> If you let modulus number N be larger than n (e.g. n = 5 in my example, just let N = 6. Then 10 + x = 0 mod 6 will yield only one possible solution that is 2)

> If you let modulus number N be larger than n (e.g. n = 5 in my example, just let N = 6. Then 10 + x = 0 mod 6 will yield only one possible solution that is 2) then I think your solution will definitely work. That was entirely what I was suggesting, but we don't need N>n. N=n would work. (btw, I defined n to be the size of the list -1, therefore in your example, the number is 4) > But I think the big problem is: you said x = (b - a) % n. That is true, but how do we get "a" (the sum of all unique numbers from 1 to n) without overflowing? The reason we do modulus summation is for avoiding overflow, granted that can get "b" easily, yet now we have to find "a" still, I find we are still stuck. Any suggestion? We can compute a using the same function that computes b. just sum all numbers from 1 to n mod n. Here is a quick demonstration in Haskell. http://lpaste.net/101868 http://lpaste.net/101868 > Even if assuming everything will work out and my previous question can be responded, still the modulus function is WAY MORE expensive to compute than xor operation. True. I just want to show how to modify another solution so there is no overflow.