Can we factor an integer N if we know some of the digits of an associated number M?

Question

What is the minimum number of digits m that is needed to build the factors of the associated number M?

We start with a number $$N=pq$$ and we try to build an associated number $$M=rs$$ that has the following format in digit terms: $$ M=rs=(pq),(p+q),(01)$$ Let's give an example to clarify what is meant by an associated number M with the right digit format. We start with $$ N=5\cdot 11=55$$ we know that $$(p+q)=5+11=16$$ so we want to build an associated number M with the following digit format $$M=55(p+q)**01$$ where ** are additional zeros if needed. At this point we don't really know that, in terms of digits of M, $$ (p+q)=16$$ or the number of zeros before the $$01$$ But we know for sure that M has the form $$M=rs=55****01$$ where the stars stand for the unknown digits of (p+q) and unknown zeros if more zeros are needed. It turned out that it is (always?) possible to build associated numbers M ( that is of course if we knew the factors ). In this particular case $$M=rs=501\cdot 1101=551601$$ In this particular case, it would have been easy to build M without knowing the factors of N by simply guessing $$(p+q)$$ from $$2sqrt(N)$$ and assuming that the last two digits of M are $$01$$ But even a small number like $$N=7*97=679$$ shows some of the difficulties related to building the associated number M. The associated number M must have the following format $$M=679*****01$$ the number of stars does not correspond to the number of missing digits of course. We know the format of M because we are working backward (we start from the solution) and it is: $$M=7001\cdot 97001=679104001$$ So we see that we were missing 4 digits out of 9 digits of M. We also see that we started with $$ N=7*97=679 $$ a 3 digit number and ended up with an associated number M $$M=679104001$$ with 9 digits.

At this point very little is known about the algorithm to build M from N beside what was said above. In particular, the relation between the number of digits of N and that of M is not known. Is-it $$3*n$$ with n the number of digits of N or some other relation?

So the real question is: can we build a 9 digit number if we only knew 5 of its digits? And of course, at this point I don't know if the proportion of unknown to known digits is constant.

Answer 1

$N = pq$

$M = (pq)(p+q)(0\dots 01) = rs$

It looks to me that you indend to have, somewhat using your notation,

$r = (p)(0\dots 01)$ and $s = (q)(0\dots 01)$

Let's say $r=p10^A+1$ and $s = q10^B + 1$.

If $(p+q)$ is going to be in the middle, then we must have $A=B$. So

$r=p10^A+1$ and $s = q10^A + 1$.

Hence $M = rs = pq10^{2A} + (p+q)10^A + 1$

The diagram below shows why $A$ needs to be the number of digits in $p+q$.

$$M = (pq) \underbrace{ \overbrace{(p+q)}^\text{A-digits} \overbrace{(0\dots 01)}^\text{A-digits}}_\text{2A-digits}$$

So $r=p10^A+1 \; \text{and} \; s = q10^A + 1 \; \text{where $A$ is the number of digits in $p+q$}$.

You asked

So the real question is: can we build a 9 digit number if we only knew 5 
of its digits? And of course, at this point I don't know if the proportion   
of unknown to known digits is constant.

Do you mean can we build a $9-$digit $N$ if we only know $5$ of its digits?

Well, if $pq$ is a $B-$digit number, then we need the first $A+B$ digits of N to find what $p$ and $q$ are. This can get tricky, however, if $p+q$ ends in some zeros; for example, $p+q = 320$.

For example, if you have $M=679104001$ but you only see $M = 67910****$, then the only possibility is

$pq = 679$ and $p+q = 10*$.

The only possibilities are $\{p,q\}=\{1, 679\}$ and $\{p,q\}=\{7, 97\}$. It isn't hard to see that the second option is the correct one.

Consider $N = 360$ and $M = 3602**1$, a $7-$-digit number.

Then we know pq = (360)(3*)(01)
So pq = 360 and p+q = 3*.

Assuming $p<q$, these are only two possibilities for p and q such that p+q = 3*:

\begin{align} p=\color{red}{15}, q=\color{blue}{24} &\implies \left\{ \begin{array}{l} p+q=\color{green}{39} \\ M=(360)(\color{green}{39})(01) \\ r=(\color{red}{15})(01) \\ s=(\color{blue}{24})(01) \\ \end{array} \right. \\ p=\color{red}{18}, q=\color{blue}{20} &\implies \left\{ \begin{array}{l} p+q=\color{green}{38} \\ M=(360)(\color{green}{38})(01) \\ r=(\color{red}{18})(01) \\ s=(\color{blue}{20})(01) \\ \end{array} \right. \\ \end{align}

In general, if you know pq, then M=(pq)(...)(0...01) where pq = ... and 0...01 have the same number of digits. Since you know what pq is, then there are only a finite number of possibilities for p and q. Hence there are only a finite number of possibilities for the value of p+q.

In general,if you are given $N = pq$ then there are as many possibilities for $r, s, M=rs$ as there are divisors $p \le q$ such that $N=pq$. For example, if you know that $pq = 1001$:

\begin{array}{|rrrr|rrr|c|} \hline N & p & q & p+q & r & s & M & \% \text{ of digits in M}\\ \hline 1001 & 1 & 1001 & 1002 & 10001 & 10010001 & 100110020001 & 50\\ 1001 & 7 & 143 & 150 & 7001 & 143001 & 1001150001 & 60 \\ 1001 & 11 & 91 & 102 & 11001 & 91001 & 1001102001 & 60 \\ 1001 & 13 & 77 & 90 & 1301 & 7701 & 10019001 & 75 \\ \hline \end{array}