2003 AMC 8 Problem 14

Attempt Problem 14 of the 2003 AMC 8 below, then check your answer against the professionally curated solution from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2003 AMC 8 solutions, or check the answer key.

All problems are used with official legal permission of the Mathematical Association of America (MAA).

14.

In this addition problem, each letter stands for a different digit. TWO+TWOFOUR\begin{array}{cccc}&T & W & O\\ +&T & W & O\\ \hline F& O & U & R\end{array} If T=7T = 7 and the letter OO represents an even number, what is the only possible value for W?W?
00
11
22
33
44
Answer: D
Concepts:cryptarithmcasework
Small hint:
Use the hundreds column after T=7.T = 7.
Big hint:
The ones column and the fact that OO is even force O.O.
Solution:
Since both TT’s are 7,7, we get that OO is either 44 or 5.5. Since OO is even, we get that O=4.O = 4.
Then, we get that R=4+4=8.R = 4 + 4 = 8. We also know that W+WW + W doesn’t carry over, since otherwise OO would be 5.5.
Therefore, WW is less than 55 and cannot be 44 or 1.1. If W=0,W = 0, then U=0,U = 0, which gives two letters the same digit. If W=2,W = 2, then U=4,U = 4, which is also not allowed.
This makes W=3.W = 3.
Thus, D is the correct answer.
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