January 2011 Brain Teaser Solution

Q: What is the largest number of consecutive integers that will add up to 2011?

A:  A good way to think about this problem is to do what Polya said "Solve a simpler problem"
So first think about  -- What is the largest number of consecutive integers that will add up to 11?

The least number of numbers would be one -- 11
You could use two consecutive numbers -- 5 and 6.
If you think about negative integers, -4, 3, -2, -1 would cancel out 1, 2, 3, and 4
so - 4, -3, -2, -1, 0, 1, 2, 3, 4, 5, 6 would give a total of 11 numbers

But we can do even better

-10 would cancel out with 10
-9 would cancel out with 9 and so on

-10, -9, -8 , -7, -6, -5, - 4, -3, -2, -1, 0, 1, 2, 3, 4, 5, 6, 7, 8 , 9, 10 would make zero if you sum them.
so just now add an 11
-10 thru -1 is 10 integers
0 is 1 integer
1 thru 10 is 10 integers and
11 is 1 integer for a total of 22.
(22 = 2 x 11)

so the number of integers is always 2 times the number itself.

For 2011,
The lowest number would be -2010 (it would cancel out with +2010)
The next number would be -2009 (it would cancel out with +2009)
The next number would be -2008 (it would cancel out with +2008)
and so on...
until -2 cancels out with +2
and -1 cancels out with +1

and then there is 0

so there would be 2010 negative numbers -2010 through -1
2010 postiive numbers 1 through 2010
plus 0
and also 2011
2010 negative integerss + 2010 positive integers + 2 more (for 0 and 2011)
for a total of 4022
4022 is the answer.