akshatgupta87 wrote:Q.) x is an integer between 1 and 999 inclusive. What is the probability that x(x+1) will be divisible by 6?
A) 1/3
B) 1/2
C) 3/5
D) 2/3
E) 5/6
Someone please explain.
OA is D
~Thanks,
Akshat
x and x+1 are consecutive integers.
Thus, one of them must be even.
To be a multiple of 6, x(x+1) must also be a multiple of 3.
Question rephrased:
What is the probability that either x or x+1 is a multiple of 3?
x, x+1, and x+2 are 3 consecutive integers.
Of every 3 consecutive integers, exactly ONE is a multiple of 3.
Thus, P(x+2 is a multiple of 3) = 1/3.
Thus, P(either x or x+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, x(x+1) is a multiple of 6.
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