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For n(n+1) to be a multiple of 3, either n or n+1 must be a multiple of 3.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
GMATGuruNY wrote:For n(n+1) to be a multiple of 3, either n or n+1 must be a multiple of 3.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
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.
or can n be any positive integer?n is a positive integer between 1 and 99
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