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Given this sequence:

[12, 13], 25, 58, 52, 98, 108, 780, 568, 306

What is the next number?

Note the square brackets.

Hint:

This sequence follows the exact same rules: [10, 10], 10, 10, 10, 10, 10, 10

Hint #2:

You only need two math operators for this series

Hint #3:

This sequence follows the exact same rules: [2, 2], 4, 8, 32

Hint #4

You only need addition and multiplication

Hint #5

Every number (except for the last one) in hint #3 sequence is one digit, and there is a clear pattern there. In what way could this pattern be applied to Hint #1?

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  • $\begingroup$ ROT13(Qbrf vg unir gb qb jvgu qvbcunagvar rdhngvbaf jurer gur svefg gjb grezf va oenpxrgf jbex nf pbafgnagf?) $\endgroup$ – Prim3numbah Jan 25 at 14:36
  • $\begingroup$ Cause I see some pattern, but not sure..? $\endgroup$ – Prim3numbah Jan 25 at 16:15
  • $\begingroup$ @Prim3numbah rot13(V qba'g xabj jung n Qvbcunagvar rdhngvba vf, ohg lrf, gur svfg gjb grezf ner pbafgnagf, zrnavat gung gurl pbhyq obgu or nalguvat.) $\endgroup$ – MilesZew Jan 25 at 21:20
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Okay, so this puzzle has been bothering me for a few days. I still don't have a good solution, but I thought I'd share my thoughts about it.

So for the two hint sequences, I have some ideas. I'm thinking along the lines of a Fibonacci sequence but with multiplication, where each digit column acts separately. So here something like $[10 10]$ becomes $(1\times 1) (0 \times 0) = (1)(0) = 10$ and that iterates. $[2 2]$ becomes $(0 \times 0) (2 \times 2) = 4$ then $2 \times 4 = 8$, $4 \times 8 = 32$, after that I suspect that only up to $32$ is given because something funny happens after that. However, we know this doesn't work for the given sequence. My only clues for this are that the first few numbers look sort of like a Fibonacci sequence [1 1 2], [2 3 5] if you split it up by digit, but this really doesn't help. I don't know, the hint says that only addition and multiplication are needed (for math operations) but as far as I can tell addition and multiplication would imply a growing sequence where here we clearly have terms that are less than previous terms, so there is some non-math operation (splitting digits?).

I'm just hoping that somehow this helps someone see something, but really I have no idea and probably nothing here is particularly useful.

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  • $\begingroup$ Just to clarify, when I said only addition and multiplication are needed, I didn't necessarily mean on the whole number, I just meant that when you do use math operators, addition and multiplication are the only ones you need. $\endgroup$ – MilesZew Jan 30 at 5:32
  • $\begingroup$ Also, you are on the right track with some of your ideas. $\endgroup$ – MilesZew Jan 30 at 5:34
  • $\begingroup$ rot13(V qba'g xabj vs vg jvyy uryc, ohg Svobanppv vf nyfb ivfvoyr va gur ynfg frdhrapr - gur arkg ryrzragf ner 2^Svo(a)) $\endgroup$ – Michał Wójcik Jan 30 at 9:00
  • $\begingroup$ To add to this, I tried nearly everything. I flipped numbers around => [02,02] -> [20,20], to help prove [10,10]. I tried splitting digits, I tried mirroring. I am stumped. $\endgroup$ – Adib Feb 4 at 19:52
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1*10+2=12

1*10+2+2^0=13

2*10+2^2+2^0=25

5*10+2^3=58

5*10+2^1=52

3^2*10+2^3=98

7*10^2+2^4*5=780

5*10^2+2^6+2^2=568

3*10^2+2^2+2^1=306

2*10^2+2^1+2^0=203

My answer is 203.

The only non-composite numbers from 1 to 10 are 1, 2, 3, 5, 7. Having this in mind, there are multiple answers.

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  • $\begingroup$ Sorry, not quite, but you are on the right track with looking at the individual digits $\endgroup$ – MilesZew Jan 26 at 19:06
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12=2^3+2^2

13=2^3+2^2+2^0

25=2^4+2^3+2^0

58=2^5+2^4+2^3+2^1

52=2^5+2^4+2^2

98=2^6+2^5+2^1

108=2^6+2^5+2^3+2^2

780=2^9+2^8+2^3+2^2

568=2^9+2^5+2^4+2^3

306=2^8+2^5+2^4+2^1

848=2^9+2^8+2^6+2^4

The answer is 848.

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  • $\begingroup$ Sorry, also not correct. I'll add a new hint that might help soon. $\endgroup$ – MilesZew Jan 26 at 22:04
  • $\begingroup$ @MilesZew we are still waiting for your Hint. $\endgroup$ – Vassilis Parassidis Jan 27 at 3:30
  • $\begingroup$ @Vassilis_Parassids New hint is up $\endgroup$ – MilesZew Jan 27 at 17:06
  • $\begingroup$ Not sure how this solution satisfies the [10, 10], 10, 10, 10, 10, 10, 10 hint $\endgroup$ – Adib Jan 28 at 0:21
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    $\begingroup$ I don't understand this answer. What is the logic behind the exponents? $\endgroup$ – user16074 Jan 28 at 12:57

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