Problem solving - 13

If n is an integer from 1 to 96, what is the probability for n*(n+1)*(n+2) being divisible by 8?

A) 25%
B) 50%
C) 62.5%
D) 72.5%
E) 75%

Answer - C
n is even - anytime n is even, it is divisible by 8
total nos using sequence theorm 96 = 2 + (#-1) 2,
hence # = 48
n is odd - again 48 no
but in 1-8, only one combination is divisible by 8, when n is 7,15, 22... hence 12 cases
probalility = possible outcomes / total outcomes = 48+12 / 96 = 60/96 = .625 = 62.5%

OR

we need to check for what values of n, n(n+1)(n+2) is divisible 8....
n(n+2) is divisible by 8 for all values of even numbers and there are 48 even nos in 1 to 96....
now the remaining part (n+1) is divisible by 8 for 12 odd numbers....such as 7, 15, 23, 31, 39.... to find this u can divide 96 by 8....
so totally 48 + 12 = 60 numbers are there between 1 to 96 for which n(n+1)(n+2) is divisible 8....
now when calculate the % it would be ( 60 / 96 ) * 100 = 62.5 %....