IIT JEE , AIEEE Forum - Mathematics , Algebra

ariyam66

The numbers x,y are chosen at random from 1,2, ................, 3n.

What is the probability that x^3 + y^3 is divisible by 3 ??

shyamkumar

Hi
The possibilities are
1.x congruent to 0(mod 3) , y congruent to 0(mod 3) (for n>=2)
2.x congruent to 1(mod 3) , y congruent to 2(mod 3)
Case 1
Favourable outcomes= nC2
Case 2
Favourable outcomes=n^2

Hence the probability= (nC2+n^2) / (3n)C2 for n>=2....for n=1 it is n^2 / (3n)C2 as the first case is not possible
Do note that i have considered 1^3 + 2^3 and 2^3 + 1^3 as the same (which has to be the case as we just need to select nos) and also x not equal to y...
Hope it is right...
Shyam

magiclko

is the answer 1/3 ???

titun

Hii,

Magico is right. The answer is 1/3.

x^3 + y^3 = (x+y)^3 - 3xy(x+y) is divisible by 3 only when (x+y)^3 is divisible by 3 i.e when x+y is divisible by 3.

Probability of x^3 + y^3 being divisible by 3

= probability of x+y being divisible by 3

= 1/3

( since, there can be only three equal no. of cases

1. x+y = 3k

2. x+y = 3k +1

3. x+y = 3k+2 )

Hence, the answer is 1/3

Cheers !!!!

magiclko

yeah, its the same thing wat i hav done, moreover we hav one more approach

shyamkumar

well if you look at my reply carefully, it turns out to be 1/3 for all values of n !!!....
Cheers!!!
Shyam

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