I feel as the though the explanation of this question is pretty hand wavey.
If k! is divisible by 64^11 where n = 11 is the largest value of n for 64^n. By picking p = 71 in the solution, the only thing that the explanation explains is that this does not contradict that n = 11 is the largest. Obviously we can’t have p = 72 because that would mean 64^12 divides k!, which contradicts the problem.
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.
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.
For us to get max P, k! must have those extra “loose” twos (5 extra to be exact)
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.
We are trying to get max P such that 2^p divides k!; It doesn’t say find max P such that it doesn’t contradict 64^11 being the largest 64^n divisor of k!. Not contradicting the problem statement is part 1. You still need to justify that if k! is divisible by 64^11 = 2^66 then this implies that k! is also divisible by 2^71.
Try to use this reasoning to solve the same problem for n = 2 instead and you’ll see that it won’t give you the right answer.
If n = 2 then we have 64^2 = 2^12 instead of 64^11 = 2^66. If you reapply the same logic, you would then say that you can have an additional 5 twos since 2^5 = 32 so your answer would be p = 17.
However, this falls apart when you realize that for n = 2, k! = 19! or 18! (you can check this). 19! and 18! has exactly 9 + 4 + 2 + 1 = 16 factors of 2 so 2^17 would actually not divide k!, even if it does not contradict n = 2.
If I understand you, you are saying that, for some values of n, adding 5 extra twos won’t work.
An example is when n=2, where we cannot get exactly 2^{17} by doing (12+5).
Hence, we need to take into consideration what k and by extension k! is or how many twos it can take.
In other words, we should have proved that there is some integer(s), k, that has 71 twos and therefore is divisible by 2^{71} without breaking the n=11 rule.
K would be 74 and 75 in this instance but your point is no such k integers exist when n = 2
Yea exactly. Like if I came across a question on the GRE, I would have to figure out exactly what K is/could be given n in order to solve something like this with certainty, which is why I thought the solution was a bit incomplete.