Sum of Consecutive Numbers (Part 2)

The sum of n consecutive positive integers is 2005. Find all possible values of n.
Source: NCTM Mathematics Teacher, December 2005

SOLUTION
Let S=a+(a+1)+(a+2)+(a+3)+\cdots+(a+(n-3))+(a+(n-2))+(a+(n-1)) be the sum of n consecutive positive integers with the first term equal a. If we add the first term to the last term, the second term to the next to the last term, etc.
a+(a+(n-1))=2a+n-1
(a+1)+(a+(n-2))=2a+n-1
(a+2)+(a+(n-3))=2a+n-1
\cdots
we notice that they sum to a constant value of 2a+n-1. This observation leads us to the formula of the sum
S=(2a+n-1)n/2
Applying the formula for 2005
2005=(2a+n-1)n/2
4010=(2a+n-1)n
4010=2\cdot 5\cdot 401
Setting n equal to one of the divisors of 4010 and solving the equation for a
n=1,a=2005
n=2,a=1002
n=5,a=399
n=10,a=196
n=401,a=\textrm{-}195
n=802,a=\textrm{-}398
n=2005,a=\textrm{-}1001
n=4010,a=\textrm{-}2004
The values of n for which a are positive are 1,2,5, and 10.

Answer: 1,2,5, and 10.

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About mvtrinh

Retired high school math teacher.
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