salopian wrote:
Bunuel DensetsuNo VeritasKarishma I have a question about applying a method from another question to this question here (
https://gmatclub.com/forum/if-an-intege ... 22078.html)
DensetsuNo gave the following quick solution for the question below:
Quote:
3-sec solution:
If an integer n is to be selected at random from 1 to 100, inclusive, what is probability n(n+1) will be divisible by 4?
Sequence of 4 numbers: 4
Numbers we have: 2
Answer: 1-2/4 = 1/2
You can see how this method works also on harder questions letting us solve them in a few seconds.
eg. 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?
Sequence of 8 numbers: 8
Numbers we have: 3
Answer 1-3/8= 5/8.
I applied the same methodology to this question and got 1-2/3 = 1/3 as the answer. Do you mind telling me where I got wrong in the application of the method as I am getting confused here. I do understand why 2/3 is the correct answer, but just want to know what is wrong with the method outlined above too.
Many thanks!
All such "methods" come with a long list of constraints which must be satisfied to apply them making them pretty much useless.
Use logic:
You have 99 integers: 1-2-3... 4-5-6 ... 7-8-9 ... 97-98-99
They form perfect sets of 3.
If you take n = 1, n(n+1) will not be divisible by 3.
If you take n = 2, n(n+1) will be divisible by 3.
If you take n = 3, n(n+1) will be divisible by 3.
Same logic will apply to each of the sets of 3 numbers. For exactly 2 numbers, the product will be divisible by 3. That is why you get the answer 2/3.
Note that you could be given a range of 1-100 numbers.
Then you have 33 sets of 3 numbers each and an extra 100 at the end which if picked will not lead to a product divisible by 3.
So out of 100 numbers, there would be 34 numbers which if picked, will not lead to a product divisible by 3.
In that case, probability of a product divisible by 3 will be 66/100.