praetorian123 wrote:

N is from 1 to 96 inclusive, what is the probability of N such that N(N+1)(N+2)is evenly divisible by

problem #1 . Eight

problem #2 . Four

problem #3 . Three

#1

EIGHT:

All "sets" of even N's are candidates

Example 2,3,4... 4,5,6 ... 6 7 8

There are 48 such "Sets"

Total possible sets = 96 (one with each N)

Probability = 48/96 =1/2 .... Is this correct.

With # 3 , i find that a set of 3 consecutive positive integers is always divisible by 3...is that right?

Can you please help me out with # 2 and # 3

Thanks

praetorian

#3 - Every 3rd number is a multiple of 3. So, one of N, (N+1), (N+2) is always divisible by 3. So, you are right. Every N satisfies that criteria. For this case, prob. should be 1.

#2 - Taking the least possible number for N i.e 1, N*(N+1)*(N+2) will be 6. For N <= 6, the following theory should hold.

As long as one of the numbers N, N+1, N+2 is divisible by N, the whole number i.e. N*(N+1)*(N+2) will be divisible by N.

I think one can follow this logic to see that 24 numbers between 1 and 96 are divisible by 4 and each such number can occur in 3 positions - N or (N+1) or (N+2) in this multiplication expression.

So, answer should be (24*3)/96 = 3/4

Amar.