1. Total number of ways to choose two different numbers

From the set $\{0,1,2,3,\dots,10\}$ there are $11$ elements.

The number of ways to choose any two different numbers is $\binom{11}{2}=55.$

2. Required condition

Let the chosen numbers be $a$ and $b$. We need both: $a+b \equiv 0 \pmod 4$ and $|a-b| \equiv 0 \pmod 4.$

Since $|a-b|$ being a multiple of $4$ means $a-b \equiv 0 \pmod 4,$ we get $a \equiv b \pmod 4.$

So both numbers must belong to the same residue class modulo $4$.

Now if $a \equiv b \pmod 4$, then $a+b \equiv 2a \pmod 4.$ For this to be divisible by $4$, we need $2a \equiv 0 \pmod 4 \implies a \equiv 0 \text{ or } 2 \pmod 4.$

So the two numbers must be chosen from the same modulo-$4$ class, and that class must be either:

• remainder $0$, or
• remainder $2$.

3. List numbers by residue modulo 4

From $0$ to $10$:

• Class $0 \pmod 4$: $\{0,4,8\}$
• Class $1 \pmod 4$: $\{1,5,9\}$
• Class $2 \pmod 4$: $\{2,6,10\}$
• Class $3 \pmod 4$: $\{3,7\}$

Only classes $0$ and $2$ work.

4. Count favorable pairs

From class $0 \pmod 4$: $\binom{3}{2}=3$ namely $(0,4),(0,8),(4,8)$.

From class $2 \pmod 4$: $\binom{3}{2}=3$ namely $(2,6),(2,10),(6,10)$.

So total favorable pairs: $3+3=6.$

5. Probability

Therefore, $P=\frac{6}{55}.$

6. Compare with stored answer

Stored correct answer is D: $\frac{6}{55}$.

This matches the derived answer.