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10 Apr 2018, 12:24
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B
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
95%
(hard)
Question Stats:
41% (02:20) correct
59%
(02:27)
wrong
based on 115
sessions
History
Date
Time
Result
Not Attempted Yet
How many values can natural number n take, if n! is a multiple of 2^20 but not 3^20?
(A) 11
(B) 21
(C) 16
(D) 20
(E) 23
(A) 11
(B) 21
(C) 16
(D) 20
(E) 23
10 Apr 2018, 16:53
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Find: Number of n's such that n! is a multiple of 2^20 and not a multiple of 3^20.
Approach: Find the min and max values of n. Min value is the smallest n such that n! is a multiple of 2^20. Once we find that, we need to find how high that n can go. It can go up to the first multiple of 3^20. So we need to find the smallest n such that n! is a multiple of 3^20 and subtract 1.
min n
First I tried n=20. How many 2s are there in 20!
\(\frac{20}{2}+\frac{20}{4}+\frac{20}{8}+\frac{20}{16}=10+5+2+1=18\)
So 20! is a multiple of 2^18, not 2^20. We need to keep looking.
Note that if n=22, then that will only increase the first term by 1, i.e. \(\frac{20}{2}\)-->\(\frac{22}{2}\). There will be no impact on the other terms. Thus 22! is a multiple of 2^19. Thus, the lowest n such that n! is a multiple of 2^20 is 24.
max n
First I tried n=40. How many 3s are there in 40!
\(\frac{40}{3}+\frac{40}{9}+\frac{40}{27}=13+4+1=18\)
So 40! is a multiple of 3^18, not 3^20. We need to keep looking.
Note that if n=42, then that will only increase the first term by 1, i.e. \(\frac{40}{3}\)-->\(\frac{42}{3}\). There will be no impact on the other terms. Thus 42! is a multiple of 3^19. Thus, the lowest n such that n! is a multiple of 3^20 is 45. So max value of n such that n! is not a multiple of 3^20 is 44.
n! ranges from 24! to 44! inclusive --> 21 possible values of n.
Answer: B
Approach: Find the min and max values of n. Min value is the smallest n such that n! is a multiple of 2^20. Once we find that, we need to find how high that n can go. It can go up to the first multiple of 3^20. So we need to find the smallest n such that n! is a multiple of 3^20 and subtract 1.
min n
First I tried n=20. How many 2s are there in 20!
\(\frac{20}{2}+\frac{20}{4}+\frac{20}{8}+\frac{20}{16}=10+5+2+1=18\)
So 20! is a multiple of 2^18, not 2^20. We need to keep looking.
Note that if n=22, then that will only increase the first term by 1, i.e. \(\frac{20}{2}\)-->\(\frac{22}{2}\). There will be no impact on the other terms. Thus 22! is a multiple of 2^19. Thus, the lowest n such that n! is a multiple of 2^20 is 24.
max n
First I tried n=40. How many 3s are there in 40!
\(\frac{40}{3}+\frac{40}{9}+\frac{40}{27}=13+4+1=18\)
So 40! is a multiple of 3^18, not 3^20. We need to keep looking.
Note that if n=42, then that will only increase the first term by 1, i.e. \(\frac{40}{3}\)-->\(\frac{42}{3}\). There will be no impact on the other terms. Thus 42! is a multiple of 3^19. Thus, the lowest n such that n! is a multiple of 3^20 is 45. So max value of n such that n! is not a multiple of 3^20 is 44.
n! ranges from 24! to 44! inclusive --> 21 possible values of n.
Answer: B
General Discussion
