B->1 point
G-> 5
P-> X
R-> 11

Lets say in the selection all 4 chips were present then we will have

1^a * 5^b * x^c * 11^d (for example lets say b =5 we will have 5*5*5*5*5 which is the same as writing 5^b)

where a-> number of blue chips selected
b-> number of green chips selected
c-> number of purple chips selected
d-> number of red chips selected

break 88000 in to prime factors = 2^6*5^3*11

What number can we fit in the point value between 5 and 11. Clearly 8(since its given purple chip carry more points than green but less than red)

2*2*2 = 8 and another 2*2*2 = 8

8^2 (= 2^6)

So we now have c=2 (number of purple chips selected)

Hope this helps.

Regards,
CR

Don't know if this is the faster approach, but here it goes.

The prime factorization of 88000 is: 2^6 * 5 * 11. We have one Green chip, one red chip, and 2^6 can be divided into 8*8 or 4*4*4. As 8 is a number that satisfies the conditions for the value of the purple chip. We know that there are 2 purple chips.

@franco a bit of correction
88000 is NOT 2^6 * 5 * 11
Rather
88000 is 2^6 * (5*5*5) * 11
hence 3 green chips

and 2^6 = 4^3 = 8^2 = 64
as no is between 5 & 11 , so only option left is 8

don't know any shorter approach , its just about factoring 88,000

Agreee.. I solved it by factoring .. solution identical to one stated above..

dumb.doofus wrote:Agreee.. I solved it by factoring .. solution identical to one stated above..

Thanks guys.. got it .... slightly different approach makes sense!