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What is the greatest possible value of integer n if 200! is divisible

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What is the greatest possible value of integer n if 200! is divisible  [#permalink]

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New post 05 Nov 2016, 13:42
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A
B
C
D
E

Difficulty:

  45% (medium)

Question Stats:

58% (01:05) correct 42% (01:34) wrong based on 202 sessions

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Re: What is the greatest possible value of integer n if 200! is divisible  [#permalink]

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New post 05 Nov 2016, 14:28
1
33^n= 3^n * 11^n

Highest prime factor will be the limiting factor.

200/11 + 200/121 = 18+1 = 19
D


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Re: What is the greatest possible value of integer n if 200! is divisible  [#permalink]

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New post 06 Nov 2016, 00:59
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stonecold wrote:
What is the greatest possible value of integer n if 200! is divisible by 33^n
A)6
B)12
C)18
D)19
E)66


33 = 3*11 i.e. to find the power of 33 in 200! we need to calculate the power of 11 and 3 in 200!
BUT
since power of 11 will be smaller (bigger prime number) than power of 3 (smaller prime number) in 200! so the power of 11 will be judgemental to find the power of 33

CONCEPT: Power of any Prime Number in any factorial can be calculated by following understanding

Power of prime x in n! = [n/x] + [n/x^2] + [n/x^3] + [n/x^4] + ... and so on
Where,
[n/x] = No. of Integers that are multiple of x from 1 to n
[n/x^2] = No. of Integers that are multiple of x^2 from 1 to n whose first power has been counted in previous step and second is being counted at this step
[n/x^3] = No. of Integers that are multiple of x^3 from 1 to n whose first two powers have been counted in previous two step and third power is counted at this step
And so on.....

Where [n/x] is greatest Integer value of (n/x) less than or equal to (n/x)
i.e. [100/3] = [33.33] = 33
i.e. [100/9] = [11.11] = 11 etc


Now, Power of prime 11 in 200! = [200/11] + [200/11^2] + [200/11^3] = 18+1+0 = 19

Answer: Option E
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Re: What is the greatest possible value of integer n if 200! is divisible  [#permalink]

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New post 06 Nov 2016, 01:48
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stonecold wrote:
What is the greatest possible value of integer n if 200! is divisible by 33^n
A)6
B)12
C)18
D)19
E)66


33 = 11 * 3

Here 3's are more frequent than 11 ( Remember we need a combination of 11 and 3 )

So, The deciding factor is 11 and not 3

200/11 = 18
18/11 = 1

Hence , correct answer will be (D) 19
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Re: What is the greatest possible value of integer n if 200! is divisible  [#permalink]

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New post 27 Nov 2017, 13:56
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Re: What is the greatest possible value of integer n if 200! is divisible &nbs [#permalink] 27 Nov 2017, 13:56
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