Problem 42

Since $WXYZ$ is a four-digit positive integer, then $WXYZ \leq 9999$. (In fact $WXYZ$ cannot be this large since all of its digits must be different.) Since $WXYZ \leq 9999$, then $TWUYV \leq 2(9999) = 19998$. Since $T \neq 0$, then $T = 1$. Next, we note that the 'carry' from any column to the next cannot be larger than 1. Thus, we make a chart of possible digits $d$ and the resulting units digit in the sum from $d + d$ with and without a carry of 1. We use this table to first determine the digits $W$ and $Y$. Since the digits in the thousands column are all the same, then the digit $W$ must be 9, since it must be at least 5 to produce a carry to the ten thousands column. We note further that this means that $X \geq 5$ to produce a carry into this column. Also, the digit $Y$ must equal 0 (since the digits $T, U, V, W, X, Y, Z$ are different). This means that there is no carry from the ones column to the tens column. We summarize what we know so far:

\begin{tabular}{r}
$9X10$ \\
$+\quad 9X0 \quad Z$ \\
\hline $19 \quad U \quad 0 \quad V$
\end{tabular}

and $X \geq 5$ and $Z \leq 4$. Since $T = 1$ and $W = 9$, then $Z$ can be 2, 3, or 4, and $X$ can be $5, 6, 7$, or 8. Note that if $X = 5$, then we have $U = 0 = Y$, which is not possible, so $X \neq 5$. If $Z = 2$, then $V = 4$. In this case, we cannot have $X = 6$ (which would give $U = 2 = Z$) or $X = 7$ (which would give $U = 4 = V$) and so $X = 8$, which gives $U = 6$. If $Z = 3$, then $V = 6$. In this case, $X$ cannot equal 6 or 8 and so $X = 7$ (which gives $U = 4$). If $Z = 4$, then $V = 8$. In this case, $X$ cannot equal 7 or 8 and so $X = 6$ (which gives $U = 2$). In summary, there are 3 possible values for $U$, namely, 2, 4, and 6. We can check that the sums $9802 + 9802 = 19604$ and $9703 + 9703 = 19406$ and $9604 + 9604 = 19208$ all satisfy the original problem.