Jtdylktuy

I have two natural numbers, $A$ and $B$, such that $A times B = AB$.

Do any such numbers exist? For example, if $20$ and $18$ were such numbers then $20 times 18 = 2018$.

Lets put aside the trivial answer $A=0$ and $B=0$ and consider both $A, B>0$.

You want numbers such that $Atimes B = Atimes10^k + B$ where $k$ is the number of digits of $B$, that is with $10^{k-1}leq B < 10^k$. So you need $B=10^k+dfrac{B}{A}$ with $B<10^k$. From which results $B>10^k$ and $B<10^k$.

So if there's no mistake there, the answer is no.

There is the pathological example $A=B=0$.

For the rest:

Let $B$ have $m$ digits. We have $AB= A*10^m+B$ We want $AB=A* B$,

We have $$A*10^m+B-A*B = A*(10^m-B)+B>B,$$
because $10^m>B$ as $B$ has $m$ digits. So you always overestimate by at least $B$.

From this its also clear, that the result is independent of the chosen
base.

Let me generealise a bit:

If we allow $A$ to be negative, we need to change the condition to $AB=A*10^m-B$.

This leads to
$$A=frac B {10^m-B}, $$
but then the right hand side is positive and again we get a contradiction.

So the last possibility is $B<0$. But then we first need to define $AB$.

A natural way to do this would be $A(-B)$.

Then we have for $A>0$: $AB=A*10^m-B$ which implies
$$A=frac B {10^m-B}. $$
This time, there is no contradiction because of sign issues.

However, now $-B>0$, hence $|10^m-B|>|B|$, so the right hand side is no integer.

The analogous argument gives also a contradiction for $A,B>0$.

So even in the more generalised setting, the answer is no.

I will use $*$ for your concatenation operation, and $cdot$ for true multiplication.

1) If you allow for leading zeroes to be ingored, then $0*0=00=0$. Notice that for all pairs of non-zero natural numbers, $acdot bleq acdot10^{lceillog_{10}(b)rceil}$ but that $a*b>acdot10^{lceillog_{10}(b)rceil}$. If $a=0$ and $bneq 0$ and we ignore leading zeroes, then we are still an overestimation, and if $aneq0$ and $b=0$ then we are still an overstimation.

2) Based on my crude estimates above, yes, there is a way of putting bounds on the size of the overestimation, but I don't know if you can do much better honestly.

3) Also based on my crude estimations, if you replace the logarithms with different bases I'm fairly sure that this shows no base greater than 1 works. Base 1 itself actually has both types of behaviour; $1*11=111>11$, but $111*111=111111<111111111$. Additionally, you have an example of equality in $11*11=1111$. There is of course the issue that $0$ is a strange object to try and work with in base 1, so let's just ignore that for now...

I can't muster the strength of will to try and prove anything for negative bases. I suspect that negative bases less than $-1$ will fail, but it is easy to see that in base $-1$ there are trivial representations that will also give you equality; $11*11=1111$ where everything in sight is $0$ in base 10.

So it was impossible with natural numbers (see other answers)
But if you willing to bend the rules a bit, by making that when decimal numbers are involved $A.a times B.b = AB.ab$

then you can find

$$x.99999999999... times 9.9999999999... = x9.99999999999...$$

or more compactly

$$x.(9) times 9.(9) = x9.(9)$$

this is possible because $x.(9) = x + 1$

and for an example if $x=4$

$5 times 10 = 50 Leftrightarrow 4.(9) times 9.(9) = 49.(9)$

(1) A slightly different method:

Let $k>1$ be an an arbitrary base. We know that $lceillog_kBrceil$ is the number of digits in $B$ in base $k$.

We want to prove if there exists a solution to $A*B=A*k^{lceil log_{k}Brceil}$+B.

Isolate $A$ and $B$, $::1-frac{1}{A}=frac{k^{lceil log_{k}Brceil}}{B}$.
For positive $A$, we have that $1-frac{1}{A} < 1$.

Recall that by the definition of logarithm, $k^{log_kB}=B$. $:$ Since$lceil xrceilgeq x$, we know $frac{k^{lceil log_{k}Brceil}}{B}geq1$. Thus, along with the other answers, we come to a contradiction. One side of the equation is less than 1, the other at least 1.

(3) Now what if $k<0$?

I will first point out that there do exist integers $A$ and $B$ for which $A * B=AB$.

We will use the following definition of negative base 10:

A number of the form $ldots d_2d_1d_0$ with numerical value $ldots+ d_2(-10)^2+d_1(-10)^1+d_0(-10)^0$ where every $0leq d_ileq9.$

Now consider $A=-4_{10}=-4_{-10}$ and $B=8_{10}=8_{-10}$. Discarding the negative sign, $AB=48$.

$A*B=-32_{10}=4(-10)^1+8(-10)^0=48_{-10}$.

Another example:

$A=-9_{10}=-9_{-10}, B=90_{10}=90_{-10}$. $AB =990$.

$A*B=-810_{10}=9(-10)^1+9(-10)^1+0(-10)^0=-990_{-10}$. This time we must include the negative sign, but regardless the digits are the same.

So for your third question, provided that we let the negative sign in the concatenation be optional, we can find examples where multiplying two numbers will yield the concatenation of their digits.

Here's my approach: take two natural numbers $n,m$ with $x,y$ number of digits respectively. Then in particular we can bound $$n cdot m leq (9 cdots text{($x$ times)} cdots 9) cdot (9 cdots text{($y$ times)} cdots 9) = (10^{x} - 1)(10^y - 1),$$
so $n cdot m leq 10^{x+y} - 10^x - 10^y + 1$. Now, we can also bound
$$nm geq (10 cdots text{($x-1$ zeros)} cdots 0)(10 cdots text{($y-1$ zeros)} cdots 0) geq 10 cdots text{($x+y-1$ zeros)} cdots 0,$$ so that $nm geq 10^{x+y-1}$.

This is as far as I've got, there are already better answers around.