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GMAT- Need your Guidance

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by MartyMurray » Sun Feb 15, 2015 6:10 am
This one has been discussed extensively on these forums. Here are the results of a search of BeattheGMAT.

https://www.beatthegmat.com/mba/search-r ... 96&x=0&y=0
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by Brent@GMATPrepNow » Sun Feb 15, 2015 7:36 am
If an integer n to be chosen randomly between 1 and 96 inclusive, what is the probability that n(n+1)(n+2) is divisible by 8 ?
A. 1/4
B. 3/8
C. 1/2
D. 5/8
E. 3/4

OA: D

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

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by [email protected] » Sun Feb 15, 2015 11:51 am
Hi Abhijit K,

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.

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by Matt@VeritasPrep » Sun Feb 15, 2015 5:35 pm
Abhijit K wrote:If an integer n is 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
Any time you have a divisibility question, think of it in terms of the factors you need. Since we need (n)(n+1)(n+2) to divide by 8, we need three factors of 2 SOMEWHERE. We can do this in a few ways:

1:: Every number is even
Here this is impossible: we can't have three consecutive integers all be even.

2:: One number is even and another is divisible by 4
This is possible: if n is even, then n and (n+2) are consecutive evens (e.g. 2 and 4, 4 and 6, 6 and 8, etc.) One of these will divide by 4, and the other will divide by 2: success!

3:: One number is divisible by 8

(2) and (3) are both possible here. If n is even, we'll satisfy (2). The only other possibility is if n is odd, but (n+1) is a multiple of 8. So the REAL question is
n is an integer from 1 to 96, inclusive. What's the probability that n is either even or one less than a multiple of 8?
... and that's a much easier question to answer :D
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by Gurpreet singh » Sun Jun 19, 2016 9:36 pm
Hi Brent,
with time constraint on GMAT can we have a easier method for solving such questions
Regards
Brent@GMATPrepNow wrote:
If an integer n to be chosen randomly between 1 and 96 inclusive, what is the probability that n(n+1)(n+2) is divisible by 8 ?
A. 1/4
B. 3/8
C. 1/2
D. 5/8
E. 3/4

OA: D

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
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by [email protected] » Mon Jun 20, 2016 8:44 am
Hi Gurpreet singh,

If you read through the other posts in this thread, you'll see that there are Number Property rules that can help you to answer this question faster. It's also worth remembering that certain questions on Test Day will take longer than average to solve (that's one of the reasons why an 'average' is an average), so you can expect to see a few questions on the Official GMAT that will take upwards of 3 minutes of work to correctly answer.

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by DavidG@VeritasPrep » Mon Jun 20, 2016 9:54 am
Abhijit K wrote:If an integer n is 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
Divisibility is a common topic. See here for some more examples:

https://www.beatthegmat.com/n-1-n-1-is-d ... 34359.html

Or

https://www.beatthegmat.com/positive-int ... 89211.html
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