# Sequence solving

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)

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

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 $$x=2^3\cdot 37853=\fbox{302824}$$.