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Is 1,000 a factor of k? a) 2,000 is a factor of 4k b) 8 is a

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Is 1,000 a factor of k? a) 2,000 is a factor of 4k b) 8 is a [#permalink] New post 09 Oct 2003, 16:00
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Is 1,000 a factor of k?

a) 2,000 is a factor of 4k

b) 8 is a factor of k
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 [#permalink] New post 10 Oct 2003, 06:08
My guess is C as well.

Reasoning:

(1) In other words 500 is a multiple of K.
(2) This says 8 is multiple of K but that has nothing to do with 1000.
(combined) If 500 is multiple of K then you can divide further and see how far you can go with a multiple of 2 in the denominator. You end up with 125, this is where (2) comes in the picture, and you can see that K has to be divisible by at least by twice the 500, which is 1000.
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[#permalink] New post 10 Oct 2003, 07:24
The answer is D. 2000 and 8 both go by 1,000.
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Re: D [#permalink] New post 10 Oct 2003, 07:44
watchdog wrote:
The answer is D. 2000 and 8 both go by 1,000.


(1) Let's say K is 500, thus 4K = 2000. 4K divisible by 2000. But 1000 is not a factor of 500.
(2) K = 320. K/8 = 40. Doesn't say anything about 1000 being a factor.
(combined) Since K has to be at least 2*500 = 1000, you can say 1000/8 = 125. And 1000/1000 = 1.
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 [#permalink] New post 14 Oct 2003, 21:53
I would pick C. Here is the reasoning:

Statement in other words: Is 2^3 x 5^3 factors of K?

1. 2^4 x 5^3 is a factor of 4k, therefore, 2^2 x 5^3 is a factor of K. -Not sufficient since we need one more 2 to affirm the statement.

2. 2^3 is a factor of K, not enough.

Having both, 2^2 x 5^3 and 2^3 as factor of k is enough to say that 2^3 x 5^3 are factors of K
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 [#permalink] New post 14 Oct 2003, 23:14
got C as well

(1) k=500p NOT SUFF
(2) k=8q NOT SUFF

Combine: k=LCM(500, 8)*r=1000r
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 [#permalink] New post 14 Oct 2003, 23:32
YES!!! :banana
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 [#permalink] New post 14 Oct 2003, 23:48
stolyar wrote:
YES!!! :banana


Stolyar

can we say ,in general, that..

If two numbers are factors of k, then their LCM is also a factor of K.
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 [#permalink] New post 14 Oct 2003, 23:57
I think yes.
If two numbers A and B are both factors of X, then X is either LCM(A, B) itself or a multiple of LCM(A, B)
  [#permalink] 14 Oct 2003, 23:57
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