1989 AMC 8 第 14 题

先试着解答 1989 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 1989 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.

把数字 22、44、55、66、99 分别填入这个减法算式的五个方框中,每个数字恰好用一次,可能得到的最小差是多少?
When placing each of the digits 2,2, 4,4, 5,5, 6,6, 99 in exactly one of the boxes of this subtraction problem, what is the smallest difference that is possible?
5858
123123
149149
171171
176176
答案:C
知识点:位值最优化
小提示:
要让差尽量小,就让三位数尽量小,同时让两位数尽量大
Make the three-digit number as small as possible and the two-digit number as large as possible
大提示:
用这些数字能组成的最大两位数使用 99 和 66;再用剩下的数字组成最小三位数
The largest two-digit number from these digits uses 99 and 6;6; build the smallest three-digit number from the digits that remain
解答:
差最小时,上面的三位数应尽量小,下面要减去的两位数应尽量大。
最大的两位数是 9696,使用数字 99 和 66。剩下的数字 22、44、55 能组成的最小三位数是 245245,所以最小差是 245−96=149245 - 96 = 149。
所以正确答案是 C。
The difference is smallest when the three-digit number (the top) is as small as possible and the two-digit number (subtracted) is as large as possible.
The largest two-digit number is 96,96, using the digits 99 and 6.6. The smallest three-digit number from the remaining digits 2,2, 4,4, 55 is 245.245. So the smallest difference is 245−96=149.245 - 96 = 149.
Thus, the correct answer is C .
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