# Turn formula with remainder

How do I turn these formulas: \begin{align} y &= \left\lfloor\frac{ x \mod 790}{10}\right\rfloor + 48 \\ z &= (x \mod 790) \mod 10 + 10\left\lfloor\frac{x}{790}\right\rfloor + 48 \end{align} so I can calculate $x$ given I only have $y$ and $z$.

e.g $x = 1002$ \begin{align} y &= \left\lfloor\frac{ 1002 \mod 790}{10}\right\rfloor + 48 \\ &= \left\lfloor\frac{ 212}{10}\right\rfloor + 48 \\ &= \left\lfloor21,2\right\rfloor + 48 \\ &= 21 + 48 \\ &= 69 \end{align} and \begin{align} z &= (1002\mod 790) \mod 10 + 10\left\lfloor\frac{1002}{790}\right\rfloor + 48 \\ &= (212) \mod 10 + 10\left\lfloor1,27\right\rfloor + 48 \\ &= 2 + 10 * 1 + 48 \\ &= 2 + 10 + 48 \\ &= 60 \end{align} How can I calculate the $x = 1002$ given that I only know that $y = 69$ and $z = 60$?

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$\int$ is completely out of context here. I guess you refer to the integer part, i.e. the floor function $\lfloor x \rfloor$, MathJaxed \lfloor x \rfloor and not \int(x) – AlexR Mar 20 '14 at 11:22
Correct. Question updated. – wittrup Mar 20 '14 at 11:41

$\def\m{\mathop{\rm mod}}\newcommand{\f}[1]{\left\lfloor #1 \right\rfloor}$ $$y = \f{\frac{x\m 790}{10}} + 48 \\ z = (x\m 790)\m 10 + 10 \f{\frac x{790}} + 48$$ Note that $(z-48) \m 10 = (x\m790)\m10$ and $y-48 = (x\m 790) \mathop{\rm div} 10$ and thus $$x\m 790 = (z-48)\m 10 + y-48$$ And also $(z - 48) \mathop{\rm div} 10 = x \mathop{\rm div} 790$ so in total $$x = 790 \cdot \left\lfloor\frac{(z - 48)}{10}\right\rfloor + (z - 48) \mod 10 + y - 48 \cdot 10$$ Where $a\mathop{\rm div} b = \f{\frac ab}$