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probability that n(n+1)(n+2) will be divisible by 8

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by GMATGuruNY » Wed Sep 28, 2016 8:14 am
If n is an integer from 1 to 96 (inclusive), what is the probability that n*(n+1)*(n+2) is divisible by 8?

A.1/4
B.1/2
C.5/8
D.7/8
E.3/4
n(n+1)(n+2) = the product of 3 consecutive integers.
WRITE IT OUT and LOOK FOR A PATTERN.

1*2*3
2*3*4
3*4*5
4*5*6
5*6*7
6*7*8
7*8*9
8*9*10

9*10*11
10*11*12
11*12*13
12*13*14
13*14*15
14*15*16
15*16*17
16*17*18

Each of the products in red is a multiple of 8.
The two examples above imply the following:
Of every 8 products, exactly 5 will be a multiple of 8.

Thus, the probability that n(n+1)(n+2) will be a multiple of 8 = 5/8.

The correct answer is C.

Alternate approach:

Case 1: n(n+1)(n+2) = even*odd*even = multiple of 8:
Since every other even integer is a multiple of 4, the product here will always include an even integer and a multiple of 4, resulting in a multiple of 8.
Thus, n can be any even integer between 1 and 96.
96/2 = 48 favorable choices for n.

Case 2: n+1 is a multiple of 8:
The product will be a multiple of 8 if n+1 is a multiple of 8.
Number of multiples of 8 between 1 and 96 = 96/8 = 12.
Thus, there are 12 favorable choices for n+1, implying that there are 12 more favorable choices for n.

Total favorable choices for n = 48+12 = 60.
Favorable choices/Total choices = 60/96 = 5/8.

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by [email protected] » Wed Sep 28, 2016 9:48 am
Hi Needgmat,

Mitch has 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 (Mitch'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.

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by Matt@VeritasPrep » Thu Sep 29, 2016 6:38 pm
You can also look for a pattern to get started, that might help you get a better feel for the problem.

Suppose n = 1. 1 * 2 * 3 doesn't work.

Suppose n = 2. 2 * 3 * 4 DOES work.

Suppose n = 3. 3 * 4 * 5 doesn't work.

Suppose n = 4. 4 * 5 * 6 DOES work.

So what's happening? As long as we get two even numbers (2 and 4, 4 and 6) we seem to be fine, so any time we have Even * Odd * Even, we should be set. (Don't worry about why this is happening on test day, just look for the pattern.)

Now we need to test for exceptions: are there numbers that work if we have Odd * Even * Odd? Well, yeah! If the Even number itself divides by 8!

So we need either n = even or (n + 1) = 8*whatever. There are 48 even n's and 12 numbers where (n + 1) = 8*whatever, so 60 numbers out of 96 will work, and our answer is 60/96, or 5/8.
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by shashank.ism » Sun Jan 15, 2017 9:45 am
Needgmat 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

OA D

Please explain


First we try to analyse the expression, n(n+1)(n+2)
Case1 : It is odd*even*odd
In this case, the expression will be divisible by 8 only if (n+1) is divisible by 8
i.e. (n+1) is 8,16,24.......96
The no. of such numbers is [96/8] = 12

Case 2: It is even*odd*even.. In such case it is always divisible by 8
The no. of such numbers is 96/2 = 48

So total numbers from 1 to 96, for which n(n+1)(n+2) is divisible by 8 = 48+12= 60

Hence, probability = 60/96 = 5/8.
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