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?
A) 1/4
B) 3/8
C) 1/2
D) 5/8
E) 3/4
For this question we start with two conditions for n(n+1)(n+2)
1st condition: if n is odd,
so n+1 is even and n+2 is odd so n(n+1)(n+2) is divisble by 8 only if n+1 is multiple of 8,
total no. of such values = 12(8,16,.....96).
2nd conditon: if n is even,so n+1 is ODD and n+2 is EVEN
so either n is divisble by 4 or not divisible by 4.
i) if n is divisible by 4, n+2 is surely divisible by 2 as it is even. (e.g. 4,5,6)
ii) if n is divisble by 2 and not by 4, then also n+2 is surely divisble by 4 (e.g. 6,7,8)
thus for all n for 2nd condition is true ..hence total no. of such values = 96/2 = 48
so total no. of n(n+1)(n+2) divisible by 8 = 48+12=60
Hence, probability = 60/98 = 5/8. so ans is D.
