2020 AMC 8 Problems/Problem 17
How many positive integer factors of have more than factors? (As an example, has factors, namely and )
Since , we can simply list its factors: There are factors; only don't have over factors, so the remaining factors have more than factors.
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 .
Let be the number of factors of n. We know by prime factorisation that . These numbers can be divided into unordered pairs where . Since , one of has or less factors and the other has or more - in to total factors of with more than 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 and divide the same factor, WLOG assumed to be , so that and , as otherwise. In the other case, and , so that . This adds one factor with more than factors, so the answer is .
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