## 2008 As Sum of Consecutive Integers

The number $2008$ can be written as a sum of consecutive integers such that the number of terms in the sum is twice the number of factors of $2008$. Find the smallest of these consecutive integers.
Source: NCTM Mathematics Teacher 2008

SOLUTION
$2008=2^3\cdot 251^1$
The exponents of the prime factors yield the number of factors of 2008
$(3+1)(1+1)=8$
Number of terms in the sum is twice the number of factors
$2\cdot 8=16$
Suppose the sum of consecutive integers starts with $a$
$2008=a+(a+1)+(a+2)+\cdots+(a+14)+(a+15)$
$=16a+\dfrac{(1+15)(15)}{2}$
$=16a+120$
$16a=1888$
$a=118$
$2008=118+119+120+\cdots+132+133$
How do we know the result is unique? Numbers of the form $2n\cdot p$, where $p$ is prime, can be written as the sum of consecutive positive integers in exactly one way. $2008=2\cdot 4\cdot251$ fits this description.

Answer: $118$