# Why does this math trick work?

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?

• $$(10a + b) \times 11 = (10a + b) \times (10 + 1) = 100a + 10(a + b) + b.$$ – T. Bongers Jan 23 at 23:23
• In 35x11, try writing out 350 on top of 35 vertically and see which places line up when you add them. – Gregory J. Puleo Jan 23 at 23:24
• Notice that it only works when the sum of the first and last digit is less than $10$ for the reasons in the previous comments. For example $11\times 37=407$. – John Douma Jan 23 at 23:26
• @DonThousand Thank you but you responded too quickly. Please refresh the page. – John Douma Jan 23 at 23:28

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 $$11$$.

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. $$76\times11=836$$.

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

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.