In the addition shown, $P$ and $Q$ each represent single $77 P$ digits, and the sum is $1 P P 7$. What is $P+Q$ ?
(A) 9
(B) 12
(C) 14
(D) 15
(E) 13
The sum of the units column is $P+P+P=3 P$.

Since $P$ is a single digit, and $3 P$ ends in a 7 , then the only possibility is $P=9$.

This gives:

![](https://cdn.mathpix.com/cropped/2024_04_20_748fdafcde2a47412608g-209.jpg?height=206&width=285&top_left_y=431&top_left_x=974)

Then $3 P=3 \times 9=27$, and thus 2 is carried to the tens column.

The sum of the tens column becomes $2+7+Q+Q$ or $9+2 Q$.

Since $9+2 Q$ ends in a 9 (since $P=9$ ), then $2 Q$ ends in $9-9=0$.

Since $Q$ is a single digit, there are two possibilities for $Q$ such that $2 Q$ ends in 0 .

These are $Q=0$ and $Q=5$.

If $Q=0$, then the sum of the tens column is 9 with no carry to the hundreds column.

In this case, the sum of the hundreds column is $7+6+Q$ or 13 (since $Q=0$ ); the units digit of this sum does not match the 9 in the total.

Thus, we conclude that $Q$ cannot equal 0 and thus must equal 5 .

Verifying that $Q=5$, we check the sum of the tens column again.

Since $2+7+5+5=19$, then 1 is carried to the hundreds column.

The sum of the hundreds column is $1+7+6+5=19$, as required.

Thus, $P+Q=9+5=14$ and the completed addition is shown below.

| 779 |
| ---: |
| 659 |
| $+\quad 559$ |
| 1997 |

ANSWER: (C)