, 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:

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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.