Question

Prove the following.

Answer 1

\[
N=10^ka_k+10^{k-1}a_{k-1}+...+10^2a_2+10a_1+a_0
\]
Prove that 11|N if and only if \(a_k-a_{k-1}+a_{k-2}-...+(-1)^ka_0\) is divisible by 11.

Answer 2

I can only proof the k is even case.
\[
\begin{align*}
N&=\sum_{r=0}^k10^ra_r\\
&=\sum_{r=0}^k(-1)^{k-r}a_r+\sum_{r=0}^k\left(10^r-(-1)^{k-r}\right)a_r\\\
\text{If k is even, then}\\
&=\sum_{r=0}^k(-1)^{-r}a_r+\sum_{r=0}^k\left(10^r-(-1)^{-r}\right)a_r\\
&=\sum_{r=0}^k(-1)^ra_r+\sum_{r=0}^k\left(10^r-(-1)^{r}\right)a_r\\
&=\underbrace{\sum_{r=0}^k(-1)^ra_r}_\text{divisible by 11 as given}+\underbrace{11\sum_{r=0}^k\left(\sum_{i=0}^{r-1}10^{r-1-i}(-1)^i\right)a_r}_\text{divisible by 11, duh!}
\end{align*}
\]

Answer 3

how did you pull out 11 in the last line ?

Answer 4

\[
a^n-b^n=(a-b)(a^{n-1}+a^{n-2}b+\ldots +a^2b^{n-3}+ab^{n-2}+b^{n-1})
\]

Answer 5

\[\begin{align} N &= \sum\limits_{r=0}^k a_r10^r \\~\\
&= \sum\limits_{r=0}^k a_r(11-1)^r \\~\\
&= \sum\limits_{r=0}^k a_r \sum \limits_{i=0}^{r}\binom{r}{i}11^{r-i} (-1)^{i} \\~\\
&= \sum\limits_{r=0}^k a_r \left[(-1)^r+\sum \limits_{i=0}^{r-1}\binom{r}{i}11^{r-i }(-1)^{i} \right] \\~\\
&= \sum\limits_{r=0}^k a_r \left[(-1)^r+11\sum \limits_{i=0}^{r-1}\binom{r}{i}11^{r-i-1} (-1)^{i} \right] \\~\\
&= \sum\limits_{r=0}^k a_r (-1)^r + 11 (stuff)\\~\\

\end{align}\]

you're working something like this ? why not simply use the congruence properties ?

Answer 6

since \(10 \equiv -1 \pmod{11}\) we have :
\[N = \sum\limits_{r=0}^k a_r10^r \equiv \sum\limits_{r=0}^k a_r(-1)^r \pmod {11}\]

end of proof.

Answer 7

...........

That's one heck of a proof.

Answer 8

binomial theorem is more natural to apply/remember compared to the identity you were using

if you use binomial theorem nobody will ask you back how/what you did ;)

Answer 9

\[a^n-b^n=(a-b)(a^{n-1}+a^{n-2}b+\ldots +a^2b^{n-3}+ab^{n-2}+b^{n-1})\]

this is also a popular formula but im strongly biased to binomial thm haha! xD

Answer 10

How can you binomial that thing?

Answer 11

Oh I didn't see your comment.

Answer 12

\(10^r = (11-1)^r\)

it is a binomial now, right ?

Answer 13

expanding it gives you terms with powers of 11 except for the last term :

\[(11-1)^r = 11(stuff) + (-1)^r\]