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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 Updated on: 21 Nov 2019, 02:23
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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

Originally posted by Kinshook on 06 Nov 2019, 11:14.
Last edited by Bunuel on 21 Nov 2019, 02:23, edited 1 time in total.
Edited the question.
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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 Nov 2019, 05:28
I think the answer should be 11.

45m = (3)^2 x 5 x m

48! can hold upto 22 3s, 10 5s and ___ Ms.

Thus the largest value that m can take will be 11, since 48! can also hold 4 11s.

Thus, 3^2 x 5 x 11 would be the factor of 48! where 11 is the largest value of m among the options.
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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 Nov 2019, 05:32
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IMO it should be \(45^m\)

\(45=3^2*5\)

highest power of \(3^2\) that can divide 48! is \([\frac{1}{2}\) \([\frac{45}{3}]+[\frac{45}{3^2}]+[\frac{45}{3^3}]]\)= 10

highest power of 5 that can divide 48! is \([\frac{45}{5}]+[\frac{45}{5^2}]\)= 9+1=10

Greatest integral value that m can have= min(10, 10)= 10

Kinshook 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 45m is a factor of 48!?

A. 1
B. 2
C. 5
D. 10
E. 11
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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 Nov 2019, 12:46
Kinshook 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 45m is a factor of 48!?

A. 1
B. 2
C. 5
D. 10
E. 11


Hi Kinshook,

Why cannot it be 11
\(\frac{48!}{45*11}=\frac{1*2*3*..5*...9*..11..*22...48}{1*3*3*5*11}\) = Integer
This shows that 48! is divisible by 45*11, hence m can be 11
Am I missing anything ?
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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 21 Nov 2019, 02:24
stne wrote:
Kinshook 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 45m is a factor of 48!?

A. 1
B. 2
C. 5
D. 10
E. 11


Hi Kinshook,

Why cannot it be 11
\(\frac{48!}{45*11}=\frac{1*2*3*..5*...9*..11..*22...48}{1*3*3*5*11}\) = Integer
This shows that 48! is divisible by 45*11, hence m can be 11
Am I missing anything ?


There was a typo: 45m instead of 45^m. Edited.
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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 21 Nov 2019, 02:37
Ans: 10; D
45=3^2+5^1
48/3+48/9+48/27= 16+5+1= 22; 3^2 occurs 11 times.
48/5+48/25=9+1= 10. 5 occurs once.
Limit is set by 5 and ans is 10.
Kudos if helped. Want to unlock gmatclub tests :)

Kinshook 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
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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 21 Nov 2019, 03:11
Bunuel wrote:
stne wrote:
Kinshook 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 45m is a factor of 48!?

A. 1
B. 2
C. 5
D. 10
E. 11


Hi Kinshook,

Why cannot it be 11
\(\frac{48!}{45*11}=\frac{1*2*3*..5*...9*..11..*22...48}{1*3*3*5*11}\) = Integer
This shows that 48! is divisible by 45*11, hence m can be 11
Am I missing anything ?


There was a typo: 45m instead of 45^m. Edited.


Thank you Sir, was wondering about the same. Wonder why Kinshook did not look into this.
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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 21 Nov 2019, 03:24
Kinshook 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 x 5)^m
Highest power of 3 by which 48! can be divided is 16+5+1 = 22
Highest power of 9 by which 48! can be divided is 11

Highest power of 5 by which 48! can be divided is 9+1 = 10

Hence, greatest value of m can be 10
D is correct.
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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 21 Nov 2019, 16:23
\(45=3^2*5\)

highest power of \(3^2\) that can divide 48! is \(\frac{1}{2}\) {\([\frac{48}{3}]+[\frac{48}{3^2}]+[\frac{48}{3^3}] \)} = (16+5+1) /2 = 11

highest power of 5 that can divide 48! is \([\frac{48}{5}]+[\frac{48}{5^2}]\)= 9+1=10

Greatest integral value that m can have is minimum of {11,10}
= 10

Answer is (D)

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For any integer P greater than 1, P! denotes the product of all the in   [#permalink] 21 Nov 2019, 16:23
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