Are there any two numbers such that multiplying them together is the same as putting their digits next to each other?

Question

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

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

From trying out a lot of different combinations, it seems as though putting the digits of the numbers together always overestimates, but I have not been able to prove this yet.

So, I have 3 questions:

1. Does putting the digits next to each other always overestimate? (If so, please prove this.)
2. If it does overestimate, is there any formula for computing by how much it will overestimate in terms of the original inputs A and B? (A proof that there's no such formula would be wonderful as well.)
3. Are there any bases (not just base 10) for which there are such numbers? (Negative bases, maybe?)

Answer 1

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 $A\times B = A\times10^k + B$ 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.

Answer 2

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.

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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.

Answer 3

If $B$ has $n$ digits then $10^{n-1} \le B <10^n$ and we want $AB = 10^nA + B$ or

But $B<10^n$ so $AB < 10^nA \le 10^nA + B$

So 1) Yes over compensation always

2) by $10^nA + B - AB = 10^{[\log_{10}(B)]+1}A + B - AB$

3) The same argument applies to any base $> 1$

Answer 4

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, $a\cdot b\leq a\cdot10^{\lceil\log_{10}(b)\rceil}$ but that $a*b>a\cdot10^{\lceil\log_{10}(b)\rceil}$. If $a=0$ and $b\neq 0$ and we ignore leading zeroes, then we are still an overestimation, and if $a\neq0$ 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.

Answer 5

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 $x=4$

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

Answer 6

(1) A slightly different method:

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

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

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

Recall that by the definition of logarithm, $k^{log_kB}=B$. $\:$ Since$\lceil x\rceil\geq x$, we know $\frac{k^{\lceil log_{k}B\rceil}}{B}\geq1$. 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.

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(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 $0\leq d_i\leq9.$

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.

Answer 7

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.