2
$\begingroup$

I slur `$z\mod L$' here to mean the only element of $\{z+nL: n\in \mathbb{Z}\}\cap [0,L).$

We are given quantities:

  • $a,b, L,$
  • $D_1 = (ax \mod L) + aw$,
  • $D_2 = bx+bw.$

We are also given the fact:

  • $D_3 =(ax + bx)\in [0,L).$, i.e. $D_3=(D_3\mod L)$

From these facts, is it possible to compute the following in terms of the given quantities? $$D_4 = ax+bx+aw+bw= D_3+aw + bw$$


Here's my try:

\begin{align} (D_1+D_2) &= ((ax \mod L) + aw + bx+bw) \\ &= ((ax \mod L)-ax+ax+bx+aw+bw)\ \\ &= ((ax \mod L)-ax)+D_4. \end{align}

So it is enough to find $(ax \mod L)-ax,$ which I am not sure is possible.

$\endgroup$
  • $\begingroup$ Ok. Comments deleted then. Now I still don't really understand how you want to derive something non-trivial from the three given equalities: the first two define $D_1$ and $D_2$, they don't really carry information. The third one does carry the information that $0\leq (a+b)x <L$. That's basically all you have. Finally, what do you mean by " is it possible to compute..."? Do you mean compute "the" representative of that modulo $L$? And how is that question related to $D_1$ and $D_2$? $\endgroup$ – Arnaud Mortier Jul 29 '18 at 22:59
  • $\begingroup$ No, I mean the quantity outright. That is, if $D_3=(a+b)\cdot x \in [0,L)$ then I want $D_3+(a+b)w \in \mathbb{R}$ $\endgroup$ – enthdegree Jul 29 '18 at 23:02
  • $\begingroup$ So to be clear, you want an answer in terms of $D_1$ and $D_2$? $\endgroup$ – 高田航 Jul 29 '18 at 23:12
  • $\begingroup$ Yes. ${}{}{}{}$ $\endgroup$ – enthdegree Jul 29 '18 at 23:14
  • $\begingroup$ You don't know $x$ and $w$ but you know $a$ and $b$ is that right? $\endgroup$ – Arnaud Mortier Jul 31 '18 at 22:02
0
$\begingroup$

The difficulty of that problem is that there is a lot of useless data. All you need is that you know $a$, $b$, and $D_2=bx+bw$.

$$D_4=D_2\left(1+\frac ab\right)$$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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