How Can You Prove That 83 Divides x in This Mathematical Series?

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SUMMARY

The mathematical series defined by the expression $\dfrac{x}{y}=1-\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{1}{4}+\cdots-\dfrac{1}{54}+\dfrac{1}{55}$ leads to the conclusion that 83 divides the integer $x$. The solution involves evaluating the series and simplifying the resulting fraction to demonstrate that the numerator is divisible by 83. The correct solution was provided by the user kaliprasad, confirming the divisibility condition.

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Here is this week's POTW:

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Let $x$ and $y$ be the positive integers such that $\dfrac{x}{y}=1-\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{1}{4}+\cdots-\dfrac{1}{54}+\dfrac{1}{55}$.

Prove that 83 divides $x$.

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Congratulations to kaliprasad for his correct solution!(Cool)

You can find the model answer below:

By using the identity $1-\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{1}{4}+\cdots-\dfrac{1}{2n}=\dfrac{1}{n+1}+\dfrac{1}{n+2}+\cdots+\dfrac{1}{2n}$, we have

$\begin{align*}\dfrac{x}{y}&=1-\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{1}{4}+\cdots-\dfrac{1}{54}+\dfrac{1}{55}\\&=1-\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{1}{4}+\cdots-\dfrac{1}{2(27)}+\dfrac{1}{55}\\&=\dfrac{1}{28}+\dfrac{1}{29}+\dfrac{1}{30}+\cdots+\dfrac{1}{54}+\dfrac{1}{55}\\&=\left(\dfrac{1}{28}+\dfrac{1}{55}\right)+\left(\dfrac{1}{29}+\dfrac{1}{54}\right)+\cdots+\left(\dfrac{1}{41}+\dfrac{1}{42}\right)\\&=\dfrac{83k}{y}\,\,\,\text{where}\,\,\,(83,\,y)=1\end{align*}$

and this completes the proof.
 

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