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

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) (Show Source): You can put this solution on YOUR website!

We want to know when the expression

is an integer.

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.

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
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