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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]

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22 Dec 2005, 05:36
This topic is locked. If you want to discuss this question please re-post it in the respective forum.

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.
Senior Manager
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22 Dec 2005, 08:25
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]

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22 Dec 2005, 08:50
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.
Director
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22 Dec 2005, 09:04
Ok, it's clear now. Thanks Laxieqv
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22 Dec 2005, 10:25
Yup... i agree w/ Laxie
Intern
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22 Dec 2005, 15:00
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
SVP
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22 Dec 2005, 19:58
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]

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23 Dec 2005, 00:58
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.
VP
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23 Dec 2005, 01:06
Agree with Laxie, D it is
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Don't be afraid to take a flying leap of faith.. If you risk nothing, than you gain nothing...

CEO
Joined: 20 Nov 2005
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Schools: Completed at SAID BUSINESS SCHOOL, OXFORD - Class of 2008
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23 Dec 2005, 02:11
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
_________________

SAID BUSINESS SCHOOL, OXFORD - MBA CLASS OF 2008

23 Dec 2005, 02:11
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# In a certain game, a large container is filled with red,

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