2003 AMC 12B 第 8 题

先试着解答 2003 AMC 12B 第 8 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 12B 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。

8.

令 ♣(x)\clubsuit(x) 表示正整数 xx 的各位数字之和。例如,♣(8)=8\clubsuit(8) = 8,且 ♣(123)=1+2+3=6\clubsuit(123) = 1 + 2 + 3 = 6。有多少个两位数 xx 满足 ♣(♣(x))=3\clubsuit(\clubsuit(x)) = 3?
Let ♣(x)\clubsuit(x) denote the sum of the digits of the positive integer x.x. For example, ♣(8)=8\clubsuit(8) = 8 and ♣(123)=1+2+3=6.\clubsuit(123) = 1 + 2 + 3 = 6. For how many two-digit values of xx is ♣(♣(x))=3?\clubsuit(\clubsuit(x)) = 3?
33
44
66
99
1010
答案:E
知识点:数字分类讨论
小提示:
令 y=♣(x)y = \clubsuit(x)。对两位数 xx,yy 介于 11 和 1818 之间
Let y=♣(x).y = \clubsuit(x). For a two-digit x,x, yy is between 11 and 1818
大提示:
♣(y)=3\clubsuit(y) = 3 迫使 y=3y = 3 或 y=12y = 12;分别数出数字和为这些值的两位数 xx
♣(y)=3\clubsuit(y) = 3 forces y=3y = 3 or y=12;y = 12; count two-digit xx with each digit sum
解答:
令 y=♣(x)y = \clubsuit(x)。因为 x≤99x \le 99,所以 y≤18y \le 18。因此 ♣(y)=3\clubsuit(y) = 3 要求 y=3y = 3 或 y=12y = 12。
数字和为 33 的两位数是 12,21,3012, 21, 30(33 个),数字和为 1212 的两位数是 39,48,57,66,75,84,9339, 48, 57, 66, 75, 84, 93(77 个),共 1010 个。
因此,正确答案是 E。
Let y=♣(x).y = \clubsuit(x). Since x≤99,x \le 99, we have y≤18.y \le 18. Then ♣(y)=3\clubsuit(y) = 3 requires y=3y = 3 or y=12.y = 12.
The two-digit numbers with digit sum 33 are 12,21,3012, 21, 30 (33 values), and those with digit sum 1212 are 39,48,57,66,75,84,9339, 48, 57, 66, 75, 84, 93 (77 values), for 1010 in all.
Thus, the correct answer is E.
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