Antimail

First of all,  I thing there is a typo, n should be a positive integer since n=0 it doesn’t make sense and for n < 0 the equality is not true.

There is an integer k, 0 <= k < n such that
(1)          floor(x) + k/n <= x < floor(x) + (k+1)/n  .      
The intuitive solution is that if we divide the interval [ floor(x), floor(x)+1 ) into n subintervals, then we can find an subinterval that contains x - if somebody wants the rigorous proof I can provide it :) .


By multiplying n to both sides of the inequality (1) we have
(2)         n* floor(x) + k <= x * n < n * floor(x) + k + 1

Therefore,
(3)         floor(x * n) = n * floor(x) + k.

Let’s consider i an integer, 0 <= i < n – k .
Adding i/n to both sides of the inequality (1) gives us:
(4) floor(x) + k/n + i/n <= x + i/n < floor(x) + (k+1)/n + i / n  .

Since 0 <= i < n – k, and 0 <= k < n; therefore,
(5) floor(x) <= floor(x) + k/n + i/n  
and
(6) floor(x) + (k+1)/n + i / n  <= floor(x)  + 1.
Using (4) , (5) and (6) we have floor(x)  <= x + i/n <= floor(x)  + 1 which provides  
(7) floor(x + i/n) = floor(x).

Similarly , for the integer i, with n –k <= i < n, we have
(8) floor(x)  + 1 <= floor(x) + k/n + i/n <= x + i/n < floor(x) + (k+1)/n + i / n  <= floor(x) + 2, which provides

(9) floor(x + i/n) = floor(x) + 1.

Considering i=0 to n-1 into (7) and (9) we have
(10) floor(x) + floor(x + 1/n) + ... + floor(x + (n-1)/n) =
           (n – k) * floor(x) + k * (floor(x) + 1) = n * floor(x) + k.

Now, from (10) and (3) follows the equality required:
floor(x * n) = floor(x) + floor(x + 1/n) + floor(x + 2/n) + ... + floor(x + (n-1)/n)

Very nice and elegant solution!

Nice puzzle; I remember having it at a contest when I was in the 6th or 7th grade.

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