3
$\begingroup$

This question already has an answer here:

Prove Divisibility test for 11

"If you repeatedly subtract the ones digit and get 0, the number is divisible by 11"

Example:

11825 -> 1182 - 5 = 1177

1177 -> 117 - 7 = 110

110 -> 11 - 0 = 11

11 -> 1-1 = 0

Therefore 11825 is divisible by 11. Note 11825 = 1075*11

I was thinking that we let x = $a_ka_{k-1}.....a_1a_0$ where the following $a_i$'s are the digits.

$\endgroup$

marked as duplicate by vadim123, Lord_Farin, Dan Rust, Arash, user61527 Oct 12 '13 at 23:56

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

3
$\begingroup$

If you split off the ones-digit and subtract it, you transform the number from $10\cdot a + b$ to $a - b$. Now, $$10\cdot a + b = (11 - 1)\cdot a + b = 11\cdot a - (a-b),$$

so $10\cdot a + b$ is divisible by $11$ if and only if $-(a-b)$ is divisible by $11$, which is the case if and only if $a-b$ is divisible by $11$. Repeat until you have a one-digit number.

$\endgroup$
0
$\begingroup$

Let $x \in \mathbb{N}$. Then there we have $x= a_0 + 10a_1 + \dots + 10^ka_k$ for some $k \in \mathbb{N}$. Is there any neat rule you can find for diving $10^n$ by $11$?

$\endgroup$

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