Question 4067
 Let the median(3rd number) of the five consecutive numbers be n, 
 then the 5 numbers must be n-2,n-1,n,n+1,n+2.
 
 We see that their sum S= n-2 + n-1 +n + n+1+ n+2 = 
  n-2 + + n+2 + n-1 + n+1 + n = 5n 

 In other words, 5 is a factor of the sum S. 
 This completes the proof.

 In general,for 2k+1 (odd) consecutive numbers, their sum must be
 2k+1 times of the median.

 How about the sum of even consecutive numbers?

 Kenny