2
$\begingroup$

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?

$\endgroup$
  • 7
    $\begingroup$ $$(10a + b) \times 11 = (10a + b) \times (10 + 1) = 100a + 10(a + b) + b.$$ $\endgroup$ – T. Bongers Jan 23 at 23:23
  • 2
    $\begingroup$ In 35x11, try writing out 350 on top of 35 vertically and see which places line up when you add them. $\endgroup$ – Gregory J. Puleo Jan 23 at 23:24
  • 1
    $\begingroup$ 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 $11\times 37=407$. $\endgroup$ – John Douma Jan 23 at 23:26
  • $\begingroup$ @DonThousand Thank you but you responded too quickly. Please refresh the page. $\endgroup$ – John Douma Jan 23 at 23:28
2
$\begingroup$

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 $11$.

$\endgroup$
2
$\begingroup$

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. $76\times11=836$.

$\endgroup$
1
$\begingroup$

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

$\endgroup$
1
$\begingroup$

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.

$\endgroup$

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.