Probability Ques,

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

It's probably easier to find the number of values, x, such that x*(x+1) is NOT divisible by 6. Since the product is gauranteed to be even, we simply find all pairs (x,x+1), such that neither is divisible by 3...

x mod 3 > 0 AND (x+1) mod 3 > 0

So, for each positive multiple of 3, there first comes a pair that satisfies the above constraints (ie. 1 and 2 preceed 3)

999/3 = 333 multiples of 3 between 1 and 999, inclusive.

Therefore, the number of values that ARE divisible by 6 is 1-1/3 = 2/3. D

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.