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by lukaswelker » Thu Apr 17, 2014 8:47 am
Hey Guys

For this one I don't even know where to start.

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?

1/9 : 1/3 : 1/2 : 2/3 : 5/6

Any help will be more then welcome
Cheers
Lukas
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by theCodeToGMAT » Thu Apr 17, 2014 9:09 am
N = 1 to 99

(N)(N+1)/3

Possible N values = 2,3,5,6 ..... 98,99 = 66 Numbers

Probability = 66/99 = 2/3
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by GMATGuruNY » Thu Apr 17, 2014 12:16 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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by ceilidh.erickson » Fri Apr 18, 2014 7:16 am
Sometimes with probability, it's more helpful to think about the opposite scenario: when will n(n + 1) NOT be divisible by 3?

As Mitch said, 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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