1987 AMC 12 第 29 题

先试着解答 1987 AMC 12 第 29 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 1987 AMC 12 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。

29.

数列递归定义如下:t1=1t_1=1;当 n>1n\gt1 时,tn=1+tn2t_n=1+t_{\frac{n}{2}}(此时 nn 为偶数),而 tn=1tn−1t_n=\frac{1}{t_{n-1}}(此时 nn 为奇数)。已知 tn=1987t_n=\frac{19}{87},则 nn 的各位数字之和为
Consider the sequence of numbers defined recursively by t1=1t_1=1 and for n>1n\gt1 by tn=1+tn2t_n=1+t_{\frac{n}{2}} when nn is even and by tn=1tn−1t_n=\frac{1}{t_{n-1}} when nn is odd. Given that tn=1987,t_n=\frac{19}{87}, the sum of the digits of nn is
1515
1717
1919
2121
2323
答案:A
知识点:recursive sequenceinverse recursionrational numbers
小提示:
大于 11 的数值来自偶数下标;介于 00 与 11 之间的数值来自奇数下标
Values greater than 11 come from even indices; values between 00 and 11 come from odd indices
大提示:
反向运用递推关系:对大于 11 的数值减去 11,对小于 11 的数值取倒数
Reverse the recursion by subtracting 11 from values above 11 and taking reciprocals of values below 11
解答:
设 N(r)N(r) 表示数值 rr 出现时的下标。反向运用递推关系可得:N(r)=2N(r−1)N(r)=2N(r-1) 适用于 r>1r\gt1,而 N(r)=N(1r)+1N(r)=N(\frac{1}{r})+1 适用于 0<r<10\lt r\lt1。
从 N(1)=1N(1)=1 开始反复运用这些关系,得到 N(2)=2N(2)=2、N(12)=3N(\frac{1}{2})=3、N(32)=6N(\frac{3}{2})=6 和 N(23)=7N(\frac{2}{3})=7。继续可得 N(53)=14N(\frac{5}{3})=14、N(83)=28N(\frac{8}{3})=28、N(38)=29N(\frac{3}{8})=29 和 N(118)=58N(\frac{11}{8})=58。
接下来,N(811)=59N(\frac{8}{11})=59、N(1911)=118N(\frac{19}{11})=118、N(1119)=119N(\frac{11}{19})=119、N(3019)=238N(\frac{30}{19})=238、N(4919)=476N(\frac{49}{19})=476,并且 N(6819)=952N(\frac{68}{19})=952。最后,N(8719)=1904N(\frac{87}{19})=1904,而 N(1987)=1905N(\frac{19}{87})=1905。因此 n=1905n=1905,其各位数字之和为 1+9+0+5=151+9+0+5=15。
因此,正确答案是 A。
Let N(r)N(r) be the index at which the value rr occurs. Reversing the recursion gives N(r)=2N(r−1)N(r)=2N(r-1) for r>1,r\gt1, and N(r)=N(1r)+1N(r)=N(\frac{1}{r})+1 for 0<r<1.0\lt r\lt1.
Starting with N(1)=1,N(1)=1, repeated use gives N(2)=2,N(2)=2, N(12)=3,N(\frac{1}{2})=3, N(32)=6,N(\frac{3}{2})=6, and N(23)=7.N(\frac{2}{3})=7. Continuing gives N(53)=14,N(\frac{5}{3})=14, N(83)=28,N(\frac{8}{3})=28, N(38)=29,N(\frac{3}{8})=29, and N(118)=58.N(\frac{11}{8})=58.
Next, N(811)=59,N(\frac{8}{11})=59, N(1911)=118,N(\frac{19}{11})=118, N(1119)=119,N(\frac{11}{19})=119, N(3019)=238,N(\frac{30}{19})=238, N(4919)=476,N(\frac{49}{19})=476, and N(6819)=952.N(\frac{68}{19})=952. Finally, N(8719)=1904N(\frac{87}{19})=1904 and N(1987)=1905.N(\frac{19}{87})=1905. Thus n=1905,n=1905, whose digit sum is 1+9+0+5=15.1+9+0+5=15.
Therefore the correct answer is A.
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