X = 61 Y = 61 So x+y = 122
x, y, z and k are four non zero positive integers satisfying 1/x + y/2 = z/3 + 4/k, minimum integral value of k for integral value of x, y and z will be
An integer x is selected in such a way that
x + x^(1/2) + x^(1/4) = 276
Then what would be the sum of all digits of the number y where
y = 100000 x^(1/4) + 1000 x^(1/2) + x
If x + 1/x = 1 then x^(1729) + x^(-1729) = ?
If we know x + 1/x =2 then what would be the value of x^2048 + 1/x^2048 +x^2047 - 1 / x^2047 + 1/x^2049 - x^2049 +2
Note: Read carefully...