Question

In this addition problem, each letter stands for a different digit. $\begin{array}{cccc}&T & W & O\\ +&T & W & O\\ \hline F& O & U & R\end{array}$ If $T = 7$ and the letter $O$ represents an even number, what is the only possible value for $W?$

$0$

$1$

$2$

$3$

$4$

Answer: D

Concepts:cryptarithm casework

Difficulty rating: 1380

Solution:

Since both $T$ 's are $7,$ we get that $O$ is either $4$ or $5.$ Since $O$ is even, we get that $O = 4.$

Then, we get that $R = 4 + 4 = 8.$ We also know that $W + W$ doesn't carry over, since otherwise $O$ would be $5.$

Therefore, $W$ is less than $5$ and cannot be $4$ or $1.$ If $W = 0,$ then $U = 0,$ which gives two letters the same digit. If $W = 2,$ then $U = 4,$ which is also not allowed.

This makes $W = 3.$

Thus, D is the correct answer.

2003 AMC 8 Problem 14

Answer: D

Difficulty rating: 1380

Solution:

Problem 14 in Other Years