6
$\begingroup$

Intuitively I would say yes but I can't find a way to prove it. I tried with small values and bruteforcing shows that there seems to only be one solution given a distinct tuplet.

For example $(1,72)$ has only $(8,9)$ as valid $a$ and $b$ values. Is there a way to do this mathematically?

Improve this question
$\endgroup$
5
  • 1
    $\begingroup$ Welcome to Puzzling.SE! For the uninitiated in the room (such as me), would you mind explaining what precisely is meant by the little symbol that looks like a Phillips head screw? Thanks, and have a great day! $\endgroup$ – Brandon_J May 4 '19 at 17:28
  • $\begingroup$ It’s bitwise addition modulo 2, @Brandon_J $\endgroup$ – El-Guest May 4 '19 at 17:30
  • $\begingroup$ Thanks for letting me know, @El-Guest . $\endgroup$ – Brandon_J May 4 '19 at 17:30
  • $\begingroup$ So essentially, I convert the number to binary (for exapmle, 3-->00011, 24-->11000), then perform theXOR operation on each pair of bits, and then convert back to a normal number (27)? $\endgroup$ – Brandon_J May 4 '19 at 17:54
  • 1
    $\begingroup$ Not sure this is a puzzle, but I'm glad you got your answer. (Please don't forget to $\color{green}{\checkmark \small\text{Accept}}$ it!) $\endgroup$ – Rubio♦ May 5 '19 at 4:34
11
$\begingroup$

By counterexample, $a,b$ pair is clearly not unique.

The pairs $(5,9)$ and $(3,15)$ both multiply to $45$, and add bitwise to $12$.

$5 \oplus 9 = 12$, $5 \times 9 = 45$
$3 \oplus 15 = 12$, $3 \times 15 = 45$

Improve this answer
$\endgroup$
3
  • 1
    $\begingroup$ @Hugh, thanks for the edits - looks nicer now! $\endgroup$ – ppgdev May 5 '19 at 3:09
  • $\begingroup$ I edited it quickly off my phone a few hours ago, so I have it another go. Hopefully it still looks good. $\endgroup$ – user46002 May 5 '19 at 4:47
  • $\begingroup$ @Hugh, sure. Thanks. $\endgroup$ – ppgdev May 5 '19 at 16:28

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.