2020 AMC 10B Problems/Problem 25
Let denote the number of ways of writing the positive integer as a productwhere , the are integers strictly greater than , and the order in which the factors are listed matters (that is, two representations that differ only in the order of the factors are counted as distinct). For example, the number can be written as , , and , so . What is ?
Note that . Since there are at most six not nexxessarily distinct factors multiplying to , we have six cases: Now we look at each of the six cases.
: We see that there is way, merely .
: This way, we have the in one slot and in another, and symmetry. The four other 's leave us with ways and symmetry doubles us so we have .
: We have as our baseline. We need to multiply by in places, and see that we can split the remaining three powers of 2 in a manner that is 3-0-0, 2-1-0 or 1-1-1. A 3-0-0 split has ways of happening (24-2-2 and symmetry; 2-3-16 and symmetry), a 2-1-0 split has ways of happening (due to all being distinct) and a 1-1-1 split has ways of happening (6-4-4 and symmetry) so in this case we have ways.
: We have as our baseline, and for the two other 's, we have a 2-0-0-0 or 1-1-0-0 split. The former grants us ways (12-2-2-2 and symmetry and 3-8-2-2 and symmetry) and the latter grants us also ways (6-4-2-2 and symmetry and 3-4-4-2 and symmetry) for a total of ways.
: We have as our baseline and one place to put the last two: on another two or on the three. On the three gives us ways due to symmetry and on another two gives us ways due to symmetry. Thus, we have ways.
: We have and symmetry and no more twos to multiply, so by symmetry, we have ways.
Thus, adding, we have .
As before, note that , and we need to consider 6 different cases, one for each possible value of , the number of factors in our factorization. However, instead of looking at each individually, find a general form for the number of possible factorizations with factors. First, the factorization needs to contain one factor that is itself a multiple of , and there are to choose from, and the rest must contain at least one factor of . Next, consider the remaining factors of left to assign to the factors. Using stars and bars, the number of ways to do this is This makes possibilities for each k.
To obtain the total number of factorizations, add all possible values for k: .
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