If $S_{100} - k + k² = 7500$ and $S_{100}$ it is the sum of $100$ consecutive positive integers, then what it is the value o k?

Question

$k$ must be one of the 100 consecutive integers.

I know the answers are $50$ and $26$.

But I got stuck into getting these numbers.

Here is what I tried:

$S_{100} = \frac{(n+n+99)\times100}{2} \implies S_{100} - k + k² = 7500 \implies \frac{(n+n+99)\times100}{2} + k(k-1) = 7500 \implies k(k-1) = 7500 - 50(2n+99)\implies k(k-1)= 50(51-2n) $.

It's easy to see that $n = 1 \implies k = 50$ is an answer. But how can I discover that $n = 19 \implies k = 26$?

Answer 1

$$\sum_{i=n}^{n+99} i = \sum_{i=1}^{n+99} i - \sum_{i=1}^{n-1} i = \dbinom{n+100}{2} - \dbinom{n}{2} = \dfrac{(n+100)(n+99)}{2} - \dfrac{n(n-1)}{2} = \dfrac{n^2+199n+9900}{2} - \dfrac{n^2-n}{2} = \dfrac{200n+9900}{2}$$

Then you have:

$$\dfrac{200n+9900}{2}+k^2-k = 7500$$

Multiplying out:

$$k^2-k+100n-2550 = 0$$

$$k = \dfrac{1\pm \sqrt{10201-400n}}{2}$$

So, $10201-400n$ must be an odd perfect square.

You can add the numbers $1$ through $100$ and $k=50$.

Or you can have the numbers $19$ through $118$ and $k=26$

Answer 2

Let's start with $$k^2-k = 50(51-2n)$$

Then we must have $k^2-k \equiv 0 \pmod{50}$.

We split this up into $k^2-k \equiv 0 \pmod 2$ and $k^2-k \equiv 0 \pmod{25}$.

\begin{align} k^2-k &\equiv 0 \pmod{25} \\ k(k-1) &\equiv 0 \pmod{25} \\ k &\equiv 0, 1 \pmod {25} \end{align}

Similarly, $k \equiv 0,1 \pmod 2$.

\begin{array}{c|ccc} & \pmod{2} & \pmod{25} \\ \hline 25 & 1 & 0 \\ 2 & 0 & 2 \\ \hline 25 & 1 & 0 \\ 26 & 0 & 1 \\ \hline \end{array}

Hence, if $k \equiv x \pmod 2$ and $k \equiv y \pmod{25}$, Then $k \equiv 25x + 26y \pmod {50}$.

\begin{array}{cc|ccc} x \pmod 2 & y \pmod{25} & k \pmod{50} & \dfrac{k(k-1)}{50} & n \\ \hline 0 & 0 & 0 & 0 & *\\ \color{red}0 & \color{red}1 & \color{red}{26} & \color{red}{13} & \color{red}{19}\\ 1 & 0 & 25 & 12 & *\\ 1 & 1 & 1 & 0 & *\\ \hline \end{array}