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I would like to redefine multiplication by adding a new step. After taking the product, what if we sum the digits. Now the definition of primes would be modified in such that primes are natural numbers (excluding 0) that cannot be generated by our new multiplication. Here is a more mathematical definition of our new multiplication M = {z | x, y $\in \mathbb{N}$, x,y $\neq$ 1 and z $\in$ $\sum_{n=0} ^m {\lfloor\frac{x*y}{10^{n}}\rfloor\mod{10}}$} where m is the number of digits inside x*y. Note: we cannot use 1 as either multiplicand and multiplier and that division would stay the same as it is traditionally meaning the method of getting the division with multiplication operands remains the same. Here's a few small examples: $$3*4 = 12 \rightarrow 1 + 2 = 3 $$ $$5*2 = 10 \rightarrow 1 + 0 = 1 $$ $$52*23 = 1219 \rightarrow 1 + 2 + 1 + 9 = 13 $$

From my intuition, it shouldn't exists a prime number in the new definition of multiplication operator. But how can I Prove or Disprove it?

Specification of the Problem: $\neg \exists$ x $\in$ $N$ s.t. x $\notin$ M.

I've thought of using Proof by Induction to maybe prove the statement which it's not rudimentary for me to applied in this situation. Maybe I'm making things to complicated for myself and such there exist a simple solution that is going over my head. How should I approach it?

NoobCoder212
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  • You shouldn't write $12=1+2$ which is obviously false. Since you are describing the steps for the calculation you defined may be you could write something like $12 \to 1+2$ – jjagmath Sep 28 '23 at 04:24
  • Thank you for pointing it out. – NoobCoder212 Sep 28 '23 at 04:32
  • I overlook that mistake HAHAHAAH – NoobCoder212 Sep 28 '23 at 04:33
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    It would be better to define a new function $f(a,b)=s(a\cdot b)$ where $s(x)$ is the sum of digits of $x$. This is not multiplication so don't call it that. To prove your claim you would have to demonstrate that all numbers can be formed that way. Induction is one approach. Another is to demonstrate a function which takes the desired result and outputs two numbers that, when fed to your function, produce the desired result. – Ross Millikan Sep 28 '23 at 04:45
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    For example, if $n$ is not a normal prime you can factor it as $c \cdot d$, then use a repunit $111111$ with $c\ 1's$ and $10000010000000100000010000001$ with $d\ 1$s and the strings of $0$s of length $c-1$. This will give you a repunit with $cd\ 1$'s from the multiplication and a final result of $cd$ – Ross Millikan Sep 28 '23 at 04:48
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    Just take $111....11 \times 10$ with digit sum $n$? – Calvin Lin Sep 28 '23 at 07:15
  • That was one heck of an oversight in my part. Takes for pointing it out. – NoobCoder212 Sep 28 '23 at 15:02

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