I will try to give one solution for this vaguely... Correct me if Im
wrong.  But, the solution suffers overflow problem if implemented.

Assume that the number which got replaced is "x" and the number which
replaced it is "y"

obtain the sum of all the numbers, say Sum
so, Sum - y + x = n(n+1) / 2          eq1

obtain the sum of the squares of all the numbers, say sqSum
so, sqSum - sqr(y) + sqr(x) = n(n+1)(2n+1) / 6        eq2

Solving the above two equations will give the actual replaced number ;


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