If 1/x + y/2 = z/3 + 4/k then what woud be minimum integral value of k for integral value of x, y and z. 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
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1/x + y/2 = z/3 + 4/kk........
Rearranging we get,.........
6k + 3xy = 2kxz + 24xx.........
Again by rearranging we get,.........
k = 24x/6 + 3xy - 2xz........
Now, for minimum integer value of k, numerator of RHS to be minimum. That means, 24x = 1, or x = 1........
Now putting x = 1 in denominator, we get.....
6 + 3y - 2z = 24..........
Hence, y = 8 and, z = 3.....,.....
Thus, minimum possible value of k = 1.........
ANSWER, k = 1.. Read more at .
Rearranging we get,.........
6k + 3xy = 2kxz + 24xx.........
Again by rearranging we get,.........
k = 24x/6 + 3xy - 2xz........
Now, for minimum integer value of k, numerator of RHS to be minimum. That means, 24x = 1, or x = 1........
Now putting x = 1 in denominator, we get.....
6 + 3y - 2z = 24..........
Hence, y = 8 and, z = 3.....,.....
Thus, minimum possible value of k = 1.........
ANSWER, k = 1.. Read more at .
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