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Probability that n(n+1) will be divisible by 3

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by gmattesttaker2 » Mon May 19, 2014 7:14 pm
Hello,

Can you please assist with this:

An integer n between 1 and 99, inclusive, is to be chosen at random. What is the probability that
n(n+1) will be divisible by 3?

A) 1/9
B) 1/3
C) 1/2
D) 2/3
E) 5/6

OA: D

Thanks a lot,
Sri
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by GMATGuruNY » Mon May 19, 2014 8:07 pm
If n is a positive integer between 1 and 99, inclusive, what is the probability that n(n+1) is a multiple of 3?

1) 1/4
2) 1/3
3) 1/2
4) 2/3
5) 5/6
For n(n+1) to be a multiple of 3, either n or n+1 must be a multiple of 3.

n, n+1, and n+2 are 3 consecutive integers.
Of every 3 consecutive integers, exactly ONE is a multiple of 3.
Thus, P(n+2 is a multiple of 3) = 1/3.
Thus, P(either n or n+1 is a multiple of 3) = 2/3.

The correct answer is D.

Another approach is to WRITE IT OUT and LOOK FOR A PATTERN:
1*2
2*3
3*4
4*5
5*6
6*7
7*8
8*9
9*10
10*11
11*12
12*13
And so on.

The products in red show that, in 2 of every 3 cases, n(n+1) is a multiple of 3.
Thus, the probability that n(n+1) is a multiple of 3 = 2/3.
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by ceilidh.erickson » Tue May 20, 2014 8:41 am
Sometimes with probability, it's more helpful to think about the opposite scenario: when will n(n + 1) NOT be divisible by 3?

n(n + 1) is classic GMAT code for the product of 2 consecutive integers. So when will the product of 2 consecutive integers NOT be divisible by 3?

We know that every product of three consecutive integers n(n + 1)(n + 2) is divisible by 3, because with any 3 numbers in a row, one of them has to be a multiple of 3. The only way for n(n + 1) NOT to be divisible by 3 is if (n + 2) is the multiple of 3 in the set. The probability of that is 1 out of 3, so the probability of that NOT happening is 2 out of 3.
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by ceilidh.erickson » Tue May 20, 2014 8:42 am
It's important to recognize common GMAT structures, like this one for the product of consecutive integers. For more on consecutive products, see:
https://www.beatthegmat.com/is-x-x-2-x-4 ... tml#718646
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by gmattesttaker2 » Tue May 20, 2014 4:45 pm
GMATGuruNY wrote:
If n is a positive integer between 1 and 99, inclusive, what is the probability that n(n+1) is a multiple of 3?

1) 1/4
2) 1/3
3) 1/2
4) 2/3
5) 5/6
For n(n+1) to be a multiple of 3, either n or n+1 must be a multiple of 3.

n, n+1, and n+2 are 3 consecutive integers.
Of every 3 consecutive integers, exactly ONE is a multiple of 3.
Thus, P(n+2 is a multiple of 3) = 1/3.
Thus, P(either n or n+1 is a multiple of 3) = 2/3.

The correct answer is D.

Another approach is to WRITE IT OUT and LOOK FOR A PATTERN:
1*2
2*3
3*4
4*5
5*6
6*7
7*8
8*9
9*10
10*11
11*12
12*13
And so on.

The products in red show that, in 2 of every 3 cases, n(n+1) is a multiple of 3.
Thus, the probability that n(n+1) is a multiple of 3 = 2/3.

Hello Mitch,

Thanks for the explanation. I was just wondering if this information is important here:
n is a positive integer between 1 and 99
or can n be any positive integer?

Thanks for all your help.


Best Regards,
Sri
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by abhasjha » Tue May 20, 2014 11:49 pm
Alternatively :

the required number = (if n is even )+ (if n is divisible by 3) - (if n is even and divisible by 3)=
=(49+33-16)/99=66/99=2/3
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by theCodeToGMAT » Wed May 21, 2014 2:08 am
(n)(n+1)/3

Lets assume that "n" is divisible by 3.. So, 3,6,9....9 ==> 33 numbers

Lets assume that "n+1" is divisible by 3.. So, n=2,5,8....98 ==> 33 numbers

Total = 66/99 = 2/3
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