How would you find how many numbers N there exist such that N consecutive positive integers add to 2013? (Assume that N=1 is a valid case whose solution is just 2013 itself).
To clarify, this, when N=2 then 1006,1007 is a solution. If N=3 then 670,671,672 is a solution and so on. How many values of N exist such that this can happen?
I got that the equation for N consecutive integers to add up to 2013 is
$Nx\; +\; N\frac{\left( N-1 \right)}{2}\; =\; 2013$
where $x$ is the first term. However, I have no idea how to find the number of solutions to this diophantine equation.
Note that $$a+(a+1)+(a+2)+\cdots+(a+(N-1)) = Na + \dfrac{N(N-1)}2 = 2013$$ Hence, we need \begin{align} N(2a+N-1) & = 1 \times 4026\\ & = 2 \times 2013\\ & = 3 \times 1342\\ & = 6 \times 671\\ & = 11 \times 366\\ & = 22 \times 183\\ & = 33 \times 122\\ & = 61 \times 66\\ \end{align} Now try out all these cases noting that $2a+N-1 > N$ for $a>1$.
You obtain the following pairs $(a,N)$ as solution: \begin{array}{|c|c|} \hline & (a,N)\\ \hline 1 & (2013,1)\\ 2 & (1006,2)\\ 3 & (670,3)\\ 4 & (333,6)\\ 5 & (178,11)\\ 6 & (81,22)\\ 7 & (45,33)\\ 8 & (3,61)\\ \hline \end{array}