If an integer n is to be chosen at random from the integers 1 to 96, inclusive, what is the probability that n(n + 1)(n + 2) will be divisible by 8?	Choose the correct choice from the following
A) 1/4	B) 3/8	C) 1/2	D) 5/8	E) 3/4
total nos of ways in which we can choose n = 96   n(n + 1)(n + 2) will be divisible by 8?   case 1: n = odd then n+2 =odd &amp; n+1 will be even i.e this needs get divided by 8, hence is a multiple of 8 so we have 8..96 = 12 multiples to fill the n+1 pos hence 12 ways   case 2: n is even then n+2 will be even &amp; the product will be divisible by 24 &amp; thus 8   so nos of values that can be used for n= 2....96 (all even nos) i.e 48 nos   total = 48+12= 
<gadget id="calculator">48+12</gadget>
<output>60</output>
60 ways   so reqd P=60/96 =5/8;   ANSWER:5/8
<result>D</result>
If an integer n is to be chosen at random from the integers 1 to 96, inclusive, what is the probability that n(n + 1)(n + 2) will be divisible by 8?