Hatred of Eleven

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 $n = \overline{ABCD}$ . This gives us the following conditions:

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

$11\mid A-B+C$

$11\mid A-B+D$

$11\mid A-C+D$

$11\mid B-C+D$

What would happen if we added the 2nd and 4th conditions together? A number divisible by 11 plus another number divisible by 11 must still be divisible by 11.

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

Because $A$ must be in the interval $[1,9]$ and $D$ must be in the interval $[0,9]$ , $A+D$ must be in the interval $[1,18]$ . The only number divisible by 11 in this interval is 11 itself, so $A+D=11$

What if we added up the bottom 4 conditions? This to should be divisible by 11.

$11\mid 3A-B-C+3D$

Let's subtract $3\times(A+D)$ from this answer:

$11\mid 3A-B-C+3D - 3(A+D) \Longrightarrow 11\mid -B-C$

By the same argument used above with $A$ and $D$ , $-B-C$ must be in the range $[-18,0]$ , meaning that we either have $B+C=0$ or $B+C=11$ .

Let's start by evaluating $B+C=0$ . Neither $B$ or $C$ can be negative, therefore we would have to have the condition $B=C=0$ . But look at our 2nd initial condition:

$11\mid A-B+C$

If we plug $0$ in for $B$ and $C$ , we get $11\mid A$ , which can't be true because A is in the interval $[1,9]$ . Therefore we must default to our other condition, $B+C=11$

We can prove that this doesn't work either. If $B+C=11$ , then $B$ and $C$ must be non-zero because if one was $0$ , the other would have to be $11$ , which is outside both of their intervals. Look at our 3rd condition:

$11\mid A-B+D$

We know that $A+D = 11$ . Taking 11 away from a number divisible by 11 should leave us with a number still divisible by 11, but if we take away $A+D$ we're left with just $11\mid -B$ , which can't be true because B is in the interval $[1,9]$ !

Because we cannot assign a value to $B+C$ , we have a contradiction, meaning that our premise that there are any 4-digit "Eleven Haters" is false in the first place, giving us a final answer of $\boxed{0}$ .