BEADS

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BEADS

by grandh01 » Tue Aug 28, 2012 7:10 pm
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

OA is D

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by alex.gellatly » Tue Aug 28, 2012 9:12 pm
grandh01 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
The correct answer should be D and here is why:

This question deals with prime factorization. We know this because all of the colors are prime numbers. So we need to prime factor out 147,000. We notice that red is 7 points... so we only care about 7. After factoring out we get 147,000= 2^3*3*5^3*7^2. Notice that there are 2 7's "in" 147,000. So that is our answer. Does this help?
A useful website I found that has every quant OG video explanation:

https://www.beatthegmat.com/useful-websi ... tml#475231

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by neelgandham » Tue Aug 28, 2012 9:20 pm
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

147000 = 7*21*1000 = 7*7*3*2*5*2*5*2*5 = 7^2 * 3 * 2^3 * 5^3.
So, 2 beads of value 7, One bead of value 3, 3 beads of value 2 and 3 beads of value 5 are removed.

The answer isD.2
Anil Gandham
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