If n is an integer and n= 2x3x5x7x11x13 77k Then which of : Quant Question Archive [LOCKED]
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# If n is an integer and n= 2x3x5x7x11x13 77k Then which of

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If n is an integer and n= 2x3x5x7x11x13 77k Then which of [#permalink]

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27 Jun 2006, 15:09
This topic is locked. If you want to discuss this question please re-post it in the respective forum.

If n is an integer and
n=
2x3x5x7x11x13
77k

Then which of the following coudl be the value of K?

a 22
b 26
c 35
d 54
e 60
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27 Jun 2006, 15:37
Yep, 26.

For me, the fastest way to get to the answer is by first reducing the fraction by dividing Num and Den by 7*11=77 and then simply start multiplying remaining numbers in the Numerator until I get one given in the answer choices (which would be 26=2*13)
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Re: Fastest way to solve [#permalink]

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27 Jun 2006, 18:08
neemo wrote:
If n is an integer and
n=
2x3x5x7x11x13
77k

Then which of the following coudl be the value of K?

a 22
b 26
c 35
d 54
e 60

B.
from numorater we get 7x11=77 and 13x2=26, since the numorater is still divisible k could be 26

sorry not sure how to explain well
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27 Jun 2006, 18:43
B

The main stem reduces to 2*3*5*7*11*13/7*11*k i.e
2*3*5*13/k

So k must be product of any combination of product of these numbers. Now look at the answers and you will see that only 26 (2*13) fits.
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Re: Fastest way to solve [#permalink]

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27 Jun 2006, 20:14
B

If N is an integer, k has to be a factor of N where N = (2x3x5x13)/k, since 77 is a factor of 7x11. So any product of the 4 numbers (2,3,5,13) besides the number themselves is considered a factor of (2x3x5x13) if N is an integer....6, 10, 26, 15, 39, 65, 30, 78, 195.
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27 Jun 2006, 20:45
B it is
(2x3x5x7x11x13)/77k-->

(2*3*5*13)/k
n is an integer only when k=26
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27 Jun 2006, 20:48
B

What I did was factorized it to the simplest form so that only K will be in the denominator.
We get n=2*3*5*13/K

Look at the choices and see which number when in denominator can be factored out.

Did not take more than 50 secs
27 Jun 2006, 20:48
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