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

+2 votes

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

+1 vote

1/x + y/2 = z/3 + 4/k

Rearranging we get,

6k + 3xy = 2kxz + 24x

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

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