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In a certain game, a large container is filled with red,

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Director
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In a certain game, a large container is filled with red, [#permalink] New post 22 Dec 2005, 04:36
In a certain game, a large container is filled with red, yellow, green, and blue beads worth, respectively, 7, 5, 3, and 2 points each. A number of beads are then removed from the container. If the product of the point values of the removed beads is 147,000, how many red beads were removed?

(A) 5
(B) 4
(C) 3
(D) 2
(E) 0

I simply don't beat it.
Thanks for your help.
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 [#permalink] New post 22 Dec 2005, 07:25
Clueless :(

Tried hard to play around with odd/even numbers but could not reach to one options of the answer.

definately not E.
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Re: Out of Practice exams (GMAT-Club) [#permalink] New post 22 Dec 2005, 07:50
allabout wrote:
In a certain game, a large container is filled with red, yellow, green, and blue beads worth, respectively, 7, 5, 3, and 2 points each. A number of beads are then removed from the container. If the product of the point values of the removed beads is 147,000, how many red beads were removed?

(A) 5
(B) 4
(C) 3
(D) 2
(E) 0

I simply don't beat it.
Thanks for your help.


The trick of this problem is to understand that the product of point values is the product of single point values. For example: if 2 red and 3 green removed--> the product= 7*7*3*3*3

Let x, y, z and t be the numbers of beads of red, yellow, green and blue removed respectively.
we have 7^x * 5^y * 3^z * 2^t = 147,000
x can't be 0 coz then LHS doesnt contain any multiple of 7.
The only choice fits here is x=2

D it is.
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 [#permalink] New post 22 Dec 2005, 08:04
Ok, it's clear now. Thanks Laxieqv
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 [#permalink] New post 22 Dec 2005, 09:25
Yup... i agree w/ Laxie
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 [#permalink] New post 22 Dec 2005, 14:00
This is how I solved it.
Since Red bead is worth 7 points,
147000/7 = 21000
21000/7 = 3000
3000/7 = 428.57 which is a fraction.
Thus there can only be 2 Red beads in the total score of 147000.

Answer = D
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 [#permalink] New post 22 Dec 2005, 18:58
147000 has three zeros so 5*2 * 5*2 * 5*2. After dividing by 1000 we get 147 the prime factors are 7*7*3

So the answer is 2.
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Re: Out of Practice exams (GMAT-Club) [#permalink] New post 22 Dec 2005, 23:58
allabout wrote:
In a certain game, a large container is filled with red, yellow, green, and blue beads worth, respectively, 7, 5, 3, and 2 points each. A number of beads are then removed from the container. If the product of the point values of the removed beads is 147,000, how many red beads were removed?

(A) 5
(B) 4
(C) 3
(D) 2
(E) 0

I simply don't beat it.
Thanks for your help.


D it is. Here is my working.

The total product value = 147000. This contains X reds with each red being 7.
So 147000/7 = 21000 implying 1 red excluded.
21000/7 = 3000 implying 1 more red exclued, and totaly 2 till now.

3000 cannot be divided by 7. Meaning that there are no more reds. Hence D.
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 [#permalink] New post 23 Dec 2005, 00:06
Agree with Laxie, D it is
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 [#permalink] New post 23 Dec 2005, 01:11
Its simple. Three zeros in the end shows that there must be 3 yellows and 3 blues. Remaining is 147 = 7*7*3 so 2 red balls.

147000 = 7*7*3*5*2*5*2*5*2
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  [#permalink] 23 Dec 2005, 01:11
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