Divisible sum
Find the minimum number such that among any positive integers there will always be at least one pair with the sum or difference divisible by 2012.
The answer is 1007.
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1 solution
Consider the sets { 0 } , { 1 , 2 0 1 1 } , . . . , { 1 0 0 5 , 1 0 0 7 } , { 1 0 0 6 } . In mod 2 0 1 2 , we can choose at most 1 element from each set, giving an answer of 1 0 0 7 .