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You might try the following approach. Think of vertex 2; it can't be joined to any odd numbered vertex so the evens are in components by themselves (as are the odds). So, look at the even points only and change the problem from the points 2,4, ..., 100 with edges determined by differences of 8 and 12 to a smaller problem of vertices labeled 1, 2, ..., 50 with edges determined by differences of 4 and 6. Iterate. Penny Amarjeet wrote back sir i am still not clear that either answer is 4 or 8. can you please help me out? Hi again, One cycle of length 25 is listed in my previous response. You can find another one that starts at vertex 2. then one starting at vertex 3, etc, until all of the vertices are in one of cycles you have. Try listing the Victoria  


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