In the addition shown, the letters $X, Y$, and $Z$ each represent a different non-zero digit. The digit $X$ is
(A) 1
(B) 2
(C) 7
(D) 8
(E) 9

$X X X$

| $Y$ | $Y$ |
| :--- | :--- | :--- |

$\begin{array}{r}+Z Z Z \\ \hline Z Y Y X\end{array}$
We start by analyzing the given sum as if we were performing the addition

by hand. Doing this, we would start with the units column.

Here, we see that the units digit of the sum $X+Y+Z$ is $X$.

Thus, the units digit of $Y+Z$ must be 0 .

|  |  | 1 |  |
| ---: | ---: | ---: | ---: |
|  | $X$ | $X$ |  |
|  | $Y$ | $Y$ | $Y$ |
| + | $Z$ | $Z$ | $Z$ |
| $Z$ | $Y$ | $Y$ | $X$ |

column.

Because $Y+Z$ are different digits between 1 and 9 , then $Y+Z$ is at most 17. Since the units digit of $Y+Z$ is 0 , then $Y+Z=10$ and there is a carry of 1 to the tens column.

Comparing the sums and digits in the tens and units columns, we see that $Y=X+1$ (since $Y$ cannot be 0 ).

Here are two ways that we could finish the solution.

Method 1

Since $Y+Z=10$, then $Y Y Y+Z Z Z=1110$.

In other words, each column gives a carry of 1 that is added to the next column to the left. Alternatively, we could notice that $Y Y Y=Y \times 111$ and $Z Z Z=Z \times 111$.

Thus, $Y Y Y+Z Z Z=(Y+Z) \times 111=10 \times 111=1110$.

Therefore, the given sum simplifies to $1110+X X X=Z Y Y X$.

If $X=9$, then the sum would be $1110+999=2009$, which has $Y=0$, which can't be the case. Therefore, $X \leq 8$.

This tells us that $1110+X X X$ is at most $1110+888=1998$, and so $Z$ must equal 1, regardless of the value of $X$.

Since $Y+Z=10$, then $Y=9$.

This means that we have $1110+X X X=199 X$.

Since $X$ is a digit, then $X+1=9$, and so $X=8$, which is consistent with the above.

(Checking, if $X=8, Y=9, Z=1$, we have $888+999+111=1998$.)

## Method 2

Consider the tens column.

The sum in this column is $1+X+Y+Z$ (we add 1 as the "carry" from the units column).

|  | 1 | 1 |  |
| :---: | :---: | :---: | :---: |
|  | $X$ | $X$ | $X$ |
|  | $Y$ | $Y$ | $Y$ |
| + | $Z$ | $Z$ | $Z$ |
| 1 | $Y$ | $Y$ | $X$ |

Since $Y+Z=10$, then $1+X+Y+Z=11+X$.

Since $X$ is between 1 and 9 , then $11+X$ is at least 12 and at most 20 .

In fact, $11+X$ cannot equal 20 , since this would mean entering a 0 in 1 Y $Y \quad X$ the sum, and we know that none of the digits is 0 .

Thus, $11+X$ is less than 20 , and so must equal $10+Y$ (giving a digit of $Y$ in the tens column of the sum and a carry of 1 to the hundreds column).

Further, since the tens column and the hundreds column are the same, then the carry to the thousands column is also 1 . In other words, $Z=1$.

Since $Y+Z=10$ and $Z=1$, then $Y=9$.

Since $Y=X+1$ and $Y=9$, then $X=8$.

As in Method 1, we can check that these values satisfy the given sum.

ANSWER: (D)

#