<s>[INST] Find, with proof, all integers $k$ satisfying the equation Find, with proof, all integers satisfying the equation
\begin{equation}
\frac{k-15}{2000} + \frac{k - 12}{2003} + \frac{k - 9}{2006} + \frac{k - 6}{2009} + \frac{k - 3}{2012} = \frac{k - 2000}{15} + \frac{k - 2003}{12} + \frac{k - 2006}{9} + \frac{k - 2009}{6} + \frac{k - 2012}{3}.
\end{equation}
This is a problem from a math competition, and here is the solution that is provided:

I can see how $k = 2015$ can be found via inspection; however, the value of $a$ is not as intuitive to me. Could someone either provide an alternate explanation, or clarify how to obtain the value of $a$?
Additionally, how is the inequality $\frac{1}{3} - \frac{5}{2000} > 0$ achieved?
 [/INST] They are just moving all the terms including $k$ to the right and all the ones that are constants to the left to make $b$.  The term $\frac {k-2000}{15}$ becomes $\frac 1{15}k-\frac {2000}{15}$ and the second half goes to the right.  The $\frac 1{15}$ in the expression for $a$ comes from the first term on the right, the $\frac 1{12}$ from the second, and so on.  All the negative ones came from the left side.
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