It's V V versus X X and Y Y

Consider 2 numbers X X and Y Y which are chosen at random from the set V = { 1 , 2 , , 3 n } V=\{1,2,\ldots,3n\} without replacement . If the probablity that X 3 + Y 3 X^{3}+Y^{3} is divisible by 3 can be written of the form n a n b + 2 c , \dfrac{n^{a}}{n^{b}+2^{c}},

where a , b , c Z a,b,c\in\mathbb{Z} and n N n\in\mathbb{N} , find ( a + c ) × b (a+c)\times b

If you feel that the probability cannot be expressed in this form then type 22 as your answer.


The answer is 0.

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1 solution

Milind Blaze
Apr 24, 2016

The entire set can be split up into 3 sets of numbers of the form 3 m + 1 , 3 m + 2 , 3 m 3m+1,3m+2, 3m ,Three numbers can be chosen in ( 3 n 2 ) {3n \choose 2} ways. Also for sum of the cubes of the nos. To be divisivle by three either the numbers must be chosen from the third set o r or one from set1 a n d and the other from set 2. This can be done in ( n 1 ) 2 {n \choose 1}^{2} + ( n 2 ) {n \choose 2} ways. Therefore the required probability is ( n 1 ) 2 + ( n 2 ) ( 3 n 2 ) \dfrac{{n \choose 1}^{2} +{n \choose 2}}{{3n \choose 2}} which simplifies to 1/3. Comparing we get a=0, b=0 and c=1. Hence the answer is 0.

Did the same. Thanks for posting

Vignesh S - 5 years, 1 month ago

Very nice solution @Milind Blaze . +1. :)

Mehul Arora - 5 years, 1 month ago

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