SOLUTION: If integer C is randomly selected from 20 to 99, inclusive, what is the probability that C^3-C is divisible by 12?

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Question 1133818: If integer C is randomly selected from 20 to 99, inclusive, what is the probability that C^3-C is divisible by 12?
Answer by greenestamps(13200) About Me  (Show Source):
You can put this solution on YOUR website!


We want to know when the expression

%28C%5E3-C%29%2F12

is an integer.

C%5E3-C+=+C%28C%5E2-1%29+=+C%28C%2B1%29%28C-1%29+=+%28C-1%29%28C%29%28C%2B1%29

So the expression is the product of three consecutive integers; we need to find the conditions under which the product of three consecutive integers is or is not divisible by 12.

12+=+%282%5E2%29%283%29

So the product of three consecutive integers will be divisible by 12 if it contains two factors of 2 and one factor of 3.

Every set of three consecutive integers contains exactly one which contains a factor of 3. So we need to determine when the product of three consecutive integers contains two factors of 2.

(1) If C is odd, then both C-1 and C+1 are even, so the product contains two factors of 2.
(2) If C is a multiple of 4, then that factor alone contains two factors of 2.
(3) If C is even but not a multiple of 4, then C-1 and C+1 are both odd; the product will contain only one factor of 2.

So only 1 out of every 4 consecutive values of C will yield a product that is NOT divisible by 12. So 3 out of every 4 WILL yield a product that is divisible by 12.

There are 80 integers from 20 to 99 inclusive; since that number is a multiple of 4, we know that exactly 3/4 of them will yield a product that is divisible by 12.

ANSWER: P(C^3-C is divisible by 12) = 3/4