MHB Min Value of $\dfrac{x}{y}-\dfrac{123}{2014}$: $\dfrac{1}{3792362}$

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The minimal value of the expression \(\left|\dfrac{x}{y}-\dfrac{123}{2014}\right|\) for positive integers \(x\) and \(y\) with \(y < 2014\) is \(\dfrac{1}{3792362}\). This is established by showing that \(\left|\dfrac{x}{y}-\dfrac{123}{2014}\right| \geqslant \dfrac{1}{1883 \times 2014}\) and that equality can be achieved specifically when \(x = 115\) and \(y = 1883\). The analysis reveals that for other values of \(x\) and \(y\), the expression exceeds \(\dfrac{1}{3792362}\). The proof relies on the co-primality of 123 and 2014, ensuring that the left side of the equality cannot be zero. Thus, the conclusion confirms that \(\dfrac{1}{3792362}\) is indeed the minimal value.
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If $x,\,y$ are positive integers with $y<2014$, show that the minimal value of $\left|\dfrac{x}{y}-\dfrac{123}{2014}\right|$ is $\dfrac{1}{3792362}$.
 
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anemone said:
If $x,\,y$ are positive integers with $y<2014$, show that the minimal value of $\left|\dfrac{x}{y}-\dfrac{123}{2014}\right|$ is $\dfrac{1}{3792362}$.
[sp]First, factorise those numbers to find that $123 = 3\times41$, $2014 = 2\times19\times53$ and $3792362 = 1883\times2014$.

So the problem is to show that $\left|\dfrac{x}{y}-\dfrac{123}{2014}\right| \geqslant \dfrac{1}{1883\times 2014}$ whenever $y<2014$, and that equality can be achieved.

Let's look first at whether equality can be achieved, so that $\left|\dfrac{x}{y}-\dfrac{123}{2014}\right| =\dfrac{1}{1883\times 2014}$. That is equivalent to $\bigl|\,2014x - 123y\,\bigr| = \dfrac y{1883}.$

Since $123$ and $2014$ are co-prime, the left side of that equation can never be zero when $y<2014$. But it is an integer, so the smallest value it can take is $1$. A routine application of Euclid's algorithm shows that for $y<2014$ this can only happen if either (i) $x=8$ and $y=131$ or (ii) $x = 115$ and $y = 1883$. In case (i), $\bigl|\,2014x - 123y\,\bigr|$ is much larger than $\dfrac y{1883}.$ But in case (ii), equality occurs.

For any other values of $x$ and $y$ (with $y<2014$), we must have $\bigl|\,2014x - 123y\,\bigr| \geqslant2$, and $\dfrac y{1883} <2$. Therefore $\bigl|2014x - 123y\bigr| > \dfrac y{1883}$ and consequently $\left|\dfrac{x}{y}-\dfrac{123}{2014}\right| > \dfrac{1}{3792362}$.

Therefore $\dfrac{1}{3792362}$ is the minimal value, and it only occurs when $x = 115$ and $y = 1883$.[/sp]
 
Very well done, Opalg!(Happy) And thanks for participating!(Smile)
 
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