If a number K is randomly chosen from the integers 20 to 9900099

Question

If a number K is randomly chosen from the integers 20 to 9900099 (both inclusive), what is the probability that K3 - K is divisible by 12?
a)3/12
b)4/12
c)3/20
d)3/4
e)None of these

eliaslatour · Answer

Well, I'm going to start by saying that I don't 100% understand the question. However, I'm going to make the assumption that K3 = K cubed or K*K*K

So what is K^3-K? It seems to me that it is K(K^2-1) and, K^2-1 is (k+1)(k-1) so finally the question is whether:

(K-1)(K)(K+1) is divisible by 12.

We are assuming that some kind of pattern will repeat every 12 or so numbers so where does that leave us?

23, 24, and 25 will all work because they will include the number 24. So the number must be at least 3/12. Additionally, every odd number divisible by 3 will work as it will be surrounded by two even numbers, generating 4*3 and those numbers occur every 6 numbers so we're up to 5/12 and numbers just above and below 18 will work (17 and 19 both work) so that gets me up to at least 7/12. So I can bring the choices down to D vs. E. I'm likely to guess D and get it right.

Not bad for the verbal guy. Let's see whether someone else can do better.

vitaliyGMAT · Answer

The hardest issue with this question is actually division.

We have K^3 - K = (K-1)K(K+1)

Above expression will be divisible by 4 (3 consecutive numbers will definitely be divisible by 3) only in two cases:

1) when K-1 will be even ---> K - odd

2) when K is a multiple of 4

Finding number of multiples of 4 and odd integers we'll geet fraction:

(4950040 + 2475020) / 9900080 = 7425060 / 9900080.

After a long time spent on factorisation we'll get 3/4. (Both numbers include primes 47 and 2633).

I guess that approximation will be most appropriate here. 75/99 is approximately 0.75 = 3/4. In case we believe that those numbers can be simplified at all.

If a number K is randomly chosen from the integers 20 to 9900099

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