2003 AMC 8 第 14 题

先试着解答 2003 AMC 8 第 14 题,然后核对你的答案与精心整理的解答,解答来自 LIVE by Po-Shen Loh。你也可以参加完整限时模拟考试、查看全部 2003 AMC 8 解答,或核对答案。

所有题目均经美国数学协会(MAA)官方合法授权使用。

14.

在这个加法题中,每个字母代表一个不同数字。TWO+TWOFOUR\begin{array}{cccc}&T & W & O\\ +&T & W & O\\ \hline F& O & U & R\end{array} 如果 T=7T = 7,且字母 OO 表示一个偶数,那么 WW 唯一可能的值是多少?
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
答案:D
知识点:数字谜分类讨论
小提示:
由 T=7T = 7 考察百位相加
Use the hundreds column after T=7.T = 7.
大提示:
个位和 OO 是偶数会确定 OO
The ones column and the fact that OO is even force O.O.
解答:
因为两个 TT 都是 77,所以 OO 是 44 或 55。由于 OO 是偶数,得到 O=4O = 4。
于是 R=4+4=8R = 4 + 4 = 8。还知道 W+WW + W 不会进位,否则 OO 会是 55。
因此 WW 小于 55,且不能是 44 或 11。如果 W=0W = 0,则 U=0U = 0,两个字母会代表同一个数字。如果 W=2W = 2,则 U=4U = 4,这也不允许。
所以 W=3W = 3。
所以正确答案是 D。
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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