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What is the greatest possible value of integer n if 100!

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What is the greatest possible value of integer n if 100!  [#permalink]

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New post Updated on: 23 Jan 2018, 03:54
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E

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Originally posted by stonecold on 05 Nov 2016, 15:00.
Last edited by stonecold on 23 Jan 2018, 03:54, edited 2 times in total.
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Re: What is the greatest possible value of integer n if 100!  [#permalink]

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New post 22 Nov 2017, 13:26
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stonecold wrote:
What is the greatest possible value of integer n if 100! is divisible by 15^n
A)20
B)21
C)22
D)23
E)24


We need to determine the largest value of n such that 100! is divisible by 15^n. For a number to be divisible by 15, it must be divisible by both 3 and 5. Thus, we need to find the largest value of n such that 100! is divisible by 3^n x 5^n.

Since we know there are fewer 5s in 100! than 3s, we can find the number of 5s and thus be able to determine the number of 5-and-3 pairs.

To determine the number of 5s within 100!, we can use the following shortcut in which we divide 100 by 5, then divide the quotient of 100/5 by 5 and continue this process until we no longer get a nonzero quotient.

100/5 = 20

20/5 = 4

Since 4/5 does not produce a nonzero quotient, we can stop.

The final step is to add up our quotients; that sum represents the number of factors of 5 within 100!.

Thus, there are 20 + 4 = 24 factors of 5 within 100!

Since there are 24 factors of 5 within 100!, we also know that there are 24 5-and-3 pairs and, thus, the largest value of n is 24.

Answer: E
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Re: What is the greatest possible value of integer n if 100!  [#permalink]

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New post 05 Nov 2016, 15:25
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stonecold wrote:
What is the greatest possible value of integer n if 100! is divisible by 15^n
A)20
B)21
C)22
D)23
E)24


15^n = 5^n * 3^n

Highest prime factor will be the limiting factor.

100/5 +100/25 =20+4 = 24

E


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Re: What is the greatest possible value of integer n if 100!  [#permalink]

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New post 06 Nov 2016, 00:44
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stonecold wrote:
What is the greatest possible value of integer n if 100! is divisible by 15^n
A)20
B)21
C)22
D)23
E)24


15 = 3*5 i.e. to find the power of 15 in 100! we need to calculate the power of 5 and 3 in 100!
BUT
since power of 5 will be smaller (bigger prime number) than power of 3 (smaller prime number) in 100! so the power of 5 will be judgemental to find the power of 15

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 5 in 100! = [100/5] + [100/5^2] + [100/5^3] = 20+4+0 = 24

Answer: Option E
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Re: What is the greatest possible value of integer n if 100!  [#permalink]

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New post 06 Nov 2016, 01:56
stonecold wrote:
What is the greatest possible value of integer n if 100! is divisible by 15^n
A)20
B)21
C)22
D)23
E)24


stonecold please change the headline of the question which says 200! instead of 100!
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Re: What is the greatest possible value of integer n if 100!  [#permalink]

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New post 06 Nov 2016, 03:07
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stonecold wrote:
What is the greatest possible value of integer n if 100! is divisible by 15^n
A)20
B)21
C)22
D)23
E)24


15 = 5 * 3

Since no of 3's will be more than no of 5's , our deciding factor will be 5

100/5 = 20
20/5 = 4

So, the highest power of 15, will be (E) 24

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Re: What is the greatest possible value of integer n if 100!  [#permalink]

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New post 03 Dec 2017, 23:31
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I tried a different route. 100! has 24 zeroes (knew this from many questions on GMATCLUB:) )

So 15's highest power is something we need --> so 15 has again only 1 '5'. Thus again, we count the number of zeroes.

But just to be sure - as we are dividing by number of '15' (15^1, 15^2..) --> 100/ 15 --> approx 6 (discarding remainder)

24 is a multiple of 6 --> so only answer that connects between 5 and 15.

Kudos, please if you found this useful :)
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Re: What is the greatest possible value of integer n if 100!  [#permalink]

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New post 22 Jan 2018, 09:15
JeffTargetTestPrep wrote:
stonecold wrote:
What is the greatest possible value of integer n if 100! is divisible by 15^n
A)20
B)21
C)22
D)23
E)24


We need to determine the largest value of n such that 100! is divisible by 15^n. For a number to be divisible by 15, it must be divisible by both 3 and 5. Thus, we need to find the largest value of n such that 100! is divisible by 3^n x 5^n.

Since we know there are fewer 5s in 100! than 3s, we can find the number of 5s and thus be able to determine the number of 5-and-3 pairs.

To determine the number of 5s within 100!, we can use the following shortcut in which we divide 100 by 5, then divide the quotient of 100/5 by 5 and continue this process until we no longer get a nonzero quotient.

100/5 = 20

20/5 = 4

Since 4/5 does not produce a nonzero quotient, we can stop.

The final step is to add up our quotients; that sum represents the number of factors of 5 within 100!.

Thus, there are 20 + 4 = 24 factors of 5 within 100!

Since there are 24 factors of 5 within 100!, we also know that there are 24 5-and-3 pairs and, thus, the largest value of n is 24.

Answer: E


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Re: What is the greatest possible value of integer n if 100!  [#permalink]

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New post 16 Feb 2018, 22:33
Re: What is the greatest possible value of integer n if 100! &nbs [#permalink] 16 Feb 2018, 22:33
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