AMC 8 2020 Problem 17
AMC8 2020 年第 17 题
整除与因数★★★☆☆
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. )
2020 有多少个正整数因子，它们各自的因子个数都大于 3?（例如，12 有 6 个因子， 即 1， 2，3，4，6，12.）
- A.6
- B.7
- C.8
- D.9
- E.10
正确答案 · Correct Answer
B
解析 · Solution
Solution 1
Since , 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 = 7 factors have more than 3 factors.
As in, we prime factorize 2020 as , 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=7.
Let be the number of factors of n. We know by prime factorization that d(2020) = 12. These 12 numbers can be divided into unordered pairs where ab = 2020. Since d(2020) = d(a)d(b), one of $d(a), d(b)$ has 3 or less factors and the other has 4 or more - in to total 6 factors of 2020 with more than 3 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 5 and 101 divide the same factor, WLOG assumed to be , so that and d(b) = 2, as otherwise. In the other case, and , so that d(a) = d(b) = 4. This adds one factor with more than 3 factors, so the answer is 7.
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