AMC8 2003

AMC8 2003 · Q14

AMC8 2003 · Q14. It mainly tests Logic puzzles.

In this addition problem, each letter stands for a different digit. If T = 7 and the letter O represents an even number, what is the only possible value for W?

在这个加法题中，每个字母代表一个不同的数字。若 T = 7 且字母 O 代表一个偶数，则 W 的唯一可能值是多少？

(A) 0 0

(B) 1 1

(D) 3 3

(E) 4 4

Answer

Correct choice: (D)

正确答案：（D）

Solution

(D) As given, $T=7$. This implies that $F=1$ and that $O$ equals either $4$ or $5$. Since $O$ is even, $O=4$. Therefore, $R=8$. Replacing letters with numerals gives \[ \begin{array}{cccc} & 7 & W & 4\\ +& 7 & W & 4\\ \hline 1& 4 & U & 8 \end{array} \] $W+W$ must be less than $10$; otherwise, a $1$ would be carried to the next column, and $O$ would be $5$. So $W<5$. $W\ne 0$ because $W\ne U$, $W\ne 1$ because $F=1$, $W\ne 2$ because if $W=2$ then $U=4=O$, and $W\ne 4$ because $O=4$. So $W=3$. The addition problem is \[ \begin{array}{cccc} & 7 & 3 & 4\\ +& 7 & 3 & 4\\ \hline 1& 4 & 6 & 8 \end{array} \]

（D）已知 $T=7$。这意味着 $F=1$，并且 $O$ 等于 $4$ 或 $5$。由于 $O$ 是偶数，所以 $O=4$。因此，$R=8$。用数字替换字母得到 \[ \begin{array}{cccc} & 7 & W & 4\\ +& 7 & W & 4\\ \hline 1& 4 & U & 8 \end{array} \] $W+W$ 必须小于 $10$；否则会向下一列进 $1$，而 $O$ 将是 $5$。因此 $W<5$。$W\ne 0$ 因为 $W\ne U$；$W\ne 1$ 因为 $F=1$；$W\ne 2$ 因为若 $W=2$ 则 $U=4=O$；并且 $W\ne 4$ 因为 $O=4$。所以 $W=3$。该加法题为 \[ \begin{array}{cccc} & 7 & 3 & 4\\ +& 7 & 3 & 4\\ \hline 1& 4 & 6 & 8 \end{array} \]

Topics

LM.001 Logic puzzles