Takes me back to 2012

If you try all even, you will find that (2+2012)*1006/2=(1000+6)(1000+7)

=1000000+13000+42=1013042.

This is 6015 short. When you move 2,4,6 to 2007,2009,2011 you make it.

3

There are 1006 even numbers from 2 to 2012. The sum of the numbers is 1006*1007.

The remaining odd numbers must add up to 1019057-1013042=6015. 6015/3=2005 so the three numbers must have an average of 2005.

We can exchange some of the even numbers with odd numbers to attain our condition. 2005+2, 2005+4, 2005+6 or 2007, 2009, and 2011.

First, lets try to see what happens when we try to form this without any odd numbers.

sum(range(2,2013,2))

The sum of all the even numbers between 1 and 2012 inclusive is $1013042$ which is $6015$ less than 1019057.

Note that we now have filled all the 1006 slots with even numbers. To increase the sum, we must remove some of the even numbers and replace them with odd numbers.

Factorise 6015.

factor(6015)

$6015 = 2005 \times 3$ Wow! Inserting 2005 three times will do the trick. We cannot however take 2005 three times. (Since all the 1006 ints are distinct.) Instead we should notice that $2005=2011-6=2009-4=2007-2$

Hence, the only set that satisfies this condition is

P As you see, we need three odd numbers.

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