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Largest possible value of exponent of prime number

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Largest possible value of exponent of prime number  [#permalink]

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New post 05 Aug 2017, 00:59
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If the greatest integer k for which \(3^{k}\) is a factor of n! is 8, what is the largest possible value of p so that \(5^{p}\) is a factor of n!?
(A) 2
(B) 3
(C) 4
(D) 5
(E) 6

The solution explained in the OG seems very complicated to me. It would be great if someone explains how they arrived at the answer too!

Please give a kudos if you like the question! :)
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Re: Largest possible value of exponent of prime number  [#permalink]

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New post 05 Aug 2017, 01:07
waiting for the solution :(
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Re: Largest possible value of exponent of prime number  [#permalink]

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New post 05 Aug 2017, 02:24
1
DeathNoteFreak wrote:
If the greatest integer k for which \(3^{k}\) is a factor of n! is 8, what is the largest possible value of p so that \(5^{p}\) is a factor of n!?
(A) 2
(B) 3
(C) 4
(D) 5
(E) 6

The solution explained in the OG seems very complicated to me. It would be great if someone explains how they arrived at the answer too!

Please give a kudos if you like the question! :)


Greatest integer k for which 3^k is a factor of n!, is a fancy way of asking how many 3's are present in n!.
Since it is given that k is 8, this means that the following numbers are present in the expansion of n!:
3,6,9,12,15,18
Each of the above number gives the following number of 3's
1,1,2,1,1,2 [3*1, 3*2, 3*3, 3*4, 3*5, 3*3*2], which sum to a total of '8'(1+1+2+1+1+2).
In the above case, we have considered n as 18, but even if we consider n as 20, the number of 3's would still be the same. It is only when we take n as 21 that another extra 3 gets added(21=3*7).
Also, as the question asks for the maximum value of p for 5^p to be a factor of n!(another fancy way of asking the number of 5's in n!), we can consider n as 20.
Numbers containing a 5 in the expansion of 20! would be 5,10,15,20(5*1,5*2,5*3,5*4) and each number is giving a single value of 5.
Therefore, the maximum value of p would be 1+1+1+1=4

Hope this helps.
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Re: Largest possible value of exponent of prime number  [#permalink]

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New post 05 Aug 2017, 07:11
nitesh181989 wrote:
DeathNoteFreak wrote:
If the greatest integer k for which \(3^{k}\) is a factor of n! is 8, what is the largest possible value of p so that \(5^{p}\) is a factor of n!?
(A) 2
(B) 3
(C) 4
(D) 5
(E) 6

The solution explained in the OG seems very complicated to me. It would be great if someone explains how they arrived at the answer too!

Please give a kudos if you like the question! :)


Greatest integer k for which 3^k is a factor of n!, is a fancy way of asking how many 3's are present in n!.
Since it is given that k is 8, this means that the following numbers are present in the expansion of n!:
3,6,9,12,15,18
Each of the above number gives the following number of 3's
1,1,2,1,1,2 [3*1, 3*2, 3*3, 3*4, 3*5, 3*3*2], which sum to a total of '8'(1+1+2+1+1+2).
In the above case, we have considered n as 18, but even if we consider n as 20, the number of 3's would still be the same. It is only when we take n as 21 that another extra 3 gets added(21=3*7).
Also, as the question asks for the maximum value of p for 5^p to be a factor of n!(another fancy way of asking the number of 5's in n!), we can consider n as 20.
Numbers containing a 5 in the expansion of 20! would be 5,10,15,20(5*1,5*2,5*3,5*4) and each number is giving a single value of 5.
Therefore, the maximum value of p would be 1+1+1+1=4

Hope this helps.




If you still consider n=18, number of 5's would then be 3. What would you suggest to understand why to choose 20, instead of 18??

Thanks in advance! :)
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Largest possible value of exponent of prime number  [#permalink]

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New post 05 Aug 2017, 10:05
1
DeathNoteFreak wrote:
nitesh181989 wrote:
DeathNoteFreak wrote:
If the greatest integer k for which \(3^{k}\) is a factor of n! is 8, what is the largest possible value of p so that \(5^{p}\) is a factor of n!?
(A) 2
(B) 3
(C) 4
(D) 5
(E) 6

The solution explained in the OG seems very complicated to me. It would be great if someone explains how they arrived at the answer too!

Please give a kudos if you like the question! :)


Greatest integer k for which 3^k is a factor of n!, is a fancy way of asking how many 3's are present in n!.
Since it is given that k is 8, this means that the following numbers are present in the expansion of n!:
3,6,9,12,15,18
Each of the above number gives the following number of 3's
1,1,2,1,1,2 [3*1, 3*2, 3*3, 3*4, 3*5, 3*3*2], which sum to a total of '8'(1+1+2+1+1+2).
In the above case, we have considered n as 18, but even if we consider n as 20, the number of 3's would still be the same. It is only when we take n as 21 that another extra 3 gets added(21=3*7).
Also, as the question asks for the maximum value of p for 5^p to be a factor of n!(another fancy way of asking the number of 5's in n!), we can consider n as 20.
Numbers containing a 5 in the expansion of 20! would be 5,10,15,20(5*1,5*2,5*3,5*4) and each number is giving a single value of 5.
Therefore, the maximum value of p would be 1+1+1+1=4

Hope this helps.




If you still consider n=18, number of 5's would then be 3. What would you suggest to understand why to choose 20, instead of 18??

Thanks in advance! :)




it says 'largest', thats why we'll have to take the maximum value of n for which 3 comes 8 times and in this case it happened to be 20 (just before 21) which also has one extra 5 (and question asks for maximum number of occurrence of 5)
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Re: Largest possible value of exponent of prime number  [#permalink]

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New post 05 Aug 2017, 11:31
1
Stem says that 3^8 is the highest power that can divide n!.
Basically we have to find the highest power of 5 less than or equal to 3^8.
So the highest power of 5 will be 4. Because if it becomes 5 then 5^5 is quite later than 3^8.
I don't know how much I could express myself but it's the gist of the problem.
Answer is 4.

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Largest possible value of exponent of prime number  [#permalink]

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New post 05 Aug 2017, 21:59
4
DeathNoteFreak wrote:
If the greatest integer k for which \(3^{k}\) is a factor of n! is 8, what is the largest possible value of p so that \(5^{p}\) is a factor of n!?
(A) 2
(B) 3
(C) 4
(D) 5
(E) 6

The solution explained in the OG seems very complicated to me. It would be great if someone explains how they arrived at the answer too!

Please give a kudos if you like the question! :)


I solved by following way. Hope it helps.

\(3!\) has one \(3 = 3^1\)
\(6!\) has two \(3 = 3^2\)
\(9!\) has four \(3 = 3^4\)
\(12!\) has five \(3 = 3^5\)
\(15!\) has six \(3 = 3^6\)
\(18!\) has eight \(3 = 3^8\)
Also \(20!\) has eight \(3 = 3^8\)
\(21!\) has nine \(3 = 3^9\)

Given greatest integer \(k\) for which \(3^{k}\) is a factor of \(n!\) is \(8\). Therefore \(n!\) cannot be \(21!\).

\(n!\) will be \(20!\).

Therefor \(n!\) with largest possible value of \(5^p\) would be \(20!\).

Lets check value of \(p\) in the above factorial to find the largest possible value of \(5^p\).

\(20!\) has four \(5 = 5^4\)

Therefore largest possible value of \(p\) so that \(5^{p}\) is a factor of \(n! = 4\)

Answer (C)...

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Re: Largest possible value of exponent of prime number  [#permalink]

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New post 05 Nov 2017, 01:41
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I solved this question in the following way:

Step 1: find the largest possible value of n! by calculating how many "3" are presented in n!
Because greatest integer k for which 3^k is a factor of n! is 8. => 3,6,9,12,15,18 are presented in n!
Since we want to find the largest possible value of n! => we can increase the value of n as long as we dont add any more "3" =>n! = 20!

Step 2: find the largest p by calculating how may "5" are presented in 20!
We have 5,10,15,20 are presented in 20! and give us four "5" => p = 4

Hence, the answer is C.
----

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Re: Largest possible value of exponent of prime number  [#permalink]

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New post 06 Nov 2017, 15:36
kumarparitosh123 wrote:
Stem says that 3^8 is the highest power that can divide n!.
Basically we have to find the highest power of 5 less than or equal to 3^8.
So the highest power of 5 will be 4. Because if it becomes 5 then 5^5 is quite later than 3^8.
I don't know how much I could express myself but it's the gist of the problem.
Answer is 4.

Sent from my Lenovo TAB S8-50LC using GMAT Club Forum mobile app


3^8 = 6561
5^4 = 625 and 5^5 is 3125
Thus why did you choose 4? can you please explain?
Re: Largest possible value of exponent of prime number &nbs [#permalink] 06 Nov 2017, 15:36
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