2 Since $24=3+5+7+9$, the number 24 can be written as the sum of at least two consecutive odd positive integers.
(a) Can 2005 be written as the sum of at least two consecutive odd positive integers? If yes, give an example of how it can be done. If no, provide a proof why not.
(b) Can 2006 be written as the sum of at least two consecutive odd positive integers? If yes, give an example of how it can be done. If no, provide a proof why not.
Solution: Let $N=(2 k+1)+(2 k+3)+\cdots+(2 k+2 n-1)$ where $n$ and $k$ are integers, $n \geq 2, k \geq 0$. Then $N=[(2 k+1)+(2 k+2 n-1)] n / 2=(2 k+n) n$, which is a product of two integers with the same parity, since adding the even number $2 k$ to the integer $n$ does not change its parity.
1. Since $2005=401 \cdot 5$, taking $n=5$ gives $2 k+5=401$, i.e. $2 k+1=397$. Checking, we see that $397+399+401+403+405=401 \cdot 5=2005$, so 2005 can be represented as the sum of at least two consecutive odd positive integers.
2. Since $2006=2 \cdot 1003=2 \cdot 17 \cdot 59$ is divisible by 2 but not by 4 , every pair of integers with a product of 2006 consists of two integers with different parity, (i.e. one odd and one even). Therefore 2006 cannot be represented as the sum of at least two consecutive odd positive integers.