More Primes
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vijaya priya
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petrifiedbutstanding
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That was awesome! Who are you!Stuart Kovinsky wrote:We know that the purple chips are worth 6, 7, 8, 9 or 10 points each.mmukher 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?
Options :
1
2
3
4
5
OA later.
The blue chips are worth 1 point each, so we can ignore those.
Let's break 88000 down to primes:
88 * 1000
11 * 8 * 10 * 10 * 10
11 * 2 * 2 * 2 * 2 * 5 * 2 * 5 * 2 * 5
so:
2^6 * 5^3 * 11
Well, we're not getting any 2s out of the 1, 5 or 11, so all the 2s have to come from x.
Therefore, x has to be 6, 8 or 10.
x can't be 6, because we don't want any 3s.
If x were 10, it would give us 2s and 5s. So to get 6 2s we'd also have to take 6 5s, which is way more than we want.
Therefore, x MUST be 8.
To get 2^6, we need two 8s: choose (b).
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deepsea13
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Stuart Kovinsky wrote:We know that the purple chips are worth 6, 7, 8, 9 or 10 points each.mmukher 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?
Options :
1
2
3
4
5
OA later.
The blue chips are worth 1 point each, so we can ignore those.
Let's break 88000 down to primes:
88 * 1000
11 * 8 * 10 * 10 * 10
11 * 2 * 2 * 2 * 2 * 5 * 2 * 5 * 2 * 5
so:
2^6 * 5^3 * 11
Well, we're not getting any 2s out of the 1, 5 or 11, so all the 2s have to come from x.
Therefore, x has to be 6, 8 or 10.
x can't be 6, because we don't want any 3s.
If x were 10, it would give us 2s and 5s. So to get 6 2s we'd also have to take 6 5s, which is way more than we want.
Therefore, x MUST be 8.
To get 2^6, we need two 8s: choose (b).
Phew! Thanks! Didn't have a clue on how to solve this problem!
Amazing!Stuart Kovinsky wrote:We know that the purple chips are worth 6, 7, 8, 9 or 10 points each.mmukher 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?
Options :
1
2
3
4
5
OA later.
The blue chips are worth 1 point each, so we can ignore those.
Let's break 88000 down to primes:
88 * 1000
11 * 8 * 10 * 10 * 10
11 * 2 * 2 * 2 * 2 * 5 * 2 * 5 * 2 * 5
so:
2^6 * 5^3 * 11
Well, we're not getting any 2s out of the 1, 5 or 11, so all the 2s have to come from x.
Therefore, x has to be 6, 8 or 10.
x can't be 6, because we don't want any 3s.
If x were 10, it would give us 2s and 5s. So to get 6 2s we'd also have to take 6 5s, which is way more than we want.
Therefore, x MUST be 8.
To get 2^6, we need two 8s: choose (b).
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bnpetteway
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I did mine a different way and I came to a conclusion a different way. I still got the right answer, but I hope it doesn't mess things up on the test.
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Mathsbuddy
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From the question:
1^b * 5^g * x^p * 11^r = 88000
Also, 88000 = 2^6 * 5^3 * 11^1 (product of prime factors)
Matching bases gives: b=1, g = 3, r = 1 and x^p = 2^6 = (2^3)^2 = 8^2 = 64^1
However, the question states that 5<x<11, therefore x = 8 and p =2
Therefore answer is 2 (B)
1^b * 5^g * x^p * 11^r = 88000
Also, 88000 = 2^6 * 5^3 * 11^1 (product of prime factors)
Matching bases gives: b=1, g = 3, r = 1 and x^p = 2^6 = (2^3)^2 = 8^2 = 64^1
However, the question states that 5<x<11, therefore x = 8 and p =2
Therefore answer is 2 (B)
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Mathsbuddy
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kswarna wrote:Nice try, but:Stuart Kovinsky wrote:We know that the purple chips are worth 6, 7, 8, 9 or 10 points each.mmukher 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?
Options :
1
2
3
4
5
OA later.
>> IT IS NOT GIVEN ANYWHERE THAT IT NEEDS TO BE IN ORDER. X CAN BE 2,3,4 TOO as purple has a value in between the highest and lowest values. This can be either the specified 5 or this can be the unspecified x. So the answer should be right applied to either. now
88,000 = 11 * 8000 *1
= 11 * (20)^3 * 1
= 11 * ( 4 * 5 ) ^ 3 *1
I.E the answer is 3 CHOICE C
x is more than 5 (so it cannot equal 0, 1, 2, 3, 4 or 5)
x is less than 11 (so it cannot equal 11 or indeed 20)
The order has to match too:
You cannot make 11 by multiplying 1 * 1 * 1 * 1 etc. It is always 1 for the product of blue points.
Therefore your correct product = 11 * (4 * 5)^3 * 1 = 11 * 4^3 * 5^3 * 1 = 11 * 8^2 * 5^3 * 1
meaning that x = 8 and number of purple = 2.
I hope that helps.
I'd say that either the problem should be rephrased in more accurate terms (to avoid people like me interpreting it this way) or there isn't one single answer.
As proof, let's start by the end and suppose values of x within the possible range (6,7,8,9,10):
If we assume the value of x=10, it is possible to find a combination of chips such that the product of their values in points = 88,000. For example:
(1 blue x 1)(2 green x 5)(16 purple x 10)(5 x 11) = 88,000
Now let's assume the value of x=8. A possible combination of chips could also yield 88,000:
(1x1)(5x5)(5x8)(8x11) = 88,000
And it is to see that for these two values of x other combinations of chips could also yield 88,000.
Cheers,
As proof, let's start by the end and suppose values of x within the possible range (6,7,8,9,10):
If we assume the value of x=10, it is possible to find a combination of chips such that the product of their values in points = 88,000. For example:
(1 blue x 1)(2 green x 5)(16 purple x 10)(5 x 11) = 88,000
Now let's assume the value of x=8. A possible combination of chips could also yield 88,000:
(1x1)(5x5)(5x8)(8x11) = 88,000
And it is to see that for these two values of x other combinations of chips could also yield 88,000.
Cheers,
Stuart Kovinsky wrote:We know that the purple chips are worth 6, 7, 8, 9 or 10 points each.mmukher 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?
Options :
1
2
3
4
5
OA later.
The blue chips are worth 1 point each, so we can ignore those.
Let's break 88000 down to primes:
88 * 1000
11 * 8 * 10 * 10 * 10
11 * 2 * 2 * 2 * 2 * 5 * 2 * 5 * 2 * 5
so:
2^6 * 5^3 * 11
Well, we're not getting any 2s out of the 1, 5 or 11, so all the 2s have to come from x.
Therefore, x has to be 6, 8 or 10.
x can't be 6, because we don't want any 3s.
If x were 10, it would give us 2s and 5s. So to get 6 2s we'd also have to take 6 5s, which is way more than we want.
Therefore, x MUST be 8.
To get 2^6, we need two 8s: choose (b).
Dear Stuart
Since there is no mention of total number of chips he picked up or the sum of the chips.
Why can't I consider some other number like
chips * their value
8(blue chips) * 1
20(green chips)*5
1 (purple chip) * 10
1(Red chip) * 11
Thanks
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Hi evs.teja,
The example that you listed does not follow the instructions given in the prompt. We're told to figure out the PRODUCT of the FACE VALUES on all the chips selected.
Part of your solution included 20 green chips worth 5 points each. In your solution, you've done...
20 x 5 = 100
But because we're dealing with a product we'd actually end up with....
(5)(5)(5).......(5) = 5^20
This goes well beyond the 88,000 total that we're trying to "hit", so your solution is not correct.
GMAT assassins aren't born, they're made,
Rich
The example that you listed does not follow the instructions given in the prompt. We're told to figure out the PRODUCT of the FACE VALUES on all the chips selected.
Part of your solution included 20 green chips worth 5 points each. In your solution, you've done...
20 x 5 = 100
But because we're dealing with a product we'd actually end up with....
(5)(5)(5).......(5) = 5^20
This goes well beyond the 88,000 total that we're trying to "hit", so your solution is not correct.
GMAT assassins aren't born, they're made,
Rich
Thanks Stuart....now I understand since it is the product of points we are looking for it can only in powers so (1)^a * (5)^b * (x)^c * (11)^d = 11 * (2)^6 * (5)^3
comparing them 8 is the only possible outcome.
got it!
Thanks again
comparing them 8 is the only possible outcome.
got it!
Thanks again
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GMATinsight
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88000 = 88 x 1000 = 2^3 x 11 x 2^3 x 5^3 = 2^6 x 5^3 x 11mmukher 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?
Options :
1
2
3
4
5
OA later.
Since the possible points are 1, 5, x and 11 therefore x must be 2 or another power of 2 only
Therefore the points represented by each purple chip = x = 2 or 4 or 8
but since 88000 has 2^6 therefore there must be 6 chips (2 points each) or 3 chips (4 points each) or 2 chips (8 points each) of purple color
but since 5<x<11 therefore x must be 8 only. Therefore 2 Chips
Answer: Option B
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stephenjmurph
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You can't divide the 88,000 by the 5x11. You change the problem and lose information. Your next equation states: BGRP x= 1600 -- that could be true with some combinations of R, G, R, P, but R would have to be 0, and you know that there is at least one R because 88,000 has a prime factor of 11.nikmahes wrote:Guys, Iam unable to understand as to how can we have a unique solution to this problem. This is how I approach -
Let, the number of Blue balls=B
Green Balls= G
Purple Balls= P
Red Balls= R
as per the problem we get,
B*5G*Px * 11R= 88000
=> BGRP x= 1600
which implies that x can be 8 or 10 as 5<x<11.
If x =8,
BGRP= 200
If x=10,
BGRP =160
Now, how can we uniquely decide the value of P? There can be multiple values for each of the variables.
Any help would be great.
Thanks,
Nikhil
Another angle: let's say BGRP = 1600. Since B=1, then GRP = 1600. Since R has to be 0 (because 11 doesn't factor into 1600), then GP=1600. 1600 factors to 2*2*2*2*2*2*5*5. If P was ten, then TWO "2*5" are removed (has to be both 5's), so then 1600 = (2*5*2*5)*2*2*2*2 = 2P's and (some number) of G's. BUT, G can't be ANY combination of only 2's because it is 5 points. 2 and 5 are primes, so there's no way to do this.
Bottom line is you are assuming you can "MAKE UP FOR THE 11" (you divide 88,000 by) by making some other combination of the other primes. primes of 2 and 5 will never equal some multiple of 11.
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akash singhal
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suppose we have one blue one red and one green chip.
Dividing their points by 88000 we get:-
(88000/1*5*11)=1600
Now, we know x can be 6,7,8,9,10
Since 6,7,9 are unable to divide 1600 we are left with 8 and 10
now both 8 and 10 can divide 1600 exactly two times....
so my answer is number of purple chips are two.
2
also no way we can find whether x is 8 or 10 thats why it is asked how many chips
not the value of x
I hope i am right.....
Dividing their points by 88000 we get:-
(88000/1*5*11)=1600
Now, we know x can be 6,7,8,9,10
Since 6,7,9 are unable to divide 1600 we are left with 8 and 10
now both 8 and 10 can divide 1600 exactly two times....
so my answer is number of purple chips are two.
2
also no way we can find whether x is 8 or 10 thats why it is asked how many chips
not the value of x
I hope i am right.....
