As Wench said. I remember you asking a similar question a while back, but my approach to solving it hasn't changed- divide 2013 by 6 (335.5) and add the 6 consecutive integers either side of your answer to get 2013. I seem to recall at least one sixer solving THAT particular problem a different way.
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Comments
Wench says,
333, 334, 335, 336, 337, 338.These are great!
thesagittarian12 says,
As Wench said. I remember you asking a similar question a while back, but my approach to solving it hasn't changed- divide 2013 by 6 (335.5) and add the 6 consecutive integers either side of your answer to get 2013. I seem to recall at least one sixer solving THAT particular problem a different way.thesagittarian12 says,
That should have read "three integers either side", but I think most readers would know I meant that.MO_Thoughts2 says,
Pure algebrax + (x+1) + (x+2) + (x+3) + (x+4) + (x+5) = 2013
6x + 15 = 2013
6x = 1998
x= 333
Therefore 333, 334, 335, 336, 337, 338