1. **Problem:** Given the 11-digit number $21A3609727B$ is divisible by 44, find the sum of all possible values of $A + B$. 2. **Recall:** A number is divisible by 44 if and only if it is divisible by both 4 and 11. 3. **Divisibility by 4:** The last two digits $7B$ must form a number divisible by 4. 4. Possible values for $B$ (0 to 9) such that $7B$ divisible by 4 are: 72 and 76 (since 72 % 4 = 0, 76 % 4 = 0). So $B$ can be 2 or 6. 5. **Divisibility by 11:** The difference between the sum of digits in odd positions and even positions must be divisible by 11. Positions (from left): Odd: 1st,3rd,5th,7th,9th,11th digits = 2, $A$, 6, 9, 2, $B$ Even: 2nd,4th,6th,8th,10th digits = 1, 3, 0, 7, 7 Sum odd = $2 + A + 6 + 9 + 2 + B = 19 + A + B$ Sum even = $1 + 3 + 0 + 7 + 7 = 18$ Difference = $(19 + A + B) - 18 = 1 + A + B$ For divisibility by 11, $1 + A + B \\equiv 0 \, (mod \, 11)$, so $1 + A + B = 11k$ for some integer $k$. Since $A$ and $B$ are digits (0-9), $1 + A + B$ can be 11 or 0 (not possible since sum is positive). So $1 + A + B = 11$ implies $A + B = 10$. 6. From step 4, $B$ can be 2 or 6. If $B=2$, then $A + 2 = 10$ so $A=8$. If $B=6$, then $A + 6 = 10$ so $A=4$. 7. Possible pairs: $(A,B) = (8,2)$ or $(4,6)$. 8. Sum of all possible $A + B$ values: $10 + 10 = 20$. **Final answer:** 20