Hatred of Eleven II

It's a well known fact that the divisibility rule for 11 is to take the alternating sum of the digits of the number, and if that number is divisible by 11 then your original number is divisible by 11.

For example, let $n = 6941$ :

$6-9+4-1 = 0 \Longrightarrow 6941 \equiv 0$ (mod $11$ )

Let us set $n = \overline{ABCDE}$ . This gives us the following conditions:

$11\nmid A-B+C-D+E$

$11\mid A-B+C-D$

$11\mid A-B+C-E$

$11\mid A-B+D-E$

$11\mid A-C+D-E$

$11\mid B-C+D-E$

What if we added up the bottom 5 conditions? This number should still be divisible by 11.

$11\mid 4A-2B+2D-4E$

Now, take away $2\times(A-B+D-E)$

$11\mid 2A-2E$

Now $2A-2E$ is clearly an even number, and because $A$ and $E$ can only assume values from $0$ to $9$ , the only number divisible by $11$ that this expression can assume is $0$

$2A-2E = 0\Longrightarrow A=E$

Now look back at our other conditions. Allowing $A=E$ , we can simplify many of them.

$11\mid A-B+C-E \Longrightarrow 11\mid -B+C$

$11\mid A-B+D-E \Longrightarrow 11\mid -B+D$

$11\mid A-C+D-E \Longrightarrow 11\mid -C+D$

These sums can only assume values in the interval $[-9,9]$ , meaning that if they are divisible by $11$ , they must all equal $0$ .

$-B+C = 0\Longrightarrow B=C$

$-B+D = 0\Longrightarrow B=D$

$-C+D = 0\Longrightarrow C=D$

Therefore, $B=C=D$

Analyzing our last condition taking into account our newfound knowledge:

$11\mid B-C+D-E\Longrightarrow 11\mid B-E \Longrightarrow B=E$

This gives us $A=B=C=D=E$ , greatly narrowing our options of 5-digit numbers to choose from. Just to be sure that all of these work, we must refer to our first condition:

$11\nmid A-B+C-D+E \Longrightarrow 11\nmid A-A+A-A+A \Longrightarrow 11\nmid A$

This condition is satisfied for all $A$ in the interval $[1,9]$ , so we're safe to assume that as long as $A=B=C=D=E$ , we have a solution. Because there are 9 integers in the interval $[1,9]$ , our solution is just that: $\boxed{9}$ .

Let us say that the number is $\overline{abcde}$ ,then we have, $\begin{aligned}\overline{abcd}\equiv 0\pmod{11}\\ \overline{abce}\equiv 0 \pmod{11}\\ \Longrightarrow \overline{abcd}-\overline{abce}\equiv 0 \pmod{11}\end{aligned}$ ,expanding the numbers and subtracting we get, $\begin{aligned}d-e\equiv 0 \pmod{11}\\ \Longrightarrow=d-e=11x[x\in {non-negative\ integers}]\\ but\ d-e<10\\ \Longrightarrow d-e=0\\ \Longrightarrow d=e\end{aligned}$ .Now,consider, $\begin{aligned}\overline{abce}\equiv 0 \pmod{11}\\ \overline{abde}\equiv 0\pmod{11}\\ \overline{abce}-\overline{abde}\equiv 0 \pmod{11}\end{aligned}$ .Again,expanding and subtracting we get, $\begin{aligned}10c-10e=11x[x\in {non-negative\ integers}]\\ \Longrightarrow x\equiv 0\mod{10}\\ but,10c-10e<110\\ \Longrightarrow 10c-10e=0\\ \Longrightarrow c=e\end{aligned}$ ,hence $c=d=e$ but, $\overline{accc}\equiv 0 \mod{11},hence\ a=c$ .So we have that our number is $aaaaa$ ,there a total of 9 such numbers.Cheers!