# 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?

• Collect terms with $k$ to the right. To deduce $a\neq 0$, took one positive term $(1/3)$(which makes it smaller when you ignore all the other positive terms) and $5$ times maximum of the negative terms.(again, makes it smaller $1/2000+1/2000+1/2000+1/2000+1/2000$) and observe $1/3-5/2000>0$. – Atbey Nov 28 '17 at 6:12

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.
• They just threw away the first four terms in the first parentheses and roundup all the terms in the second to $\frac 1{2000}$. Each of those reduces the value of the expression, which is where the $\gt$ came from. They just need to prove $a \neq 0$ and this shows a positive lower bound. To show $\frac 13-\frac 5{2000} \gt 0$, just put them over a common denominator, which is $6000$, and subtract. – Ross Millikan Nov 28 '17 at 6:10