Bunuel wrote:
If an integer n is chosen from the integers 1 through 72, 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
Are You Up For the Challenge: 700 Level QuestionsWe are given that an integer n is to be chosen at random from the integers 1 to 72, 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 three factors of 2 and will be divisible by 8.
Since there are 72 integers between 1 and 72, inclusive, and half of those integers are even, there are 36 even integers (i.e., 2, 4, 6, …, 72) from 1 to 72, inclusive. Thus, when n is even, there are 36 instances in which n(n + 1)(n + 2) will be divisible by 8.
Case 2: n is odd
f n is odd, then n(n + 1)(n + 2) still can be divisible by 8, but only if the even factor (n + 1) is a multiple of 8. So, let’s determine the number of multiples of 8 between 1 and 72, inclusive.
Number of multiples of 8 = (72 - 8)/8 + 1 = 64/8 + 1 = 9. Thus, when n is odd, there are 9 instances in which n(n + 1)(n + 2) will be divisible by 8.
In total, there are 36 + 9 = 45 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: 45/72 = 5/8.
Answer: D
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