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For any integer P greater than 1, P! denotes the product of all the in

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For any integer P greater than 1, P! denotes the product of all the in  [#permalink]

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New post 07 Mar 2018, 10:44
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Question Stats:

57% (01:10) correct 43% (01:35) wrong based on 14 sessions

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For any integer P greater than 1, P! denotes the product of all the integers from 1 to P, inclusive. What is the greatest integer m for which 45^m is a factor of 48!?

A) 1
B) 2
C) 5
D) 10
E) 11

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For any integer P greater than 1, P! denotes the product of all the in  [#permalink]

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New post 07 Mar 2018, 11:12
1
saswata4s wrote:
For any integer P greater than 1, P! denotes the product of all the integers from 1 to P, inclusive. What is the greatest integer m for which 45^m is a factor of 48!?

A) 1
B) 2
C) 5
D) 10
E) 11


\(45^m=(3^2*5)^m=3^{2m}*5^m\)

so to know the value of \(m\) we need to know how many powers of \(5\) are possible in \(48!\), this can be done as

\(\frac{48}{5}+\frac{48}{5^2}=9+1=10\). Hence \(m=10\)

Option D
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Re: For any integer P greater than 1, P! denotes the product of all the in  [#permalink]

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New post 08 Mar 2018, 04:20
niks18 wrote:
saswata4s wrote:
For any integer P greater than 1, P! denotes the product of all the integers from 1 to P, inclusive. What is the greatest integer m for which 45^m is a factor of 48!?

A) 1
B) 2
C) 5
D) 10
E) 11


\(45^m=(3^2*5)^m=3^{2m}*5^m\)

so to know the value of \(m\) we need to know how many powers of \(5\) are possible in \(48!\), this can be done as

\(\frac{48}{5}+\frac{48}{5^2}=9+1=10\). Hence \(m=10\)

Option D


Why you did not consider powers of 3?
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For any integer P greater than 1, P! denotes the product of all the in  [#permalink]

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New post 08 Mar 2018, 04:39
1
MrCleantek wrote:
Why you did not consider powers of 3?


Hey MrCleantek ,

3 is not considered because out of 3 and 5, we will always have atleast as many number of 3s as 5s whereas the converse is not true.

Consider an example here: Let's say we have 20! and we need to find out number of 10s. Now every 10 will be composed of one 2 and one 5.

If you find the number of 2's, you will get 18

Number of 5's = 4. Hence, you can see we cannot have more than four 10s because we have only four 4s. Thus finding the number of 5s would do.

Does that make sense?

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For any integer P greater than 1, P! denotes the product of all the in &nbs [#permalink] 08 Mar 2018, 04:39
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