Onthemove wrote:
Hi Bunue/ Raihanuddin
Can you use the same theory for n(n+1) divisible by 8. For example take (1,2,3,4,5,6,7,8), only 7 and 8 work so prob is 2/8 = 1/4?
Also how would you do it or even n(n+1)(n-1) divisible by 8 ? Is there a general rule/trick, might help a lot to reduce time spent on these questions.
If it is by 8, then you have to do the above plus an additional step.
if n = 97, n(n+1) not divisible by 8
if n= 98 still not divisible by 8 and
if n= 99 also not divisible by 8
Now notice that from 1 to 96, total 24 values of n is divisible by 8. We have 3 more values as shown above. So out of 99 values 24 are divisible.
Probability = 24/99 = 8/33
You may think that where I have got 96. from 1 to 99 we can't divide 99 by 8. The highest number we can divide is 96. So we have three more numbers left. That's why we have to do the additional step. We are using cycle of (1-8), (9-16) (17-24) ..........
You can see that 2 numbers from (1-8) are divisible by 8 if you consider n(n+1). Similarly 2 numbers from (9-16) and so on
How many such pairs do we have from 1-99.? 12 pairs of 8. So 2 numbers from each pair are divisible. Therefore total 2*12= 24 are divisible out of 96 numbers. And from remaining number none is divisible by 8. So, out of 99 numbers 24 are divisible.
You can do the next question similarly.
Very importantI think in official question you will not get such number as 8 by which you can't divide last number, which is 99. So, don't worry for that. The given question in the main post is an official question and I have always got that GMAT gives you last number divisible by the divisor.