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?)
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    $\begingroup$ If $B$ has $b$ digits then $B < 10^b$ but $AB > 10^b A$. This argument applies in any (positive) base. $\endgroup$ Dec 8, 2018 at 0:43
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    $\begingroup$ Yes. I admit that was a little unclear given that you're using it to mean concatenation. $\endgroup$ Dec 8, 2018 at 0:48
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    $\begingroup$ @TobyMak that's still overestimating, just like 20 * 18 < 2018 $\endgroup$
    – Pro Q
    Dec 8, 2018 at 0:48
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    $\begingroup$ Just a side note: as multiplication of numbers is commutative, those we would need A=B, because otherwise the concatenations A.B and B.A would yield different results whereas AB and BA are equal. $\endgroup$ Dec 8, 2018 at 18:31
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    $\begingroup$ @PhilippImhof Not necessarily, since you're not asking that the same holds for the reversed order. For instance change a bit the quest and suppose to look for $A, B$ such that $2A\cdot B=A*B$, where * is the concatenation and $\cdot$ the multiplication. Then your reasoning should apply again, but $A=3, B=6$ is a solution. $\endgroup$
    – Del
    Dec 9, 2018 at 11:16

7 Answers 7


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.

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    $\begingroup$ This is really nice. $\endgroup$
    – Randall
    Dec 8, 2018 at 1:34

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.


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 $> 1$

  • $\begingroup$ By AB for part (2), do you mean A*B? $\endgroup$
    – Pro Q
    Dec 8, 2018 at 7:55
  • $\begingroup$ Also, I believe (1) should include the case where A = B = 0 $\endgroup$
    – Pro Q
    Dec 8, 2018 at 8:02
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    $\begingroup$ I don't think we should include $A=B=0$ because $00$ isn't a valid expression. I took it as a given that $A \ge 1$. But if you wish to include that as a case you may. And yes $AB$ is the product of $A$ and $B$. $\endgroup$
    – fleablood
    Dec 8, 2018 at 16:59
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    $\begingroup$ It's perfectly consistant. I always used AB to mean A times B and i never used it to mean anything else. I never used any notation for concatenation. $\endgroup$
    – fleablood
    Dec 9, 2018 at 5:01
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    $\begingroup$ Of course. The left hand side is $AB$ is multiplying the two digits. And $10^nA + B$ the right hand side is concatinating. $\endgroup$
    – fleablood
    Dec 10, 2018 at 2:22

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.

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    $\begingroup$ Those are not base-1 numbers, as in any base b, the allowed digits are from 0 to b-1 $\endgroup$
    – Ben Voigt
    Dec 9, 2018 at 3:45
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    $\begingroup$ @BenVoigt: In unary numeral system, you can use any arbitrary symbol for tallying. If you insist, you can use $0$ as repeated symbol. So zero would be represented with an empty string (not very convenient...), one with $0$, two with $00$ and four with $0000$. $\endgroup$ Dec 9, 2018 at 16:13
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    $\begingroup$ @EricDuminil: That's a fine system and "unary" is a good name for it, but not "base-1", since it has no relationship to positional number systems. It does not qualify as an answer to OP's final question "Are there any bases for which there are such numbers?" $\endgroup$
    – Ben Voigt
    Dec 10, 2018 at 0:31

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 $\lceil\log_kB\rceil$ 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.

(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, $AB=48$.


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.

  • $\begingroup$ Absolutely right - fixed it now. Probably would've been more standard to swap the roles of $x,y$ and $n,m$ but meh shrugs $\endgroup$ Dec 9, 2018 at 21:08

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