## 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 $10$.