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My question is: Given that the equation $\,\frac{8}{3}x - a = \frac{9}{4}x + 123\,$ has positive integral solution where a is also positive integer, find the minimum possible value of a.

Please any guidance to solve this question would be helpful as even after trying a lot I was unable to solve it.

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Do you mean $$\frac{8}{3}\,x$$ or $$\frac{8}{3x}$$ (and similarly with "9/4x")? – Zev Chonoles Jun 2 '12 at 2:45
@user1296721 Please check I edited your question correctly – DonAntonio Jun 2 '12 at 2:49
@– Zev Chonoles I meant the first one. – mgh Jun 2 '12 at 2:51
@DonAntonio:Thanks the edit you made is correct. – mgh Jun 2 '12 at 2:53

I'll assume you meant $$\frac{8}{3}x-a=\frac{9}{4}x+123\Longleftrightarrow\frac{5}{12}x=123+a\Longleftrightarrow x=\frac{12\cdot 123+12a}{5}$$ As $\,12\cdot 123\,$ ends with a $\,6\,$, we need $\,12a\,$ to end with a $\,4\,$ (to have a multiple of $\,5\,$), and that happens for $\,a=2\,$

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I would like to raise a minor point. Would prefer getting to $5x=12(123+a)$, then observe that the left side is divisible by $5$, so we need the right side to be. So $5$ must divide $123+a$. Smallest positive $a$ that works is $2$. – André Nicolas Jun 2 '12 at 2:54
@Andre That indeed looks both more elegant and clear. – DonAntonio Jun 2 '12 at 2:56

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