \cdot 101, we can simply list its factors:

1, 2, 4, 5, 10, 20, 101, 202, 404, 505, 1010, 2020.

There are 12 factors; only 1, 2, 4, 5, 101 don't have over 3 factors, so the remaining 12-5 = \boxed{\textbf{(B) }7} factors have more than 3 factors.

 

Solution 2

As in Solution 1, we prime factorize 2020 as 2^2\cdot 5\cdot 101, and we recall the standard formula that the number of positive factors of an integer is found by adding 1 to each exponent in its prime factorization, and then multiplying these. Thus 2020 has (2+1)(1+1)(1+1) = 12 factors. The only number which has one factor is 1. For a number to have exactly two factors, it must be prime, and the only prime factors of 2020 are 2, 5, and 101. 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 2020 is 4. Thus, there are 5 factors of 2020 which themselves have 1, 2, or 3 factors (namely 1, 2, 4, 5, and 101), so the number of factors of 2020 that have more than 3 factors is 12-5=\boxed{\textbf{(B) }7}.