If an integer n is to be chosen at random from the integers 1 to 88, inclusive, what is the probability that n(n + 1)(n + 2) will be divisible by 8? Select the correct option:	A) 1/4	B) 3/8	C) 1/2	D) 5/8	E) 33/44
n(n+1)(n+2) will be divisible by 8 when n is a multiple of 2 or when (n+1) is a multiple of 8.   Thus when n is even, this whole expression will be divisible by 8.   from 1 to 88, there are 44 even integers.   Now when (n+1) is multiple by 8, we have 12 such values for (n+1)   probability that n(n+1)(n+2) will be divisible by 8   = (44 + 12)/88   = 66/88   = 33/44   Ans is 33/44
<result>E</result>
If an integer n is to be chosen at random from the integers 1 to 88, inclusive, what is the probability that n(n + 1)(n + 2) will be divisible by 8?