Incomplete/poor explanation of Question 6 from "PrepSwift: # of Numbers in Factorials II" [12371]

[Kweku]

Hi, @user12176

@user12176

However, this does not say anything about whether k! would be divisible by the loose twos we would have remaining. There is no proof or explanation that those additional loose twos would divide k! evenly after dividing by 2^66

Think of it this way,

We are to count 11 bundles of 6.

At exactly 66, we have those 11 bundles and like you said, we can’t use 72 as we’d have 12 bundles instead.

Because we want to maximize P, we must add 5 more to 66 to get 71, knowing that, 5 twos can’t form a single extra bundle of 6 to “contradict” the problem.

@user12176

However, this does not say anything about whether k! would be divisible by the loose twos we would have remaining.

For us to get max P, k! must have those extra “loose” twos (5 extra to be exact)

@user12176

There is no proof or explanation that those additional loose twos would divide k! evenly after dividing by 2^66.

We are trying to get the max P, hence k! must also have the max number of twos that will not “contradict” the problem, and that is 71 twos.