In the sum shown, $P$ and $Q$ each represent a digit. The value of $P+Q$ is
(A) 3
(B) 5
(C) 7
(D) 6
(E) 4

$P Q Q$

$P P Q$

$+Q Q Q$

876
The sum of the units column is $Q+Q+Q=3 Q$.

Since $Q$ is a single digit, and $3 Q$ ends in a 6 , then the only possibility is $Q=2$.

Then $3 Q=3 \times 2=6$, and thus there is no carry over to the tens column.

The sum of the tens column becomes $2+P+2=P+4$, since $Q=2$.

Since $P$ is a single digit, and $P+4$ ends in a 7 , then the only possibility is $P=3$.

Then $P+4=3+4=7$, and thus there is no carry over to the hundreds column.

We may verify that the sum of the hundreds column is $3+3+2=8$, since $P=3$ and $Q=2$.

The value of $P+Q$ is $3+2=5$, and the final sum is shown.