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How many values can natural number n take, if n! is a multiple of 2^20

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How many values can natural number n take, if n! is a multiple of 2^20  [#permalink]

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New post 10 Apr 2018, 12:24
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Question Stats:

32% (02:37) correct 68% (02:22) wrong based on 41 sessions

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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

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Re: How many values can natural number n take, if n! is a multiple of 2^20  [#permalink]

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New post 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
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Re: How many values can natural number n take, if n! is a multiple of 2^20  [#permalink]

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New post 19 Apr 2018, 21:58
24! Contain 2^20
45! contain 3^20

So numbers are be range 24! and 44!

So total numbers are 44-24+1=21

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Re: How many values can natural number n take, if n! is a multiple of 2^20  [#permalink]

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New post 19 Apr 2018, 22:21
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itisSheldon wrote:
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


The problem in this question is the starting point.

What value should we try for n such that it has around 20 2s? Alternatively, we can start the count from 2 , 4, 6, 8 ... counting the number of 2s as we go along but that is a rather painful process. So let's try to guess a value for n.

We get at least one 2 from each multiple of 2. Then there are those multiples that give more than one 2 such as 4 and 12 etc. So let's try to figure out the number of 2s when n is about the 13th multiple of 2. Say when n = 26

26! will have 13 + 6 + 3 + 1 = 23 2s
So we need to go back a few steps and check when we get the 20th 2. We get it at n = 24.

Next, we repeat the process for 20 3s. Since 3s are rarer than 2s, let's try the 15th multiple of 3 i.e. 45
45! will have 15 + 5 + 1 = 21 3s. 45 adds two 3s to the mix so till 44, we have only 19 3s. That is perfect.

So n can take values from 24 to 44 which is 44 - 24 + 1 = 21 values

Answer (B)
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Re: How many values can natural number n take, if n! is a multiple of 2^20  [#permalink]

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Re: How many values can natural number n take, if n! is a multiple of 2^20   [#permalink] 21 Apr 2019, 07:43
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