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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.

+2 votes
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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

posted Apr 1, 2015 by anonymous

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2 Solutions

+3 votes

Takin x,y,z as 1,8,3 respectively
We get minimum value as 1

solution Apr 2, 2015 by Jaspalsingh Parmar
+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

solution Aug 18, 2015 by anonymous
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