LUANDATO wrote: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

First recognize that n, n+1 and n+2 are 3 CONSECUTIVE INTEGERS.

Now let's make some observations:

When n = 1, we get: (1)(2)(3), which is NOT divisible by 8
n = 2, we get: (2)(3)(4), which is DIVISIBLE BY 8
n = 3, we get: (3)(4)(5), which is NOT divisible by 8
(4)(5)(6), which is DIVISIBLE BY 8
(5)(6)(7), which is NOT divisible by 8
(6)(7)(8), which is DIVISIBLE BY 8
(7)(8)(9), which is DIVISIBLE BY 8
(8)(9)(10), which is DIVISIBLE BY 8
-----------------------------
(9)(10)(11), which is NOT divisible by 8
(10)(11)(12), which is DIVISIBLE BY 8
(11)(12)(13), which is NOT divisible by 8
(12)(13)(14), which is DIVISIBLE BY 8
(13)(14)(15), which is NOT divisible by 8
(14)(15)(16), which is DIVISIBLE BY 8
(15)(16)(17), which is DIVISIBLE BY 8
(16)(17)(18)which is DIVISIBLE BY 8
-----------------------------
.
.
.
The pattern tells us that 5 out of every 8 products is divisible by 8.
So, [spoiler]5/8[/spoiler] of the 96 products will be divisible by 8.
This means that the probability is 5/8 that a given product will be divisible by 8.

Answer: D

Cheers,
Brent

LUANDATO wrote: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

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) 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, 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

Hi LUANDATO,

Brent's already listed out the pattern in the sequence, so I won't rehash any of that here. Instead, I'll focus on WHY that pattern exists.

For a number to be evenly divisible by 8, it has to include at least three 2's when you prime factor it.

For example,
8 is divisible by 8 because 8 = (2)(2)(2).....it has three 2s "in it"
48 is divisible by 8 because 48 = (3)(2)(2)(2)(2).....it has three 2s "in it" (and some other numbers too).

20 is NOT divisibly by 8 because 20 = (2)(2)(5)....it only has two 2s.

In this question, when you take the product of 3 CONSECUTIVE POSITIVE INTEGERS, you will either have....

(Even)(Odd)(Even)

or

(Odd)(Even)(Odd)

In the first option, you'll ALWAYS have three 2s. In the second option, you'll only have three 2s if the even term is a multiple of 8 (Brent's list proves both points). So for every 8 consecutive sets of possibilities, 4 of 4 from the first option and 1 of 4 from the second option will give us multiples of 8. That's 5/8 in total.

Final Answer: D

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