Easy
x + y + z = 0 x 2 + y 2 + z 2 = 6 x 3 + y 3 + z 3 = − 6 . x 2 0 1 6 + y 2 0 1 6 + z 2 0 1 6 = 2 B + C , where 0 ≤ C < 2 B and B is a positive integer.
Find 2 + B + C .
The answer is 2020.
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2 solutions
Yes, answer can be both 2016 and 2020.
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ans can be a 1014 too as a=4 b=1008 and c=2 and there are so many such possibilities though i got to understand that why it has been reported by multiple users and got to know the possible mistake but you should accept that it is wrongly stated.
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Thanks. I have edited the problem for clarity. Those who answered 2020 and 1014 have been marked correct. The correct answer is now 2020.
In future, if you spot an error with the problem, please report it so that we can take the necessary action.
Let x 3 + y 3 + z 3 = − ( x 2 + y 2 + z 2 ) + ( x + y + z ) ⇒ ( x 3 + x 2 − x ) + ( y 3 + y 2 − y ) + ( z 3 + z 2 − z ) = 0 . If we now add − 3 to both sides, we now obtain:
( x 3 + x 2 − x − 1 ) + ( y 3 + y 2 − y − 1 ) + ( z 3 + z 2 − z − 1 ) = − 3 ;
or [ x 2 ( x + 1 ) − ( x + 1 ) ] + [ y 2 ( y + 1 ) − ( y + 1 ) ] + [ z 2 ( z + 1 ) − ( z + 1 ) ] = − 3 ;
or ( x 2 − 1 ) ( x + 1 ) + ( y 2 − 1 ) ( y + 1 ) + ( z 2 − 1 ) ( z + 1 ) = − 3 ;
or ( x − 1 ) ( x + 1 ) 2 + ( y − 1 ) ( y + 1 ) 2 + ( z − 1 ) ( z + 1 ) 2 = − 3
which has triplet solutions ( x , y , z ) = ( − 2 , 1 , 1 ) ; ( 1 , − 2 , 1 ) ; ( 1 , 1 , − 2 ) , and x 2 0 1 6 + y 2 0 1 6 + z 2 0 1 6 = 2 + 2 2 0 1 6 ⇒ B = 2 0 1 6 , C = 2 . Hence, 2 + B + C = 2 + 2 0 1 6 + 2 = 2 0 2 0 .
Solve the three equations (x,y,z)=(1,1,-2) in any order Which is (1)^(2016)+(1)^(2016)+(-2)^(2016) =(-2)^(2016)+2 A=-2,B=2016 and C=2 Hence the answer is 2016 But the important thing to note here is that (-2)^(2016)=(2)^(2016) So A=±2 Hence 2020 can be another answer