Question 4067: Show that the sum of any 5 consecutive counting numbers must have a factor of 5
Answer by khwang(438)   (Show Source):
You can put this solution on YOUR website! 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
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