BTGModeratorVI wrote: ↑Wed Dec 16, 2020 12:51 pm
If an integer n is to be chosen at random from the integers 1 to 96, inclusive, what is the probability that n(n + 1)(n + 2) will be divisible by 8?
A. 1/4
B. 3/8
C. 1/2
D. 5/8
E. 3/4
Answer:
D
Source: official guide
Solution:
We are given that an integer n will be chosen at random from the integers 1 to 96, inclusive, and we need to determine the probability that n(n + 1)(n + 2) will be divisible by 8.
We should recall that when a number is divisible by 8, it is divisible by 2^3, i.e., three factors of 2. We should also recognize that n(n + 1)(n + 2) is the product of three consecutive integers.
Case 1: n is even. Any time that n is even, n + 2 will also be even. Moreover, either n or n + 2 will be divisible by 4, and thus n(n + 1)(n + 2) will contain at least three factors of 2 and will be divisible by 8.
Since there are 96 integers between 1 and 96, inclusive, and half of those integers are even, there are 48 even integers (i.e., 2, 4, 6, …, 96) from 1 to 96, inclusive. Thus, when n is even, there are 48 instances in which n(n + 1)(n + 2) will be divisible by 8.
Case 2: n is odd. If n is odd, then n(n + 1)(n + 2) still can be divisible by 8 if the factor (n + 1) (which is the only even number among n, n + 1 and n + 2) is a multiple of 8. So, let’s determine the number of multiples of 8 between 1 and 96 inclusive.
Number of multiples of 8 = (96 - 8)/8 + 1 = 88/8 + 1 = 12. Thus, when n is odd, there are 12 instances in which n(n + 1)(n + 2) will be divisible by 8.
There is no overlap for the results of Case 1 and Case 2, so we can simply add the two results. In total, then, there are 48 + 12 = 60 outcomes in which n(n + 1)(n + 2) will be divisible by 8.
Thus, the probability that n(n + 1)(n + 2) is divisible by 8 is: 60/96 = 10/16 = 5/8.
Answer: D
