A strange property about modulus

Question

I came across this strange property :

A = 1365636947370912401969102170896050049833536043036222945024106106365095957448673476595717951882200
B = 1353221295111631852153727244194823890765967481385223615864962219103729677003203245224034685374611

1. Why $0\notin \{(B \times 11^n \bmod A) \bmod 11 \mathrel: n \in \mathbb N \} = \{ 1,2,3,\dots,10\} \text{?}$
2. Is there a more general property behind it?

Answer 1

As it was guessed in the comments, the important feature of the given $A$ is that $A$ is composed of prime powers, modulo each of which the multiplicative order of $11$ belongs to the set formed by first 9 primes $p_1=2$, ..., $p_9=23$. Together they give a huge multiplicative order of $11$ modulo $A$ equal the primorial $2\cdot 3\cdots 23=223092870$, and it is the period length of base-11 expansion of the fraction $B/A$. The period itself as an integer is given by $$T:=B\cdot \frac{11^{223092870}-1}{A}.$$ The "strange property" referred to in the question can be stated as all base-11 digits of $T$ are nonzero.

It may be surprising that one can achieve nonzero digits in the whole period with $A,B$ of relatively small base-11 length (as compared to period length $223092870$), and here I give a possible construction for such numbers. I did not reverse engineer the OP's approach, but I suspect it was something similar.

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The idea is to construct $\frac{B}{A}$ as the sum of fractions $$\sum_{i=1}^9 \frac{N_i}{11^{p_i}-1},$$ where numerators $N_i$ are to be determined and should be coprime to the corresponding denominators.

The period length of base-11 expansion of $\frac{N_i}{11^{p_i}-1}$ is $p_i$, and they are clearly pairwise coprime. Hence, the elements of the resulting period (i.e. the digits of $T$) are the sums of all possible combinations of nine elements taken one from each of these small periods plus a carry, taken modulo 11.

We fix base-11 digits of each of the small periods be from the set $\{9,10\}$, which "magically" gives a fixed carry equal 8, and sum of any nine such digits plus the carry results in the interval $[89,98]$. Note that the numbers in this interval do have a fixed quotient 8 (the carry) from division by 11, while there is no residue 0 modulo 11.

More specifically, we can set $$N_i := \frac{11^{p_i}-1}{10}\cdot 9 + 1,$$ which have all base-11 digits equal 9, except the last one being 10. They give the following $A,B$ as the denominator and numerator of the fractional part of the sum stated above:

A = 1365636947370912401969102170896050049833536043036222945024106106365095957448673476595717951882200
B = 148979346996371790012285143790831164050922165954621623661554497897875876190337579031255061695809

Just in case, I have computationally verified that the base-11 digits of the corresponding $T$ are nonzero.