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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]
can someone please explain the last step in detail please. Thanks
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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]
 
Apeksha2101 wrote:
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?

­Since 18! contains \(2^{16}\) and \(3^8\), and \(18^n\) can be written as \((2*3^2)^n\). Therefore n has to be 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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