, 2026

Since $P4R$ is a three-digit
integer, then $P\neq0$. $$\begin{array}{ccccc} &&\!\!\!P\!\!\!&\!\!4\!\!&\!\!\!R\!\!\! \\ + &&\!\!\!7\!\!\!&\!\!\!Q\!\!\!&\!\!\!S\!\!\! \\ \hline &&\!\!\!T\!\!\!&\!\!U\!\!\!&\!\!\!1\!\!\! \end{array}$$ If $\text{P gt;2}$,
then $\text{P+7 gt;9}$ and so $\text{T gt;9}$. In this case, $TU1$ would be a four-digit integer which
is not possible, and so $P=2$.

In the hundreds column $P+7=2+7=9$.

So, there can be no carry from the tens column to the hundreds column
for the same reason as just described, and thus $T=9$. The remaining digits are $0$, $3$, $5$, $8$.

The ones (units) digit of the sum is $1$, and so the only possibilities for
$R$ and $S$ are $3$ and $8$, in some order.

Since $3+8=11$, there is a carry of
$1$ from the ones column to the tens
column.

Thus, the sum in the tens column is $1+4+Q=5+Q$.

Since there is no carry from the tens column to the hundreds column,
then $5+Q=U$, which gives $Q=0$ and $U=5$.

The two possible sums are shown below.

$\begin{array}{ccccc} &&\!\!\!2\!\!\!&\!\!4\!\!&\!\!\!3\!\!\! \\ + &&\!\!\!7\!\!\!&\!\!\!0\!\!\!&\!\!\!8\!\!\! \\ \hline &&\!\!\!9\!\!\!&\!\!\!5\!\!\!&\!\!\!1\!\!\! \end{array}$   
$\begin{array}{ccccc} &&\!\!\!2\!\!\!&\!\!4\!\!&\!\!\!8\!\!\! \\ + &&\!\!\!7\!\!\!&\!\!\!0\!\!\!&\!\!\!3\!\!\! \\ \hline &&\!\!\!9\!\!\!&\!\!\!5\!\!\!&\!\!\!1\!\!\! \end{array}$