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# Q 4. If n is the product of the integers from1 to 20

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SVP
Joined: 28 May 2005
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Location: Dhaka
Q 4. If n is the product of the integers from1 to 20 [#permalink]

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13 Oct 2005, 13:51
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Q 4.
If n is the product of the integers from1 to 20 inclusive, what is the greatest integer k for which 2k is a factor of n?
A. 10
B. 12
C. 15
D. 18
E. 20
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VP
Joined: 22 Aug 2005
Posts: 1112
Location: CA

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13 Oct 2005, 14:21
D.

1 * 2 * 3 * 4 *....*9*...*19*20

is divisible by: 36
k = 18
Director
Joined: 21 Aug 2005
Posts: 789

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13 Oct 2005, 14:25
E

1*4*...*10*...*20 is divisible by 40 = 2k. So K=20
Senior Manager
Joined: 02 Oct 2005
Posts: 297

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13 Oct 2005, 14:59

Duttsit: I think you took only 19 & 20 into consideration. You can take any even and 20 we should be able to get 2 * 20 as factor.

Actually all the below are factors and we need to get only the 'greatest'
SVP
Joined: 28 May 2005
Posts: 1705
Location: Dhaka

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14 Oct 2005, 14:21
But OA is D.
Can somebody explain that. or is it OA is wrong?
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Senior Manager
Joined: 02 Oct 2005
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14 Oct 2005, 14:54
Duttsit is the person who got the answer D.

Director
Joined: 14 Oct 2003
Posts: 583
Location: On Vacation at My Crawford, Texas Ranch

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14 Oct 2005, 15:30
nakib77 wrote:
But OA is D.
Can somebody explain that. or is it OA is wrong?

OA probably wrong, where is this question from?
SVP
Joined: 03 Jan 2005
Posts: 2233

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14 Oct 2005, 17:58
Is the question correct? I mean n is a factor of n itself right? So n/2 would be the biggest k (since we know n is even and thus n/2 is an integer). In other words k would be 20!/2. The question may be asking about k as a product of prime numbers or something ...
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Current Student
Joined: 28 Dec 2004
Posts: 3357
Location: New York City
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14 Oct 2005, 19:10
Oh I get it the question is asking

2^k is a factor of N

so then
lets see...

20/2=10

then you get another 2 from 4, you get 2 2s from 8, 12, you get another 2, from 16 you get 3 2s... and one more 2 from 20, so lets count all the 2s then

10+1+2+1+3+1=18

K can be 18....D it is...
Senior Manager
Joined: 02 Oct 2005
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14 Oct 2005, 21:13
To be more clear ... Since we are asked 2 power of k, lets take only multiples of 2.

2 * 4 * 6 * 8 ... * 20

2, 6, 10, 14, 18 = 2^1 * odd (2^1 occurs 5 times)
4, 12, 20 = 2^2 * odd (2^2 occurs 3 times)
8 = 2^3
16 = 2^4

5 * 1 + 3 * 2 + 3 + 4 = 18. Hence D.
SVP
Joined: 28 May 2005
Posts: 1705
Location: Dhaka

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15 Oct 2005, 06:04
Thank you. I didn't see that exponent. I was so mad while I was solving it.

this question is from paper based question from ETS.
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15 Oct 2005, 06:04
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