1. **State the problem:** We have the addition of three 3-digit numbers where $x$ and $y$ are single digits, and the sum is a 4-digit number: $$ \begin{array}{cccc} &7 & 7 & x \\ +&6 & y & x \\ +& & y & y & x \\ \hline 1 & x & x & 7 \\ \end{array} $$ We need to find the value of $x + y$. 2. **Analyze the units column:** The units digits add up to the units digit of the sum. Units column: $x + x + x = 3x$. The units digit of the sum is 7, so: $$3x \equiv 7 \pmod{10}$$ Try values of $x$ from 0 to 9: - $3 \times 9 = 27$ ends with 7, so $x=9$. 3. **Analyze the tens column:** The tens digits add up plus any carry from the units column. Tens column digits: $7 + y + y = 7 + 2y$ plus carry from units column. Since $3x = 27$, carry from units column is 2. Sum in tens column: $$7 + 2y + 2 = 9 + 2y$$ The tens digit of the sum is $x = 9$, so the tens digit of the sum is 9. Therefore: $$9 + 2y \equiv 9 \pmod{10}$$ Simplify: $$2y \equiv 0 \pmod{10}$$ Possible $y$ values satisfying $2y \equiv 0 \pmod{10}$ are $y=0,5$. 4. **Analyze the hundreds column:** The hundreds digits add up plus carry from tens column. Hundreds digits: $7 + 6 + y = 13 + y$ plus carry from tens column. Carry from tens column is: $$\frac{9 + 2y}{10}$$ Since $9 + 2y$ is either 9 (if $y=0$) or 19 (if $y=5$), carry is 0 or 1 respectively. Sum in hundreds column: - If $y=0$: $13 + 0 + 0 = 13$ (carry 1, digit 3) - If $y=5$: $13 + 5 + 1 = 19$ (carry 1, digit 9) The hundreds digit of the sum is $x = 9$. So the hundreds digit of the sum is 9, which matches $y=5$ case. 5. **Analyze the thousands column:** The thousands digit of the sum is 1. Carry from hundreds column is 1 (from previous step). Sum in thousands column is carry only: $$1 = 1$$ This is consistent. 6. **Conclusion:** We found $x=9$ and $y=5$. Therefore: $$x + y = 9 + 5 = 14$$ **Final answer:** 14