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Prime factoring, remove multiple signs and find a prime!

We all know that prime factoring is to factorize a number to multiplications of primes, for example, \(4=2 \times 2\). (Note: arrange the primes from small to big.) Remove the multiple signs, you'll get a number: 22. Repeat the steps, \(22= 2 \times 11\). Remove the multiple signs to get 211, which is a prime. Now, lets do this for 6, which you can find a prime in just one step: \(6=2 \times 3\), get 23, a prime. 8 is a bit tricky, you'll need to do a number of times to get an 18 or 19 digit number (I lost my draft paper). How about the other numbers?

Note by Takeda Shigenori
3 years, 10 months ago

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I got an idea. I guess this is where programming is going to help us. Maybe I can try making a program which will factorize the number and then add the values. If I get any answers then I'll tell you. If anyone of you have any idea about how it would be then plz reply.

Also (of-course) there are 2 exceptions, which are naturally 0 and 1. Aryan Gaikwad · 3 years, 4 months ago

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The number is 311 for 9. Frodo Baggins · 3 years, 10 months ago

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8=3331113965338635107 Sharky Kesa · 3 years, 4 months ago

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64 = 2 x 2 x 2 x 2 x 2 x 2 = 222222

now 222222 = 2 × 3 × 7 × 11 × 13 × 37

= 237111337

237111337 = 29 × 101 × 80953 = 2910180953

2910180953 = 853 × 3411701 = 8533411701

8533411701 = 3 x 181 x 367 x 42821 = 318136742821

318136742821 = 127 x 2505013723 = 1272505013723

1272505013723 is prime. Awesome. So what's the logic? Aryan Gaikwad · 3 years, 4 months ago

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That's crazy!

Can we prove that it's always possible to make a prime froma composite? I tried and it seems pretty complicated. It works for all the values iI tried though. Nathan Ramesh · 3 years, 4 months ago

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@Nathan Ramesh Yea, that's what I'm thinking too. It has worked for every single case I've tried, but I can't seem to prove it. I've been trying to find a case where the successive numbers turn into a recursive loop, therefore disproving the conjecture of all composites turn prime, but no luck so far.

The tricky part is the concatenation, can't seem to find a mathematically rigorous way to define it. Daniel Liu · 3 years, 4 months ago

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@Daniel Liu Yeah, it's hard to do it by just randomly looking for one.

I had some questions.

What is the maximum number of steps to get to a prime?

Do numbers ever get smaller?

Will this produce every prime if we do every number? Which specific numbers will it not produce? Nathan Ramesh · 3 years, 4 months ago

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@Nathan Ramesh Be aware! Maybe this will become Takeda's Conjecture and it may be remain unsolve for centuries! Haha just kidding... Christopher Boo · 3 years, 4 months ago

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@Christopher Boo Takeda's conjecture :) But I don't think this conjecture will actually be proved as there seems to be no way to define the concatenation part Tan Li Xuan · 3 years, 4 months ago

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In the case of 12=2x3x3 and 233 is prime... Heder Oliveira Dias · 3 years, 4 months ago

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I don't think you can even prove that there are infinitely many composite numbers that can be reduced to a prime... Also, I don't think you can prove there are infinitely many numbers that can't be reduced to a prime... This looks like an operation that, no matter what question you ask, it will probably be an open one, and very difficult to solve. Unless I'm missing some trivial solution here, but I doubt it. David Austen · 3 years, 6 months ago

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I dont understand...what is the purpose of doing this?? Tanya Gupta · 3 years, 4 months ago

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3 = \(1 \times 3\), get 13. That's very interesting! Siao Chi Mok · 3 years, 9 months ago

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@Siao Chi Mok actually, \( 3 = 3 \) and that's the end of it. The prime factorization doesn't include 1's, in that case you could put as many 1's as you like. David Austen · 3 years, 6 months ago

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@Siao Chi Mok If 1 could be included in a prime factorization then \( 3 = 1 \times 3 , 3 = 1 \times 1 \times 3 , ...... \) so 1 cannot be included in a prime factorization ( Also 1 is not a prime ) Tan Li Xuan · 3 years, 4 months ago

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