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Probability that n(n + 1) is a multiple of 3

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Probability that n(n + 1) is a multiple of 3

by gmattesttaker2 » Mon Feb 17, 2014 6:11 pm
Hello,

Can you please assist with this:

If n is a positive integer between 1 and 99, inclusive, what is the probability that
n(n + 1) is a multiple of 3?

(A) 1/4
(B) 1/3
(C) 1/2
(D) 2/3
(E) 5/6


OA: D


I tried to solve this as follows:

n = 1 => n(n+1) = 1(1+1) - Not a multiple of 3
n = 2 => n(n+1) = 2(2+1) - Is a multiple of 3
n = 3 => 3(3+1) - Is a multiple of 3
n = 4 => 4(4+1) - Not a multiple of 3

However, I was not sure how to get the correct answer from here. Thanks for your help.

Best Regards,
Sri
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by Brent@GMATPrepNow » Mon Feb 17, 2014 6:24 pm
gmattesttaker2 wrote:Hello,

Can you please assist with this:

If n is a positive integer between 1 and 99, inclusive, what is the probability that
n(n + 1) is a multiple of 3?

(A) 1/4
(B) 1/3
(C) 1/2
(D) 2/3
(E) 5/6
n and (n + 1) are consecutive integers.

The product n(n + 1) will be a multiple of 3 if n is a multiple of 3 OR (n + 1) is a multiple of 3

So, P(the product is a multiple of 3) = P(n is a multiple of 3 OR n+1 is a multiple of 3)
= P(n is a multiple of 3) + P(n+1 is a multiple of 3)
= 1/3 + 1/3
= [spoiler]2/3[/spoiler]
= D

Aside: Every 3rd integer from 1 to 99 inclusive is a multiple of 3. So, P(n is a multiple of 3) = 1/3
Likewise, P(n+1 is a multiple of 3) = 1/3

Cheers,
Brent
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by [email protected] » Mon Feb 17, 2014 9:12 pm
Hi Sri,

Your idea to decipher the pattern behind the sequence was a good option. Sequence prompts on the GMAT will follow type of pattern; your ability to figure out that pattern might be essential to the question (such as this question) or helpful in finding an easy approach to answering whatever specific question that the prompt asks for.

You figured out the first 4 numbers in the sequence...

N=1; 1(1+1) is not a multiple of 3
N=2; 2(2+1) IS a multiple of 3
N=3; 3(3+1) iS a multiple of 3
N=4; 4(4+1) is not a multiple of 3

Now go just a little further....
N=5; 5(5+1) IS a multiple of 3
N=6; 6(6+1) IS a multiple of 3

Now we can see the sequence: NOT, IS, IS, NOT, IS, IS

So, 2 out of every 3 numbers in the sequence will give you a multiple of 3. This pattern repeats. Working up to 99, we would have 33 "sets" of NOT/IS/IS.

It will clearly be 2/3.

Final Answer: D

Keep this logic/approach in mind, especially if you don't immediately see the pattern behind a question. There's NO time to stare on Test Day; you're better served brute-forcing your way through this question and discovering the pattern than hoping that it will just "come to you."

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by GMATGuruNY » Mon Feb 17, 2014 10:05 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.
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