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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.

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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}

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  • $\begingroup$ How do you try out the cases? $\endgroup$ – 1110101001 Dec 7 '13 at 3:01
  • $\begingroup$ Do you just set N equal to the smaller of the two divisors and see if there exists an integer solution to a such that N(2a + N - 1) = the product of the 2 divisors? $\endgroup$ – 1110101001 Dec 7 '13 at 3:08
  • $\begingroup$ @user2612743 Yes. Note that these are the only possibilities. $\endgroup$ – user17762 Dec 7 '13 at 3:10
  • $\begingroup$ What would you do if instead of 2013 you were provided a larger number? The list of factors would get very large -- what alternate method (if there is one) could you use? $\endgroup$ – 1110101001 Dec 7 '13 at 3:12
  • $\begingroup$ @user2612743 Even for a large number, the same method works in principle. If you have a large number of factors, then note that $N$ and $2a+N-1$ must be of opposite parity. So you can remove some of the cases, but still there could be potentially a large number of cases and almost all of them will give rise to a valid solution. $\endgroup$ – user17762 Dec 7 '13 at 3:14

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