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In a certain game, a large bag is filled with blue, green, purple and [#permalink]
3
Choose a red coin first,
88000 / 11 = 8000, As 8000 is not divisible by 11, so can't consider red chip anymore.

Now we will choose green coin as much as possible,
8000 / 5 = 1600
1600 / 5 = 320
320 / 5 = 64, we can't consider green chips anymore.

But, to make 88000 value, we still need some chips that will have the product of 64.

The possible different combination for making product of 64 are listed below,
64 = 1 X 64
64 = 2 X 32
64 = 4 X 16
64 = 8 X 8

But, we had a constrain for the purple coin, its value needed to be greater than 5 but less than 11.
So, the only option that fits that criteria is 8 and choosing 2 purple coins of value 8 makes the product of 64 (8 X 8).

P:S- you don't need to choose a red coin first, even if you choose a green coin or more than one green coin, then choose a red coin, no matter what you will ultimately end up at 64 as remaining.
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Re: In a certain game, a large bag is filled with blue, green, purple and [#permalink]
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GeminiHeat wrote:
In a certain game, a large bag is filled with blue, green, purple and red chips worth 1, 5, x and 11 points each, respectively. The purple chips are worth more than the green chips, but less than the red chips. A certain number of chips are then selected from the bag. If the product of the point values of the selected chips is 88,000, how many purple chips were selected?

A. 1

B. 2

C. 3

D. 4

E. 5


This question begs for some prime factorization.

88,000 = (2)(2)(2)(2)(2)(2)(5)(5)(5)(11)

First, we can see that there must be one (11-point) red chip.
Now, what role do these 2's play? Since there are no 2's hiding among the 5-point chips or the 11-point chips, the 2's must be associated with the x-point chips.
Since we know that each purple chip is worth 6,7,8,9 or 10 points, we know that x must equal 6, 8 or 10.

x cannot equal 6, because we don't have any 3's in the prime factorization.
If x were to equal 10, we'd need six 5's to go with our six 2's. Since we don't have six 5's in the prime factorization of 88,000, we can rule out the possibility that x equals 10.

By the process of elimination, x MUST equal 8.
Since 8 = (2)(2)(2), we can see that the six 2's can be used to create two products of 8.

Answer: B

Cheers,
Brent
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Re: In a certain game, a large bag is filled with blue, green, purple and [#permalink]
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Prime factored. And got it in a reasonable time. Under 4 min for a tough one!
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In a certain game, a large bag is filled with blue, green, purple and [#permalink]
Possible factors: 1, 5, x, 11
5<x<11, x=6,7,8,9,10

The product of all chips is 88,000 -> we need to factor this into point values
88,000=8*11*10^3
Notice that, we can't decompose the 8 further - if we did, we would have 2^3 or 2^2(2), but none of the chips are worth 2 or 4. 10 does not work either - because none of the chips would fill in for the value of 8. This implies that our x value must be 8.
8*11*10^3=8*11*5^3*2^3 =8^2*11*5^3
So, we drew 2 purples (2 eight chips)
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