Problem Solving

Hi SaraLotfy,

Most of your logic is right-on, and you're close to solving this question. I'm going to give you a nudge and then you can reattempt it.

C^3 - C = C(C^2 - 1)

BUT you're not done yet; you can go one step further with (C^2 - 1)....

C(C^2 - 1) = C(C + 1)(C - 1)

Now, using the logic you presented before, what would you do next?...

GMAT assassins aren't born, they're made,
Rich

RBBmba@2014 wrote:If integer C is randomly selected from 20 to 99, inclusive. What is the probability that C^3 - C is divisible by 12 ?

A. 1/2
B. 2/3
C. 3/4
D. 4/5
E. 1/3

We are given that an integer C is to be chosen at random from the integers 20 to 99 inclusive, and we need to determine the probability that C3 - C will be divisible by 12.

We should recall that when a number is divisible by 12, it is divisible by 4 and 3. We should also recognize that C3 - C = C(C2 - 1) = C(C - 1)(C + 1) = (C - 1)(C)(C + 1) is a product of three consecutive integers. Furthermore, we should recognize that any product of three consecutive integers is divisible by 3! = 6; thus, it's divisible by 3. We have to make sure it's also divisible by 4.

Case 1: C is odd

If C is odd, then both C - 1 and C + 1 will be even. Moreover, either C - 1 or C + 1 will be divisible by 4. Since C(C - 1)(C + 1) is already divisible by 3 and now we know it's also divisible by 4, C(C - 1)(C + 1) will be divisible by 12.

Since there are 80 integers between 20 and 99 inclusive, and half of those integers are odd, there are 40 odd integers (i.e., 21, 23, 25, ..., 99) from 20 to 99 inclusive. Thus, when C is odd, there are 40 instances in which C(C - 1)(C + 1) will be divisible by 12.

Case 2: C is even

If C is even, the both C - 1 and C + 1 will be odd. If C is a multiple of 4 but not a multiple of 3, then either C - 1 or C + 1 will be divisible by 3 (for example, if C = 28, then C - 1 = 27 is divisible by 3, and if C = 44, then C + 1 = 45 is divisible by 3). In this case, C(C - 1)(C + 1) will be divisible by 12.

If C is a multiple 4 and also a multiple of 3, then C is a multiple of 12 and of course C(C - 1)(C + 1) will be divisible by 12. Therefore, if C is even and a multiple of 4, then C(C - 1)(C + 1) will be divisible by 12. So, let's determine the number of multiples of 4 between 20 and 99 inclusive.

Number of multiples of 4 = (96 - 20)/4 + 1 = 76/4 + 1 = 20. Thus, when C is even, there are 20 instances in which C(C - 1)(C + 1) will be divisible by 12.

In total, there are 40 + 20 = 60 outcomes in which C(C - 1)(C + 1) will be divisible by 12.

Thus, the probability that C(C - 1)(C + 1) will be divisible by 12 is: 60/80 = 6/8 = 3/4.

Answer: C