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
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
#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