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very hard number property, help, help, pls

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by tanviet » Thu Oct 29, 2009 5:29 am
this is from 15 testset.

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

a,1/4
b,3/8
c,1/2
d,5/8
e,3/4
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by NikolayZ » Thu Oct 29, 2009 5:45 am
Hey !

96-1+1=96 - is the quantity that represents all possibilities of n.

we have n(n+1)(n+2) (call it T) - consecutive integers!
btw, also notice that there are 12 multiples of 8 at 1-96.
For T to be divisible by 8, it is necessary that at least one of its factors be the multiple of 8.
let's figure it out.
for n- we have 12 possibilities
for (n+1) - also 12 possibilities
for (n+2) - 12 possibilities
if n is odd, such as 7, n(n+1) might be the multiple of 8. so, also 12 possibilities.
If N is even, n(n+2) also might be a multiple.

So, we have 12*5=60 , possible solutions for T in order to be divisible by 8.

The probability will then be - 60/96- or 10/16 or 5/8.
I hope i didn't mix smth up.
What is the OA?
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by mp2437 » Thu Oct 29, 2009 6:23 am
This has been solved a few days ago by a number of different methods, check the forum please. Answer is 5/8.
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