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the value f n

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by [email protected] » Mon Mar 07, 2016 10:21 pm
Hi didieravoaka,

There's a nice discussion of this question here:

https://www.beatthegmat.com/probability- ... 76490.html

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by Matt@VeritasPrep » Thu Mar 17, 2016 9:14 pm
You need either n to be divisible by 3 or (n + 1) to be divisible by 3; in other words, you need n to have remainder 0 or remainder 2 when divided by 3.

If we consider all the integers (negative, positive, 0, whatever), exactly (1/3) of them have remainder 0 when divided by 3 and exactly (1/3) of them have remainder 2. So (2/3) of all integers fit our criteria.

In this case, the same idea holds: we have 99 integers, (1/3) of which have remainder 0 and (1/3) of which have remainder 2, so (2/3) of our numbers do the trick.
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