I need help to find the x:

6 3412 x 8605648

I've noticed that the end doubles everytime so we have 6 then 12, then we should have 24 then 48 , I have also noticed that the first 2 numbers multiplied make the last 2 of the number so if we use 3412 we can see that 3*4=12 . Now take a look at the last number 8605648 we can see that 8*6=48 but we have now 3 numbers 0,5,6 that I don't understand from where they come ,I assumed that the 0 is used as a delimitator and then saw that the other 2 numbers 5+6=11 and 4+8=12 so 1 more than 1,that's why I tought the answer would have been something like this: 24 at the end cuz of 6,12,24,48 46 at the beginning cuz 4*6=24 then i added a 0 as a delimiter and i get 460xx24 2+4=6 so x+x should be equal to 5, there are 2 numbers that give 5 and those are 2+3 and 4+1, i don't know which one i should use, please help me.

The puzzle was designed by Mislav Predavec, who has given permission to post (which was via email to Jonathan Allan) - it's from his Algebrica (and was found at http://zagaza.ru/za548.htm)

  • $\begingroup$ Note also that $4 \times 3=12$ and $6 \times 8=48$ so you'll also need to consider if it might be $64 \cdots 24$. $\endgroup$ – Jonathan Allan Jan 12 at 20:48
  • $\begingroup$ yea i have writtet 46 because it ends 24 and it has 4 as final number,i'm using this criteria because in 8605648 it ends with 8 and starts with. $\endgroup$ – alnesi Jan 12 at 21:04
  • $\begingroup$ Root source, I believe: news.generiq.net/Trilogica/algebrica.html (should this have permission? - It's marked as copyright Mislav Predavec) $\endgroup$ – Jonathan Allan Jan 12 at 21:19
  • $\begingroup$ hmm i don't really know if it is close the post $\endgroup$ – alnesi Jan 12 at 21:28
  • 1
    $\begingroup$ I emailed Mislav Predavec and he said it was fine. $\endgroup$ – Jonathan Allan Jan 12 at 22:59

I generally don't like these types of puzzles because there's so many possible answers motivated by an almost infinite amount of justifications. Here's one such solution:

Observe that the prime factorisations of our numbers: $$6=2\cdot 3 \\ 3412=2^2\cdot 853 \\ x=? \\ 8605648=2^4\cdot 537853$$ seem to follow a trend - they're all of the form $2^a\cdot p$, where $p$ is prime. Also, each prime earlier in the list seems to be contained in each prime later in the list (e.g. $537853$ contains $853$). Furthermore, if we observe the number of digits in each of our numbers $(6,3412\dots)$ we find that the pattern goes roughly like $1,4,\_,7$ where $\_$ is a placeholder for the number of digits of $x$. We might guess that $\_$ should be replace with $6$, since $1+\boldsymbol{3}=4$, $4+\boldsymbol{2}=6$, and $6+\boldsymbol{1}=7$.

Looking at the powers of $2$, it seems evident that $x$'s prime factorization should be of the form $2^3\cdot p$ where $p$ is a prime contained by $537853$. Keeping in mind our prediction that $x$ has $6$ digits, we find that the only prime that satisfies all these conditions is $37853$. Thus $x=2^3\cdot 37853=\fbox{302824}$.


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