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Nice question.

by logitech » Sun Dec 07, 2008 2:51 pm
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?

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by kris610 » Sun Dec 07, 2008 3:29 pm
Factorize 88,000 as 11, 5^3 and 2^6. The purple chip has a value between 5 and 11 and you want an answer of the form n^m, where n is the value of the purple chip.

Split 2^6 in 2^3 and 2^3 you get two purple chips, each with a value of 8.
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by Abdulla » Sun Dec 07, 2008 5:56 pm
kris610 wrote:Factorize 88,000 as 11, 5^3 and 2^6. The purple chip has a value between 5 and 11 and you want an answer of the form n^m, where n is the value of the purple chip.

Split 2^6 in 2^3 and 2^3 you get two purple chips, each with a value of 8.
everything was okay until you splite the 2^6 ... could you explain more please..
Abdulla
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by dmateer25 » Sun Dec 07, 2008 6:13 pm
Abdulla wrote:
kris610 wrote:Factorize 88,000 as 11, 5^3 and 2^6. The purple chip has a value between 5 and 11 and you want an answer of the form n^m, where n is the value of the purple chip.

Split 2^6 in 2^3 and 2^3 you get two purple chips, each with a value of 8.
everything was okay until you splite the 2^6 ... could you explain more please..
Think of it like this. We have 2^6 = 64

Possible value of chips:

If the chip had a value of 64 only 1 would be selected.
If the chip had a value of 8 then 2 would be selected (8x8)
If the chip value of 4 then 3 would be selected (4x4x4)
If the chip value was 2 then 6 would be selected. (2x2x2x2x2x2)

We know from the problem that the chip value must be between 5 and 11 so the value of the chip must be 8. Therefore, 2 purple chips were selected.


Hope this helps
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by pbanavara » Sun Dec 07, 2008 6:18 pm
I got a value of either 10*1 or 1*10

11*80*10*10

1 r*11, 16 g*5, 10b and 1p*10 ( because p has to be between 5 and 11 ) .. blue can be either 1 or 10
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by kris610 » Sun Dec 07, 2008 6:42 pm
Hi Abdulla,
Let me try to make it clearer.

You have 2^6, which you need to express as a power of 6,7,8,9 or 10.

Now, 6 and 9 are ruled out as you do not have 3 in the factors and likewise for 7.

If you want to express a power of 10, you need to use one or more of the 5s (from 5^3), in which case the remaining 2s cannot be expressed as a power of 8 or 10 -- you can try this out.

So, you are left with the option of splitting 2^6 into two 2^3s i.e. you have two 8s for the two purple chips.
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by cramya » Mon Dec 08, 2008 6:40 am
Think of it like this. We have 2^6 = 64

Possible value of chips:

If the chip had a value of 64 only 1 would be selected.
If the chip had a value of 8 then 2 would be selected (8x8)
If the chip value of 4 then 3 would be selected (4x4x4)
If the chip value was 2 then 6 would be selected. (2x2x2x2x2x2)

We know from the problem that the chip value must be between 5 and 11 so the value of the chip must be 8. Therefore, 2 purple chips were selected.


Very good explanation Dmateer!
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by penumbra547 » Tue Dec 09, 2008 10:26 am
kris610 wrote:Factorize 88,000 as 11, 5^3 and 2^6. The purple chip has a value between 5 and 11 and you want an answer of the form n^m, where n is the value of the purple chip.

Split 2^6 in 2^3 and 2^3 you get two purple chips, each with a value of 8.
very well done!
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by 720dreaming » Thu Dec 11, 2008 2:05 pm
Got a value of 10 as well...but do like the reasoning presented for value of 8. Whats the OA?
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