Let $A$ and $B$ be integers with the same number of digits (like 35 and 92). Why is it that $$\underbrace{AA\ldots A}_{W\ A's} \cdot \underbrace{BB\ldots B}_{Z\ B's} = \underbrace{AA\ldots A}_{Z\ A's}\cdot \underbrace{BB\ldots B}_{W\ B's}?$$ (Here the $A$'s and $B$'s are being concatenated, not multiplied.) For example A=3 and B=8, W=2 and Z=4, why would 33*8888=3333*88? Why does this equation work?

I was drinking coffee when I randomly came up with this so I was not quite successful in finding an answer. I tried explaining this trend using the logarithmic graph below. I was trying to see whether changing the difference between W and Z would have an effect on the trend and it didn't. What I noticed did have a trend was the actual smaller number between W and Z. The larger the smallest of the two was, the more the A_W/A_Z function began to behave like a regular logarithm. Can someone try to explain 1- why the expression even works in the first place and 2- why the division of AW/AN approaches a 10^n where n= the difference between W and N enter image description here


The reason is that $$33\cdot8888 = (3\cdot11)(8\cdot1111) = (8\cdot11)(3\cdot1111) = 88\cdot3333.$$ In general, if $A$ and $B$ have $N$ digits, we have $$\underbrace{AA\ldots A}_{W\ A's} = A\cdot \underbrace{1\underbrace{00\ldots0}_{N-1\ 0's}1\underbrace{00\ldots0}_{N-1\ 0's}100\ldots\ldots01\underbrace{00\ldots0}_{N-1\ 0's}1}_{W\ 1's} = A\cdot X.$$ Let $X$ be the number on the right that is multiplied by $A$. Similarly we define $Y$ as the number on the right of the below equation that is multiplied by $B$. $$\underbrace{BB\ldots B}_{Z\ B's} = B\cdot \underbrace{1\underbrace{00\ldots0}_{N-1\ 0's}1\underbrace{00\ldots0}_{N-1\ 0's}100\ldots\ldots01\underbrace{00\ldots0}_{N-1\ 0's}1}_{Z\ 1's}= B\cdot Y.$$ Putting everything together, we get $$\underbrace{AA\ldots A}_{W\ A's} \cdot \underbrace{BB\ldots B}_{Z\ B's} = (A\cdot X)(B\cdot Y) = (A\cdot Y)(B\cdot X) = \underbrace{AA\ldots A}_{Z\ A's}\cdot \underbrace{BB\ldots B}_{W\ B's}.$$

  • $\begingroup$ Thank you for the great response-and edits . I honestly could not have found this myself. Do you know what a problem like this might be called? I haven't seen problems dealing with the actual number of digits in a number. $\endgroup$
    – Michael
    May 21 '17 at 5:37
  • $\begingroup$ @Michael I'm glad I could help! As far as I know there isn't a name for this specific type of problem -- I think it just falls under the category of elementary number theory. $\endgroup$ May 22 '17 at 1:48

(This is too long to post as a comment on the answer by @fractal1729 .)

It might be worth mentioning that the numbers $X$ and $Y$ are given by geometric series: $$X=10^{N(W-1)}+10^{10(W-2)}+...+10^{1N}+10^{0N}=\frac{10^{NW}-1}{10^N-1}\\ Y=10^{N(Z-1)}+10^{10(Z-2)}+...+10^{1N}+10^{0N}=\frac{10^{NZ}-1}{10^N-1}.$$

Let $A^W$ denote the numeral formed by concatenating $W$ copies of $A$, and let $N$ be the length of $A$. By inspection we have $$\begin{align}A^W &=A^{W-1}10^N + A \end{align}$$ and by applying this recursively, we find $$\begin{align}A^W &=A^{W-1}10^N + A\\ &=(A^{W-2}10^N+A)10^N+A\\ &=A^{W-2}10^{2N}+A\,10^N+A\\ &...\\ &=A\cdot(10^{(W-1)N}+10^{(W-2)N}+...+10^{1N}+10^{0N})\\ &=A\cdot\frac{10^{NW}-1}{10^N-1} \end{align}$$

and similarly for $B^Z$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.