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When I was doing some hexadecimal calculation, I found that a trick that I've known that it exists in decimal arithmetic, also exists in hexadecimal one, that is, the shortcut for multiplying by 11.

Here are some examples, multiply some decimals by 11, giving that: \begin{align*} 11_{10} \times{} 11_{10} &:= 1\ (1+1)\ 1 = 121_{10} \\ 62_{10} \times{} 11_{10} &:= 6\ (6+2)\ 2 = 682_{10} \\ 65_{10} \times{} 11_{10} &:= 6\ (6+5)\ 5 = 715_{10} \\ 78_{10} \times{} 11_{10} &:= 7\ (7+8)\ 8 = 858_{10} \\ \end{align*}

While treating them as hexadecimals, then multiply them by hexadecimal $11_{16}$ as well, giving that: \begin{align*} 11_{16} \times{} 11_{16} &:= 1\ (1+1)\ 1 = 121_{16} \\ 62_{16} \times{} 11_{16} &:= 6\ (6+2)\ 2 = 682_{16} \\ 65_{16} \times{} 11_{16} &:= 6\ (6+5)\ 5 = 6\text{B}5_{16} \\ 78_{16} \times{} 11_{16} &:= 7\ (7+8)\ 8 = 7\text{F}8_{16} \\ \end{align*}

I don't think this is of a coincidence, there must be a rigorous mathematical proof that supports it. However I have no idea how to prove this kind of things, could anyone give some suggestions? Thanks in advance!

Bill Dubuque
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Michael Lee
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    Just think about what $11_b$ means; it represents $b + 1$. So if you multiply $(\alpha \beta)_b$ by $11_b$, you get precisely $(\alpha b + \beta)(b + 1) = \alpha b^2 + (\alpha + \beta) b + \beta$. – Jakob Streipel Nov 26 '21 at 03:50
  • Can you come up with the rigorous proof for base $10$? If the answer is yes, then you will be able to extend that proof to any base : just change $10$ to the desired base. You can also extend this pattern to $111,1111$ and so on. – Sarvesh Ravichandran Iyer Nov 26 '21 at 03:51
  • @prets Thanks! A short but thorough explanation, it really helps me getting out of this thing. Could you post it as answer? :) – Michael Lee Nov 26 '21 at 04:09
  • @TeresaLisbon Thank you for the suggestion! I'm trying to generalize it to other base/factor, just as you said. – Michael Lee Nov 26 '21 at 04:11
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    Because radix notation is a polynomial function of the radix, you are essentially observing the effect of multiplying a polynomial by $x+1$ (when the digit sums don't carry). For a more interesting example see here on the multiplicativity of reversing the digits, e.g. $,12\cdot 13 = 156\Rightarrow 21\cdot 31 = 651\ \ $ – Bill Dubuque Nov 26 '21 at 04:50
  • @BillDubuque Thanks for clarifying the things. And appreciate that earlier answer for polynomial reversal, just give me a new insight into the daily-used numbers! – Michael Lee Nov 26 '21 at 06:59
  • @MichaelLee I have done so. – Jakob Streipel Nov 26 '21 at 15:22

2 Answers2

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The following is in the base that you are working with

When you take a $2$ digit number $\overline{AB}$ and multiply it with $11$, this is the same as $$\overline{AB}\cdot (10+1)$$ $$=\overline{AB0}+\overline{AB}$$ $$=A\cdot 100+(A+B)\cdot 10+B$$

You can also get similar results with more digits.

Alan Abraham
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Comment turned answer, on OP's request.

What $11_b$ means in a given base $b$ is the number $b + 1$. The two "digit" number $\alpha\beta$ in the same base, let's denote it $(\alpha \beta)_b$ similarly means $$ (\alpha \beta)_b = \alpha b + \beta. $$ Now multiplying these two quantities together and distributing, we get $$ (\alpha \beta)_b \times 11_b = (\alpha b + \beta)(b + 1) = \alpha b^2 + (\alpha + \beta) b + \beta $$ which, back in base $b$ notation, is then $(\alpha(\alpha+\beta)\beta)_b$ (unless the central "digit" overflows, as noted in some of OP's calculations).

Jakob Streipel
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