I am given a number $N$. I need to find $M\le N$ such that $M$ can be written as a sum of consequent numbers from 1 to x.
I know this formula $1+2+\cdots +n = \frac{n(n+1)}{2}$.
How can I find such M?
Find M <= N such that sum of consequent numbers from 1 to x is M.
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sequences-and-series
elementary-number-theory
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3Quadratic equation: $M^2+M=2N$ – 2017-01-22
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0Thanks. Is there a ready formula that can find such M? – 2017-01-22
1 Answers
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$$ \begin{align} &\, M \,=\, \sum_{k=1}^{n}k = \frac{n(n+1)}{2} \le N \\[2mm] &\, \frac{n(n+1)}{2}=N \,\Rightarrow\, n^2+n-2N=0 \,\Rightarrow\, n=\frac{-1\pm\sqrt{1+8N}}{2} \\[2mm] &\, n\in{\mathbb N}^{+} \,\Rightarrow\, \color{red}{n=\left\lfloor \frac{\sqrt{8N+1}-1}{2} \right\rfloor} ,\,\, \color{blue}{M=\frac{n(n+1)}{2}} \end{align} $$
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0Do these brackets mean round down? – 2017-01-26
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0@BaimyrzaShamyr: Yes. See [Floor Function](https://en.wikipedia.org/wiki/Floor_and_ceiling_functions). – 2017-01-26
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0Bravo! Simple and Impressive solution. – 2017-01-26