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Re: What is the greatest value of n such that 18^n is a factor of 18! ? [#permalink]
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18 is 2*3*3 which is 3^2 Now using the factorial formula 18/2=9 18/2^2=4 18/2^3=2 18/2^4=1 we have 16 twos and 8 threes in 18 factorial so maximum we can have 8 threes since it's 3^2 already the highest power which would divide the number is 4
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Re: What is the greatest value of n such that 18^n is a factor of 18! ? [#permalink]
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answer is D, 4
18 can be written as 2*3^2
18^n= 2^n*3^2n
following the factorial manipulation method here https://gmatclub.com/forum/variations-i ... l#p2189076
max power of 2 divisible by 18! is 16 = 9+4+2+1
18/2= 9
18/4=4
18/8=2
18/16=1
max power of 3 divisible by 18! is 8= 6+2
18/3=6
18/9= 2
(any higher power will give value<1 so we stop here)
=> max power of 3 is the constraint here, max power of 3 is 8 , 3^8 written as 3^2(4)
i.e n=4
hope this helps!
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Re: What is the greatest value of n such that 18^n is a factor of 18! ? [#permalink]
Aryan19852 wrote:
18 is 2*3*3 which is 3^2 Now using the factorial formula 18/2=9 18/2^2=4 18/2^3=2 18/2^4=1 we have 16 twos and 8 threes in 18 factorial so maximum we can have 8 threes since it's 3^2 already the highest power which would divide the number is 4

­I do understand how we got 16 twos and 8 threes, but  I don't understand how we got 4 as highest power?
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Re: What is the greatest value of n such that 18^n is a factor of 18! ? [#permalink]
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