Divide a googolplex into two numbers with no prime factor under 100

Question

Very likely, a googolplex won't be Goldbached into 2 provably prime numbers in my lifetime. But it seems possible to do a split that can avoid small primes. The first 24 odd primes have a product of $k=1152783981972759212376551073665878035$, and it's easy to calculate that a googolplex mod $k$ is $x =929084311939231970567712621104570410$, which is prime when 509 is subtracted. I'm not sure what the next steps would be.

Can you divide a googolplex into two odd numbers where neither is divisible by any odd prime 97 or under? If so, how high up can you push that before it starts being computationally challenging?

Answer 1

Definitions:

• googolplex: $g = 10^{10^{100}}$;
  
• $k$-rough number: a positive integer whose prime factors are all greater than or equal to $k$ (see Rough number);
  
• primorial $p_n\#$: $\quad p_n\# = \prod\limits_{k=1}^n p_k$, where $p_k$ is $k$-th prime number (see Primorial).

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Suppose we want to write $$g = a+b,\tag{1}$$ where $a,b$ are $20$-rough numbers.

Consider all remainders $g (\bmod p_k)$, where $p_k\le 20$.
And choose remainders for $a$ and $b$, such that

• $a,b \not \equiv 0 (\bmod p_k)$;
• $a+b\equiv g (\bmod p_k)$.

\begin{array}{|c|c|c|c|} \hline p_k & g (\bmod p_k) & a (\bmod p_k) & b (\bmod p_k) \\ \hline 2 & 0 & 1 & 1 \\ 3 & 1 & 2 & 2 \\ 5 & 0 & 1 & 4 \\ 7 & 4 & 1 & 3 \\ 11 & 1 & 2 & 10 \\ 13 & 3 & 1 & 2 \\ 17 & 1 & 2 & 16 \\ 19 & 9 & 1 & 8 \\ \hline \end{array} Note that there are many cases for $a,b$ remainders; here is one of them.

Then (to reconstruct $a$ from its remainders) applying Chinese remainder theorem, one can enter

ChineseRemainder[{1,2,1,1,2,1,2,1},{2,3,5,7,11,13,17,19}]

in Wolfram Alpha and get that

$$a \equiv 8835191 (\bmod 19\#);\tag {2}$$

and

ChineseRemainder[{1,2,4,3,10,2,16,8},{2,3,5,7,11,13,17,19}]

gives us

$$b \equiv 1054679 (\bmod 19\#),\tag{3}$$

where $19\# = p_8\# = 2\cdot 3 \cdot 5 \cdot 7 \cdot 11 \cdot 13 \cdot 17 \cdot 19 = 9699690$.

So we can choose $a = 8835191$, $b = g-a$.

Or $a = 9699690 k + 8835191$, where $k$ is appropriate non-negative integer number.

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For writing googolplex in the form $(1)$ (at least one of examples), where $a,b$ are $100$-rough, we expand previous table to the form: \begin{array}{|c|c|c|c|} \hline p_k & g (\bmod p_k) & a (\bmod p_k) & b (\bmod p_k) \\ \hline 2 & 0 & 1 & 1 \\ 3 & 1 & 2 & 2 \\ 5 & 0 & 1 & 4 \\ 7 & 4 & 1 & 3 \\ 11 & 1 & 2 & 10 \\ 13 & 3 & 1 & 2 \\ 17 & 1 & 2 & 16 \\ 19 & 9 & 1 & 8 \\ 23 & 13 & 1 & 12 \\ 29 & 24 & 1 & 23 \\ 31 & 5 & 1 & 4 \\ 37 & 10 & 1 & 9 \\ 41 & 1 & 2 & 40 \\ 43 & 24 & 1 & 23 \\ 47 & 9 & 1 & 8 \\ 53 & 46 & 1 & 45 \\ 59 & 48 & 1 & 47 \\ 61 & 47 & 1 & 46 \\ 67 & 10 & 1 & 9 \\ 71 & 45 & 1 & 44 \\ 73 & 1 & 2 & 72 \\ 79 & 52 & 1 & 51 \\ 83 & 10 & 1 & 9 \\ 89 & 16 & 1 & 15 \\ 97 & 35 & 1 & 34 \\ \hline \end{array}

Then, applying

 ChineseRemainder[
    {1, 2, 1, 1, 2, 1, 2, 1, 1, 1, 1, 1, 2, 1, 1, 1, 1, 1, 1, 1, 2, 1, 1, 1, 1},
    {2, 3, 5, 7,11,13,17,19,23,29,31,37,41,43,47,53,59,61,67,71,73,79,83,89,97}]

we have $$ a \equiv 1\: 744\: 949\: 340\: 521\: 974\: 476\: 409\: 807\: 187\: 663\: 679\: 281 (\bmod 97\#), $$ where $$97\# = p_{25}\# = 2\: 305\: 567\: 963\: 945\: 518\: 424\: 753\: 102\: 147\: 331\: 756\: 070.$$