, 2013

The sum of the units column is $P+P+P=3P$. Since $P$ is a single digit, and $3P$ ends in a 7, then the only possibility is $P=9$. This gives: [[IMAGE0]] Then $3P=3\times 9=27$, and thus 2 is carried to the tens column. The sum of the tens column becomes $2+7+Q+Q$ or $9+2Q$. Since $9+2Q$ ends in a 9 (since $P=9$), then $2Q$ ends in $9-9=0$. Since $Q$ is a single digit, there are two possibilities for $Q$ such that $2Q$ 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.