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Official Guide Problem 234

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Official Guide Problem 234

by sssssss » Sun Apr 12, 2009 10:32 pm
I saw the posting here https://www.beatthegmat.com/og-ps-234-t8849.html, but had a question on it, so I'm starting a new topic. Hope that's ok.

The problem reads: 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 official answer states that the red beads represent factors of 7 in the final point score so all you have to do is figure out how many times you can divide 147 000 by 7 evenly. Which is twice. So there must be 2 red beads removed.

My problem with this answer is that would assume that the number of yellow, green or blue beads removed are not multiples of 7.2 is a possible answer, but why is it the ONLY possible answer?

As I understand the problem, we are saying that 147000 = 7R5Y3G2B, where R is the number of red removed, Yis the number of yellow removed, G is green and B is blue.
So 147000=210RYGB.
RYGB = 700

All we need are 4 numbers that have a product of 700. R cannot be 0 and it cannot be 3 because those numbers do not divide evenly into 700. But why R=5, Y=2, G=10, B=7 or R=4, Y=5, G=5, B=7? Or, if we are not restricted by the answer choices: R=700, Y=1, G=1, B=1?

Thanks in advance for the help, it's driving me crazy.
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by Feep » Sun Apr 12, 2009 11:36 pm
It is indeed the only possible answer, because this question is a prime factorization question in disguise.

If you pull out blue, green, or yellow beads in multiples of 7, this is not going to mean that those beads, when their point values are multiplied, is going to be divisible by 7. In fact, it's impossible. If you pull seven blue beads, their product is not 2 * 7, but rather 2^7, which is 128. The only way anything is ever going to be divisible by a prime number is if that prime number itself is included in the product.

To illustrate, consider one of your sample "other" answers, R = 700, Y = 1, G = 1, B = 1. This is 2*3*5*SEVEN HUNDRED SEVENS, that is to say, 7^700, which is some ungodly number. Not 7*700.

I hope this answers your question.
I tutor GMAT/GRE level mathematics privately in the Los Angeles region, as well as via Skype for a discounted rate. Send me a message if you're interested.
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Thanks

by sssssss » Mon Apr 13, 2009 7:16 pm
I kind of see it. I see why if we are talking about exponents, the answer has to be 2. I'm not entirely clear why we are talking about 2^7 and not 2*7. But I'll play around with it and try to figure it out. Thanks
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factorization

by g2000 » Mon Apr 13, 2009 7:32 pm
As Feep mentioned, it's a factorization problem.

First, factorize 147000 and you get the following

147000 = 2*2*2 * 3 * 5*5*5 * 7*7

The problem says the product of the points is equivalent of 147000. Therefore what we need to do is to count the number of 7.

In our case, it is 2(because there are 2 sevens in our factorization).
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by gmat740 » Mon Apr 13, 2009 7:58 pm
RYGB = 700
RYGB = 7*5*5*4

So which number is corresponding to which bead??

please help me understand this!!
I am not able to get how we can surely say that 7 corresponds to red or green?

Karan
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by sssssss » Thu Apr 16, 2009 5:27 am
I see my mistake. The product of the point value means you have to multiply everything together. So Karan, ignore my equation. I was totally wrong. It sounds like you're making the same mistake I was.
Simply put, this question is asking for the product of point values. So imagine a simpler version. You have 2 red beads worth 7 each. And 2 yellow beads worth 5 each. The PRODUCT of their values would be 7*7*5*5, or 7^2*5^2 or Point of beads^number of that kind of bead.

So the correct equation for the original problem would be, as someone posted already:
7^R*5^Y*3^G*b^2=147000.

Then it's just figuring out how many times 7 is a factor of 147000.
Thanks everyone. I see this now!
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