In the addition shown, $P, Q$ and $R$ each represent a single digit, and the sum is 2009 . The value of $P+Q+R$ is
(A) 9
(B) 10
(C) 11
(D) 12
(E) 13

![](https://cdn.mathpix.com/cropped/2024_04_20_6027bc27089ed4fc493cg-060.jpg?height=154&width=249&top_left_y=2292&top_left_x=1296)
In the ones column, $P+Q$ ends in 9. So $P+Q=9$ or $P+Q=19$.

Since $P$ is at most 9 and $Q$ is at most 9 , then $P+Q$ is at most 18 .

Therefore, $P+Q=9$ since $P+Q$ cannot equal 19 .

In the tens column, since $Q+Q$ ends in zero, $Q+Q$ equals 0 or 10 .

Therefore, either $Q=0$ or $Q=5$.

If $Q=0$, there would be no "carry" to the hundreds column where $P+Q$ (plus no carry) ends in a zero.

This is not possible since we already determined that $P+Q=9$.

Therefore, $Q=5$ and $P=4$, giving a 1 carried from the tens column to the hundreds column. In the hundreds column, we have $1+4+5$ which gives a 1 carried from the hundreds column to the thousands column.

Then $R$ plus the 1 carried equals 2 , so $R=1$. Thus, $P+Q+R=4+5+1=10$.

| $P Q P$ |
| ---: |
| $+\quad R Q Q Q$ |
| 2009 |$\quad$| 454 |
| ---: |
| 2055 |
| 2009 |

ANSWER: (B)