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

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

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New post 22 Nov 2009, 18:05
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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
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Re: Number Properties Question  [#permalink]

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New post 22 Nov 2009, 18:13
shanewyatt wrote:
If n is the product of the integers from1 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


IMO D

1*2 *3 * 4 * ....20

if we take 2 common out of all even numbers from 1 to 20, we have \(2^{18}\). Hence k=18 because \(2^{18}\)is the max factor in form \(2^K\)
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Re: Number Properties Question  [#permalink]

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New post 22 Nov 2009, 18:25
1
shanewyatt wrote:
If n is the product of the integers from1 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


\(n=20!\) and we need to find greatest integer \(k\), for which \(2^k*n=20!\). Obviously \(k\) would be highest when \(n=1\). So basically we are asked to determine the highest power of \(2\) in \(20!\).

Finding highest power of prime in \(n!\): everything-about-factorials-on-the-gmat-85592.html

Hence the power of \(2\) in \(20!\) would be: \(\frac{20}{2}+\frac{20}{4}+\frac{20}{8}+\frac{20}{16}=10+5+2+1=18\)

Answer: D.
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Re: Number Properties Question  [#permalink]

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New post 22 Nov 2009, 21:01
There are 10 numbers divisible by 2
There are 5 numbers divisible by 4
There are 2 numbers divisible by 8
There is 1 number divisible by 16.

Hence the total number of 2’s in 20! are 10+5+2+1=18
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Re: If n is the product of the integers from 1 to 20 inclusive,  [#permalink]

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New post 20 Apr 2014, 00:49
n= 20!

And

2^K * A = n = 20!

=> 20! = 2^K * A

Mathematically , we need to find the powers of 2 contained in the factorial given.

2/4/6/8/10/12/14/16/18/20

Total contribution of '2's' is 18

Hence the answer
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Re: If n is the product of the integers from 1 to 20 inclusive,  [#permalink]

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New post 04 Nov 2016, 08:32
1
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 the integers from 1 to 20 inclusive = 20!

Highest value of \(2^k\) is

20/2 = 10
10/2 = 5
5/2 = 2
2/2 = 1

Hence the value of k will be 10 + 5 + 2 + 1 = 18

So, The correct answer will be (D) 18
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Re: If n is the product of the integers from 1 to 20 inclusive,  [#permalink]

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New post 16 Jan 2018, 10:37
Bunuel wrote:
shanewyatt wrote:
If n is the product of the integers from1 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


\(n=20!\) and we need to find greatest integer \(k\), for which \(2^k*n=20!\). Obviously \(k\) would be highest when \(n=1\). So basically we are asked to determine the highest power of \(2\) in \(20!\).

Finding highest power of prime in \(n!\): http://gmatclub.com/forum/everything-ab ... 85592.html

Hence the power of \(2\) in \(20!\) would be: \(\frac{20}{2}+\frac{20}{4}+\frac{20}{8}+\frac{20}{16}=10+5+2+1=18\)

Answer: D.


Hey bunuel, this is a very useful formula in finding the highest power for which 2 is a factor of n.

Could you share some possible variations of this? how could it become more difficult?
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Re: If n is the product of the integers from 1 to 20 inclusive,  [#permalink]

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New post 16 Jan 2018, 10:46
destinyawaits wrote:
Bunuel wrote:
shanewyatt wrote:
If n is the product of the integers from1 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


\(n=20!\) and we need to find greatest integer \(k\), for which \(2^k*n=20!\). Obviously \(k\) would be highest when \(n=1\). So basically we are asked to determine the highest power of \(2\) in \(20!\).

Finding highest power of prime in \(n!\): http://gmatclub.com/forum/everything-ab ... 85592.html

Hence the power of \(2\) in \(20!\) would be: \(\frac{20}{2}+\frac{20}{4}+\frac{20}{8}+\frac{20}{16}=10+5+2+1=18\)

Answer: D.


Hey bunuel, this is a very useful formula in finding the highest power for which 2 is a factor of n.

Could you share some possible variations of this? how could it become more difficult?


Everything about Factorials on the GMAT
Power of a Number in a Factorial Problems
Trailing Zeros Problems

For more check Ultimate GMAT Quantitative Megathread

Hope it helps.
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Re: If n is the product of the integers from 1 to 20 inclusive,  [#permalink]

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New post 03 Apr 2018, 08:00
It's always pleasing to get a question like this. The answer must be D, integer 2 will be having a power of 18 in 20!.
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Re: If n is the product of the integers from 1 to 20 inclusive,   [#permalink] 03 Apr 2018, 08:00
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