probability

This topic has expert replies
Legendary Member
Posts: 876
Joined: Thu Apr 10, 2008 8:14 am
Thanked: 13 times

probability

by ketkoag » Fri Apr 24, 2009 2:59 am
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

OA: D please explain how to solve it..
Source: — Problem Solving |

User avatar
Community Manager
Posts: 1049
Joined: Sun Apr 06, 2008 5:15 pm
Location: Pittsburgh, PA
Thanked: 113 times
Followed by:27 members
GMAT Score:710

by dmateer25 » Fri Apr 24, 2009 5:46 am
If n is an even it will be divisible by 8.

For example: 2 x 3 x 4 (has 2^3 in it, so it is divisible by 8)
4x5x6 (has 2^3 in it, so it is divisible by 8)


so that gives us 48 that are divisible by 8.

Now you have to look at when n is an odd number. When n is odd and the n+1 is divisible by 8, then these will be divisible by 8 as well.


so for example... 7 x8 x9..... 15x16x17.. These will all be divisible by 8 as well.

96/8 = 12


(48+12)/96 = 60/96 = 5/8