2020 AMC 10 B
Complete problem set with solutions and individual problem pages
Let denote the number of ways of writing the positive integer as a product ,where , 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 6 can be written as , and , so . What is ?(2020 AMC 10B, Question #25)
- A.
- B.
- C.
- D.
- E.
Solution 1:
Note that . Since there are at most six not necessarily distinct factors multiplying to 96 , we have six cases. . Now we look at each of the six cases. : we see that there is 1 way, merely 96 . : This way, we have the 3 in one slot and 2 in another, and symmetry. The four other 2 's leave us with 5 ways and symmetry doubles us so we have 10 . : We have as our baseline. We need to multiply by 2 in 3 places, and see that we can split the remaining three powers of 2 in a manner that is or . A split has ways of happening and symmetry; and symmetry), a split has ways of happening (due to all being distinct) and a split has 3 ways of happening ( and symmetry) so in this case we have ways. : We have as our baseline, and for the two other 2 s, we have a or split. The former grants us ways and symmetry and and symmetry) and the latter grants us also ways and symmetry and 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 5 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 6 ways. Thus, adding, we
Solution 2:
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 3 , and there are to choose from, and the rest must contain at least one factor of 2 . Next, consider the remaining factors of 2 left to assign to the factors. Using stars and bars, the number of ways to do this makes To obtain the total number of factorizations, add all possible values for k:
