- #1
ruud
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I'm getting confused and can't seem to wrap my head around this problem. Prove that the sum of the squares of any 3 consecutive odd numbers when divided by 12 gives a remainder of 11.
I'm not sure how to set this up or proceed I figured that
(n^2 + (n +2)^2 + (n+4)^2)/12 = x + 11
Where n is any odd integer
then I got
3n^2 +12n + 20 = 12x +123
3n(n+4) = 12x + 103
I'm not sure where to take it from here
Can someone start out on an alternate solution or set it up differently for me?
Thanks
I'm not sure how to set this up or proceed I figured that
(n^2 + (n +2)^2 + (n+4)^2)/12 = x + 11
Where n is any odd integer
then I got
3n^2 +12n + 20 = 12x +123
3n(n+4) = 12x + 103
I'm not sure where to take it from here
Can someone start out on an alternate solution or set it up differently for me?
Thanks