If integer C is randomly selected from 20 to 99, inclusive.

BTGmoderatorDC 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

OA C

Source: Manhattan Prep

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

BTGmoderatorDC 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

OA C

Source: Manhattan Prep

We have to determine whether C^3 – C is divisible by 12.

C^3 – C = C(C^2 – 1) = C(C – 1)(C + 1) = (C – 1)C(C + 1) = Product of three consecutive integers

This question can be solved taking few sample values of C from 20 to 99, inclusive. Note that from 20 to 99, inclusive, there are 80 integers -- 40 even and 40 odd.

1. Say C = 20; thus, (C – 1)C(C + 1) = 19*20*21 = 19*4*5*3*7 = Multiple of 12;

2. Say C = 21; thus, (C – 1)C(C + 1) = 20*21*22 = 4*5*3*7*2*11 = Multiple of 12;

3. Say C = 22; thus, (C – 1)C(C + 1) = 21*22*23 = 3*7*2*11*23 = Multiple of 6, not 12;

4. Say C = 23; thus, (C – 1)C(C + 1) = 22*23*24 = 2*11**23*24 = Multiple of 12;

If you take another set of four values of C: {23, 24, 25, 26}, you would find that three values would ensure that C^3 – C is divisible by 12 and one would not. So out of 80 integer values of C, 3/4th values would ensure that C^3 – C is divisible by 12 and 1/4th would not.

The correct answer: C

Hope this helps!

-Jay
_________________
Manhattan Review GMAT Prep

Locations: GRE Manhattan | ACT Tutoring Tampa | GRE Prep Courses Seattle | Boston IELTS Tutoring | and many more...

Schedule your free consultation with an experienced GMAT Prep Advisor! Click here.