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saswata4s
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GMAT 1: 730 Q49 V41
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For any integer P greater than 1, P! denotes the product of all the in 07 Mar 2018, 10:44
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Difficulty:

65% (hard)

Question Stats:

52% (01:56) correct 48% (01:08) wrong based on 21 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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niks18
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For any integer P greater than 1, P! denotes the product of all the in 07 Mar 2018, 11:12
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saswata4s
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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\(45^m=(3^2*5)^m=3^{2m}*5^m\)

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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For any integer P greater than 1, P! denotes the product of all the in 08 Mar 2018, 04:20
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saswata4s
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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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 08 Mar 2018, 04:39
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MrCleantek

Why you did not consider powers of 3?
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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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