Query about Question 6 from "PrepSwift: # of Numbers in Factorials II" [12371] (half the answer)

[user9656]

https://prepswift.gregmat.com/quizzes/quiz/of-numbers-in-factorials-ii-quiz (Question 6)

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Hey, was with the solution till we got 66, lost the answer from there, would love some help, thank you

[Kweku]

Say the student found 66 total 2s ie 2⁶⁶, there’d be exactly eleven 64s or eleven 2⁶s.

If the student found 67 total 2s, ie 2⁶⁷, there’d be eleven 64s and an extra 2¹ (which is not enough to form 64).

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If the student found 71 total 2s, ie 2⁷¹, there’d be eleven 64s and 5 extra 2s or 2⁵ leftover ( 2⁵ is not enough to form a 64)

If the student found 72 total 2s, ie 2⁷², there’d be exactly twelve 64s (this goes beyond the eleven bundles of 64).

So from 2⁶⁶ to 2⁷¹, there are always eleven 64s (2⁶s) with extra 2s (in some instances) that are not enough to form 64 so we don’t count.

Another Approach

Another way to see it is to ask yourself what the quotient is when 66, 67, 68, 69, 70, and 71 are divided by 6? In each instance, the quotient will be 11. Since 71 is the biggest, that’s the answer.