Scoretop 4 - Q9 - Number properties/Probability

akay

I have no clue on this one...

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

beny · Posted: Tue Aug 28, 2007 8:14 pm Post subject:

These are three consecutive integers. For the product to be divisible by 8, at least two of these integers need to be even (the lowest combination is 2,3,4 and the highest combination is 96,97,98 ). Notice that as long as the first number (n) is even, you will have at least 2 even numbers out of the three, thus, the product is divisible by 8. Now you just need to know how many even integers there are between 1 and 96 inclusive (there are 48... you can either find this using counting methods or just know that the sequence starts odd, ends even, thus exactly half of the numbers must be even).

48/96 = 1/2

Answer is C.

givemeanid · Posted: Wed Aug 29, 2007 6:04 am Post subject:

When (n+1) is straight divisible by 8, the product will be divisible by 8 (n will be odd in this case). There are 96/8 = 12 such numbers.

So, total = 48 + 12 = 60 numbers.

Prob = 60/96 = 5/8