Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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 $z = 60$?

share|improve this question
    
$\int$ 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 at 11:22
    
Correct. Question updated. –  wittrup Mar 20 at 11:41
add comment

1 Answer 1

up vote 1 down vote accepted

$\def\m{\mathop{\rm mod}}\newcommand{\f}[1]{\left\lfloor #1 \right\rfloor}$ $$y = \f{\frac{x\m 790}{10}} + 48 \\ z = (x\m 790)\m 10 + 10 \f{\frac x{790}} + 48$$ 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 $a\mathop{\rm div} b = \f{\frac ab}$

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.