Unusual multiplication system

It is always nice to solve a problem. But even more interesting to come up with it. For example such.

In addition to the usual, there is an "unusual" multiplication system. Here are some examples from this system.

$$

$$3 \ cdot4 = 148 \\ 3 \ cdot5 = 185 \\ 3 \ cdot7 = 259 \\ 3 \ cdot8 = 296 \\ 3 \ cdot9 = 333 \ $$

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Question. What is equal to $$$1 \ cdot1$in the "unusual" multiplication system?
The task has a unique solution in the decimal system of calculus. I am not sure that these equalities are found every day. But many received them. And this is exactly the multiplication.

Admittedly, the reaction to the question was similar to the one that O. Bender met, entering the imported market.

In his hand a young man held an astrolabe. “Oh, bayadere, ty-ri-ri, ty-ri-ra!” He sang as he approached the imported market.

There was a lot to do for him. He squeezed himself into a line of vendors who traded on the collapse, put the astrolabe ahead, and began to shout in a serious voice:

- To the astrolabe? Cheap astrolabe for sale! For delegations and women's departments discount.

Unexpected supply for a long time did not give rise to demand.

Moreover, the author was suspected that he secretly hates others. That is absolutely not true. The author does not avoid any wine or table talk. I repeat once again, many people met the given equalities in the study of mathematics. Only they were recorded differently. For example:

$$

$$\ frac {1} {3} \ cdot \ frac {5} {9} = \ frac {5} {27}$$

Or so:

$$

$$. (3) *. (5) =. (185)$$

If you build a multiplication table for numbers $$$\ frac {x} {9} and \ frac {y} {9}$where x and y lie in the range 1 ... 9, then for most pairs the length of the period is 9 and its value is determined by the product $$$12345679 \ cdot x \ cdot y$. Exceptions are pairs, when the numerals in the numerator are divided by 3. For such numbers, the period length will be shorter, and it is such cases that were given as examples of "unusual" multiplication. If we multiply the products from the left part of the examples at the beginning of the article and the number 12345679, then we get the "beautiful periodic" numbers:

$$

$$12 \ cdot 12345679 = 148 148 148 \\ 15 \ cdot 12345679 = 185 185 185 \\ 21 \ cdot 12345679 = 259 259 259$$

Now I understand what is equal to $$$1 \ cdot 1$in the "unusual" multiplication system. Answer $$$012345679$. I note that determining the length of the period of the decimal representation of a simple fraction is not a made-up task. Both F. Gauss and one of the Bernoulli brothers were engaged in this.
In continuation of the topic I will quote Kozma Prutkov:

When throwing pebbles into the water, look at the circles they form: otherwise, such a throwing will be empty fun.

How can the study of periods of inverse prime numbers be useful? In 1903, American mathematician Frank Nelson Cole made a famous report at a meeting of the American Mathematical Society, presenting the number divisors of Mersenne. $$$2 ^ {67} -1$or M67. The minimum factor in the decomposition it finds is 193,707,721. The primepi function for the argument 193,707,721 is 10,749,692. Can you reduce the number of candidates?
One of the methods suggests the above example with “unusual” multiplication. Since, according to the Fermat Minor Theorem $$$2 ^ {p-1} -1$divided by p, where p is a prime number, and the same number is a divisor for $$$2 ^ {67} -1$, we can limit the list of prime numbers to numbers $$$67 \ cdot {K} + 1$. What will it give us a win?
Below are statistics on the number of primes of this type for different upper bounds:
1000 -> 1
10,000 -> 20
100,000 -> 142
1 000 000 -> 1 195
Suppose that for 200,000,000 such numbers will be approximately 200,000, which reduces the number of candidates by 50 times. But this is a small gain. Cole showed that the number of candidates have the form $$$1608 \ cdot {K} + 1$and $$$1608 \ cdot {K} + $ 120$. For an interval of 2 ... 100,000 such numbers is 38, for 2 ... 1,000,000 their number is 300.
Below is the code that I used in my experiments with an “unusual” piece.

Function UnUsualMult (A, B)
denumA = 0;
n = A; r = n% 10;
while n <> 0 do
denumA = denumA * 10 + 9;
n = (nr) / 10;
r = n% 10;
enddo;

denumB = 0;
n = B; r = n% 10;
while n <> 0 do
denumB = denumB * 10 + 9;
n = (nr) / 10;
r = n% 10;
enddo;

AB = 1; denum = 1;
if a <> denumA then
denum = denum * denumA;
AB = AB * A;
endif;

if B <> denumB then
denum = denum * denumB;
AB = AB * B;
endif;

if denum = 1 then
result = "" + AB;
return result;
endif;


result = ""; num = AB * 10;
while TRUE do

while num <denum cycle
num = num * 10;
result = result + "0";
if num = ab then
break;
endif;
end loop;

r = num% denum;
if r = AB then
result = result + (num-r) / denum;
break;
else
result = result + (num-r) / denum;
endif;
num = r * 10;
if num = ab then
result = result + "0";
break;
endif;
enddo;

return result;
Endfunction