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If k is an integer and 33!/22! is divisible by 6^k

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Manager
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Joined: 05 Nov 2012
Posts: 138
Re: If k is an integer and 33!/22! is divisible by 6^k  [#permalink]

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New post 23 Dec 2012, 10:51
Marcab wrote:
I guess, we have deviated a bit. Can anyone please let me know how to solve 33!/22! with alternative method.

did you check out my post and daviesj post above?
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Joined: 11 Sep 2013
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Concentration: Finance, Finance
Re: If k is an integer and 33!/22! is divisible by 6^k  [#permalink]

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New post 23 Apr 2014, 19:15
We need numbers which are divisible by 3 only
4 numbers are divisible by 3
33= one 3
30= one 3
27= three 3's
24= one 3
Total six 3's
Its easy
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GRE 1: Q169 V154
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Re: If k is an integer and 33!/22! is divisible by 6^k  [#permalink]

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New post 22 Nov 2016, 03:39
A 6 is formed via a two and a three
Hence we ned to get the number of three's as number of 2's are enough
33=> one
30=> one
27=> three
24=> one
hence 6 three's
Hence 6^k is a factor

So D
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Re: If k is an integer and 33!/22! is divisible by 6^k  [#permalink]

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New post 26 Nov 2019, 08:07
Amateur wrote:
6^k =2*3 ^k

now lets take 33!
number of power of 2 is 33/2 + 33/4 + 33/8 + 33/16 + 33/32 = 16+8+4+2+1=31
number of power of 3 is 33/3 + 33/9 + 33/27 = 11+3+1=15

Similarly for 22!
number of power of 2 is 22/2 + 22/4 +22/8 + 22/16 = 11+5+2+1=19
number of power of 3 is 22/3 + 22/9 = 7+2=9
33!/22! contains 2^12 * 3^6
since we need maximum power of 6 we are limited by power of 3 above which is 6.... so k=6
let me know if you want me to be more clear.....


neatest way to do this problem
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Re: If k is an integer and 33!/22! is divisible by 6^k   [#permalink] 26 Nov 2019, 08:07

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