# Solve 2 equations with 2 variables and the floor() function

I have two equations and two unknowns:

$R = \lfloor M * V\rfloor$

$N = \lfloor\frac{X - M}{R}\rfloor + 1$

$V$, $N$, $X$ are known. I need to solve for $R$ and $M$. I'm not sure how to handle the floor function.

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$V=M/R$, so $N=\lfloor \frac{X}{R}-V\rfloor +1$. – Daryl Jan 5 '13 at 7:25
Doesn't $V = R / M$? Also, I missed that $R$ is similarly the result of the floor function. – mjibson Jan 5 '13 at 7:29
$N = \lfloor\frac{X}{R} - \frac{1}{V}\rfloor + 1$ is correct, but I'm still unsure how to solve for $R$ in that case. – mjibson Jan 5 '13 at 7:33
Yes, when I replied, there we no floor functions on the first equation though, so you will need a strategy change. – Daryl Jan 5 '13 at 7:46

Your equations imply that $R\le MV < R+1$ and $N-1 \le (X-M)/R < N$. Therefore, if $R$ can positive, the solution $(M,R)$ with $R>0$ can be any point inside the 2D region bounded by five straight lines: \begin{align} R &\le MV,\\ R &> MV - 1,\\ (N-1)R &\le X-M,\tag{1}\\ X-M &< NR,\tag{2}\\ R&>0,\tag{3}\\ \end{align} with the additional constraint that $R$ is an integer. If $R$ is allowed to be negative, the solution $(M,R)$ with $R<0$ can be found in the analogous way, with the inequality signs in $(1)-(3)$ flipped.
Why do the equations imply that $MV < R + 1$? Considering the original problem, that is not the case. The constraint is $1 <= R <= M$. If the equations imply that perhaps I stated them incorrectly. – mjibson Jan 5 '13 at 7:50
Foy any real number, you have $\lfloor x\rfloor \le x<\lfloor x\rfloor+1$. Hence the inequality $R\le MV<R+1$. The strict inequality $MV<R+1$ does not contradict $1\le R\le M$ if $V$ can be a number much smaller than $1$. – user1551 Jan 5 '13 at 8:02