1994 AMC 8 第 23 题

先试着解答 1994 AMC 8 第 23 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 1994 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。

23.

如果 XX,YY 和 ZZ 是不同数字,那么下列加法中可能得到的最大 33 位和
XXXYX+X\begin{array}{cr} & XXX \\ & YX \\ + & X \\ \hline \end{array}
具有哪种形式?
If X,X, Y,Y, and ZZ are different digits, then the largest possible 33-digit sum for
XXXYX+X\begin{array}{cr} & XXX \\ & YX \\ + & X \\ \hline \end{array}
has the form
XXYXXY
XYZXYZ
YYXYYX
YYZYYZ
ZZYZZY
答案:D
知识点:数字谜位值
小提示:
按列写出加法;如果 XX 太大,百位会进位,结果就变成四位数
Write the addition in columns; if XX is too large the hundreds place carries and the answer gains a fourth digit.
大提示:
先取能使和仍为三位数的最大 XX,再选 YY 使总和尽可能大,并看结果数字对应哪些字母
Take the largest XX that keeps the sum to three digits, then choose YY to make the total as large as possible and see which letters the resulting digits match.
解答:
和的百位来自 XX 加上可能的进位,所以若 X=9X = 9,和会变成四位数。允许的最大值是 X=8X = 8。
为了使和最大,取 Y=9Y = 9:此时 888+98+8=994888 + 98 + 8 = 994。它的数字是 99、99、44;因为 9=Y9 = Y,而 44 是新数字 ZZ,所以和的形式是 YYZYYZ。
所以正确答案是 D。
The hundreds digit of the sum comes from XX plus any carry, so if X=9X = 9 the sum would spill over into four digits. The largest allowed value is X=8.X = 8.
To make the sum as large as possible, take Y=9:Y = 9: then 888+98+8=994.888 + 98 + 8 = 994. Its digits are 9,9, 9,9, 4;4; since 9=Y9 = Y and 44 is a new digit Z,Z, the sum has the form YYZ.YYZ.
Thus, the correct answer is D .
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