Why does this math trick work?

Question

35 by 11 is 385 because 3+5 is 8, so it's the digit in the middle.

Same for:

72 by 11 is 792 because 7+2 is 9, so it's the digit in the middle.

I see it works because 35 by 10 is 350, or 72 by 10 is 720. The 0 is replaced with the extra digit. The last digit is 5 by 1 or 2 by 1, so it stays the same.

But why should the middle digit be the sum of the first and last?

Answer 1

Let "$ab$" be a two digit number.

Then $ab = 10a + b$ and $ab\times 11 = (10a+b)(10 + 1) =$

$ 10a(10 + 1) + b(10+1) =$

$ (100a + 10a)+ (10b + b) =$

$100a + (10a + 10b)+b =$

$100a + 10(a+b) + b$.

And if $a+b < 10$ we get $100a + 10(a+b) + b = a(a+b)b$.

Not it doesn't work if $a+b \ge 10$. Example $84\times 11= 924$ and $9+4 =13$ and not $2$. But notice that $9+4 -11 = 2$.

If $a+b \ge 10$ you get:

$100a + 10(a+b) + b = 100a + 10([a+b-10] + 10)+b$

$=100a + 100 + 10[a+b-10] + b = (a+1)(a+b-10)b$.

If we write this as $cde$ we have $c+e = a+b + 1 = (a+b-10) + 11 = d+11$.

So you can modify to rule to if $cde = K\times 11$ then either $c +e =d$ or $c+e = d+11$

We can extend this further:

if $M$ is a multiple of eleven then if you add the even position digits together and add the odd position digits together the sums are equal or off by a multiple of $11$.

Answer 2

Let's give the details on an example: $72$ (in base $10$) denotes $7\cdot 10+2$. So, since $11=10+1$, $$72\times 11=(7\cdot10+2)\times10+(7\cdot 10+2)=7\cdot 10^2+(\color{red}{2+7})\cdot10+2=7\cdot 10^2+9\cdot 10+2.$$

Answer 3

When performed in written calculation,

$$ab\times11$$ is

$$\ \ \ \ ab\\ab\\\ \ \overline{acb}$$

so that the digits are $a,a+b,b$. This breaks when there is a carry, e.g. $76\times11=836$.

Answer 4

Let $\,\ \ \begin{align}x&=10\\[.3em] n&=1\end{align}\ \ \,$ in $\,\ \ \begin{align}&\,(\color{#c00}1+\color{#0a0}x)\,(a_n x^n\! +\!\cdots\!+ a_1x+ a_0)\ =\\[.3em] &\color{#0a0}{a_n} x^{n+1}\! +\! (\color{#c00}{a_n}\!+\!\color{#0a0}{a_{n-1}})x^n\! +\!\cdots\!+\! (\color{#c00}{a_1}\!+\!\color{#0a0}{a_0})x\! +\! \color{#c00}{a_0}\end{align}$

i.e. adding a polynomial $\,\color{#c00}f\,$ to its left-shift $\,\color{#0a0}{xf}\,$ adds successive interior coefficients $\,\color{#c00}{a_k}+\color{#0a0}{a_{k-1}}$

Radix arithmetic is a special case simply because radix notation has polynomial form.