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If n is the product of the integers from 1 to 20 inclusive, what is th

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If n is the product of the integers from 1 to 20 inclusive, what is th  [#permalink]

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New post 24 Jan 2020, 21:59
shanewyatt wrote:
If n is the product of the integers from 1 to 20 inclusive, what is the greatest integer k for which \(2^k\) is a factor of n?

A. 10
B. 12
C. 15
D. 18
E. 20


Product of all integers from 1 to 20 = 1*2*3*---*20 = 20!

Power of any Prime Number in any factorial can be calculated by following understanding

Power of prime x in n! = [n/x] + [n/x^2] + [n/x^3] + [n/x^4] + ... and so on

Where [n/x] is greatest Integer value of (n/x) less than or equal to (n/x)
i.e. [100/3] = [33.33] = 33
i.e. [100/9] = [11.11] = 11 etc.
Where,
[n/x] = No. of Integers that are multiple of x from 1 to n
[n/x^2] = No. of Integers that are multiple of x^2 from 1 to n whose first power has been counted in previous step and second is being counted at this step
[n/x^3] = No. of Integers that are multiple of x^3 from 1 to n whose first two powers have been counted in previous two step and third power is counted at this step
And so on.....



So, Power of 2 in 20! = [20/2] + [20/2^2] + [20/2^3] + [20/2^4] = 10+5+2+1 = 18

Answer: Option D
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Re: If n is the product of the integers from 1 to 20 inclusive, what is th  [#permalink]

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New post 28 Mar 2020, 04:35
shanewyatt wrote:
If n is the product of the integers from 1 to 20 inclusive, what is the greatest integer k for which \(2^k\) is a factor of n?

A. 10
B. 12
C. 15
D. 18
E. 20



Power of 2 in 20! = 10 + 5 + 2 + 1 = 18

IMO D
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Re: If n is the product of the integers from 1 to 20 inclusive, what is th   [#permalink] 28 Mar 2020, 04:35

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