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

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Director
Joined: 17 Dec 2005
Posts: 547
Location: Germany
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.
Senior Manager
Joined: 11 Nov 2005
Posts: 328
Location: London

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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.
SVP
Joined: 24 Sep 2005
Posts: 1885
Re: Out of Practice exams (GMAT-Club) [#permalink]

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22 Dec 2005, 08:50
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.

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
Joined: 17 Dec 2005
Posts: 547
Location: Germany

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22 Dec 2005, 09:04
Ok, it's clear now. Thanks Laxieqv
Senior Manager
Joined: 05 Oct 2005
Posts: 485

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22 Dec 2005, 10:25
Yup... i agree w/ Laxie
Intern
Joined: 11 Sep 2005
Posts: 19
Location: Framingham, MA, USA

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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.

SVP
Joined: 16 Oct 2003
Posts: 1801

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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

Senior Manager
Joined: 15 Apr 2005
Posts: 415
Location: India, Chennai
Re: Out of Practice exams (GMAT-Club) [#permalink]

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23 Dec 2005, 00:58
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.

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
Joined: 06 Jun 2004
Posts: 1053
Location: CA

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23 Dec 2005, 01:06
Agree with Laxie, D it is
_________________

Don't be afraid to take a flying leap of faith.. If you risk nothing, than you gain nothing...

CEO
Joined: 20 Nov 2005
Posts: 2894
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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