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Nice little problem !

Call the consecutive integers ' x ' and ' x+1 '.

The sum is [ x + 2(x+1) ] = A

x + 2x + 2 = A

3x + 2 = A

Subtract 2 from each side:

3x = A - 2

Divide each side by 3:

x = (A-2)/3 and that's the smaller of the two integers.

The larger one is (x+1) = (A+1)/3

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