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

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About mvtrinh

Retired high school math teacher.
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