GMATmission wrote:

If n is an integer from 1 to 96, what is the probability for n*(n+1)*(n+2) being divisible by 8?

a. 25%

b. 50%

c. 62.5%

d. 72.5%

e. 75%

There is another way I approach such questions if I am short of time (otherwise I prefer the logical approach given above) - Brute Force/Pattern Recognition/Intuition - whatever you may want to call it.

We need to find the numbers in which the product is a multiple of 8. I know I get a multiple of 8 after every 8 numbers. I will also get an 8 when I multiply 4 by an even number. In first 8 numbers, I have exactly two multiples of 4.

Basically, I figure that I should look at the first 8 cases. In all other cases, the pattern will be repeated. It helps that n can be from 1 to 96 i.e. a multiple of 8:

1*2*3 N

2*3*4 Y

3*4*5 N

4*5*6 Y

5*6*7 N

6*7*8 Y

7*8*9 Y

8*9*10 Y

5 of the first 8 products are divisible by 8 so my answer would be 5/8 = 62.5%

_________________

Karishma

Veritas Prep GMAT Instructor

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