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- Feb 14, 2012

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Solve the equation $\left\lfloor \dfrac{25x-2}{4} \right\rfloor=\dfrac{13x+4}{3}$.

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Solve the equation $\left\lfloor \dfrac{25x-2}{4} \right\rfloor=\dfrac{13x+4}{3}$.

- Mar 31, 2013

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Solve the equation $\left\lfloor \dfrac{25x-2}{4} \right\rfloor=\dfrac{13x+4}{3}$.

$\lfloor\frac{25x-6}{4}\rfloor = \frac{13x+1}{3}$

So $\frac{13x + 1}{3}$ = integer say n

So x = $\frac{3n-1}{13}$

So we get

$\lfloor\frac{25\frac{3n-1}{13}-6}{4}\rfloor = n$

Or

$\lfloor\frac{75n-103}{52}\rfloor = n$

Or 23n-103= > 0 and < 52

Or 103 <= 23n < 155

Or $\frac{103}{23} <=n < \frac{155}{23}$

So n = 5 or 6

Hence x = $\frac{14}{13}$ or $\frac{17}{13}$

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kaliprasad!

It is not required. I did it to make the RHS simpler but it did not help

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I see. Thanks for the reply,It is not required. I did it to make the RHS simpler but it did not help