gmatnmein2010 wrote: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

If the number divided by 8 leaves a remainder of

a) 1

n is odd, so is n+2; n+1 is not divisible by 8 => n(n + 1)(n + 2) is not divisible by 8

b) 2
n is even; so is n+2; n+2 is a multiple of 4 => n(n + 1)(n + 2) is divisible by 8

c) 3
n is odd, so is n+2; n+1 is not divisible by 8 => n(n + 1)(n + 2) is not divisible by 8

d) 4
n is even; so is n+2; n is a multiple of 4 => n(n + 1)(n + 2) is divisible by 8

e) 5
n is odd, so is n+2; n+1 is not divisible by 8 => n(n + 1)(n + 2) is not divisible by 8

f) 6
n is even; so is n+2; n+2 is a multiple of 8 => n(n + 1)(n + 2) is divisible by 8

g) 7
n is odd, so is n+2; n+1 is divisible by 8 => n(n + 1)(n + 2) is divisible by 8

f) 8
n is even; so is n+2; n is a multiple of 8 => n(n + 1)(n + 2) is divisible by 8

Now since 96 is a multiple of 8 there is an equal probability of one these cases happening

Probability = 5/8

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.