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What is the maximum possible value of $k$ for which $2013$ can be written as a sum of $k$ consecutive positive integers?

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inoxb911
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    Maybe this helps you. – mag Apr 15 '21 at 06:16
  • You need only to find out the minimum possible positive integer $m$ for which $s = 4m^2 - 4m + ((4 \times 4026) + 1) = (2m-1)^2 + (4 \times 4026) $ is a perfect square. Then the maximum possible value of $k$ is given by $$\frac {{\sqrt s} - (2m - 1)} {2}.$$ – Anil Bagchi. Apr 15 '21 at 06:43
  • So it reveals that $m = 3$ will do the job for you and for this value of $m$ we have $\sqrt s = 127.$ So the maximum possible value of $k$ is given by $\frac {127 - 5} {2} = \frac {122} {2} = 61.$ – Anil Bagchi. Apr 15 '21 at 06:55

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Hint: The sum of $m$ consecutive positive integers is given by $\frac{m(m+1)}{2}$. So what you want is to find the highest value of $k$ for which exist a natural number $m$ so that $$2013=\frac{(m+k)(m+k+1)}{2}-\frac{m(m+1)}{2}$$ holds.

I think meanwhile you found an answer by yourself. This is why I now give a full answer to this for all who are asking this or a similar question in future:

The equation above is equivalent to $$4026=k(k+2m+1)=k^2+2km+k.$$ By the first representation you see that $k$ has to be a divisor of $4026$. All divisors of $4026$ are given by (you can either calculate it by yourself or use W|A)$$1, 2, 3, 6, 11, 22, 33, 61, 66, 122, 183, 366, 671, 1342, 2013, 4026. $$ By the second representation you see that $k^2<4026$ holds. Thus the largest candidate for $k$ is $61$ (as $66^2>4026$). So we now check for this candidate if there exists such a natural number $m$. By solving the equation above with $k=61$ we get $m=2\in\Bbb N$. Since we were only looking for the largest possible value of $k$ we are done.

mag
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