2015 AMC 12B 第 20 题

先试着解答 2015 AMC 12B 第 20 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 2015 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。

20.

对每个正整数 nn,令 mod⁡5(n)\operatorname{mod}_5(n) 表示 nn 除以 55 的余数。递归定义函数 f:{0,1,2,3,…}f : \{0, 1, 2, 3, \ldots\} ×{0,1,2,3,4}\times \{0, 1, 2, 3, 4\} →{0,1,2,3,4}\to \{0, 1, 2, 3, 4\} 如下:
f(i,j)={mod⁡5(j+1)当 i=0 且 0≤j≤4,f(i−1,1)当 i≥1 且 j=0, 且f(i−1,f(i,j−1))当 i≥1 且 1≤j≤4。 \tiny f(i, j) = \begin{cases} \operatorname{mod}_5(j + 1) & \text{当 } i = 0 \text{ 且 } 0 \le j \le 4, \\ f(i - 1, 1) & \text{当 } i \ge 1 \text{ 且 } j = 0, \text{ 且} \\ f(i - 1, f(i, j - 1)) & \text{当 } i \ge 1 \text{ 且 } 1 \le j \le 4 \end{cases}\text{。}
f(2015,2)f(2015, 2) 是多少?
For every positive integer n,n, let mod⁡5(n)\operatorname{mod}_5(n) be the remainder obtained when nn is divided by 5.5. Define a function f:{0,1,2,3,…}f : \{0, 1, 2, 3, \ldots\} ×{0,1,2,3,4}\times \{0, 1, 2, 3, 4\} →{0,1,2,3,4}\to \{0, 1, 2, 3, 4\} recursively as follows:
f(i,j)={mod⁡5(j+1)if i=0 and 0≤j≤4,f(i−1,1)if i≥1 and j=0, andf(i−1,f(i,j−1))if i≥1 and 1≤j≤4. \tiny f(i, j) = \begin{cases} \operatorname{mod}_5(j + 1) & \text{if } i = 0 \text{ and } 0 \le j \le 4, \\ f(i - 1, 1) & \text{if } i \ge 1 \text{ and } j = 0, \text{ and} \\ f(i - 1, f(i, j - 1)) & \text{if } i \ge 1 \text{ and } 1 \le j \le 4. \end{cases}
What is f(2015,2)?f(2015, 2)?
00
11
22
33
44
答案:B
知识点:递推找规律模运算
小提示:
对小的 ii,逐行建立 f(i,j)f(i, j) 的表
Build the table of f(i,j)f(i, j) row by row for small ii
大提示:
跟踪 j=2j = 2 这一列;当 ii 足够大后它会变成常数
Track the column j=2;j = 2; it becomes constant once ii is large enough
解答:
在每一行中从左到右应用递推,得到 i\j01234012340123401230241303410431313511111 \begin{array}{c|ccccc} i\backslash j&0&1&2&3&4\\ \hline 0&1&2&3&4&0\\ 1&2&3&4&0&1\\ 2&3&0&2&4&1\\ 3&0&3&4&1&0\\ 4&3&1&3&1&3\\ 5&1&1&1&1&1 \end{array} 如果某一行全是 11,递推就会使下一行也全是 11。因此由归纳法可知,对每个 i≥5i\ge5 都有 f(i,2)=1f(i,2)=1。
因为 2015≥52015 \ge 5,所以 f(2015,2)=1f(2015, 2) = 1。
因此,正确选项是 B。
Applying the recursion from left to right in each row gives i\j01234012340123401230241303410431313511111 \begin{array}{c|ccccc} i\backslash j&0&1&2&3&4\\ \hline 0&1&2&3&4&0\\ 1&2&3&4&0&1\\ 2&3&0&2&4&1\\ 3&0&3&4&1&0\\ 4&3&1&3&1&3\\ 5&1&1&1&1&1 \end{array} If a row consists entirely of 11s, the recursion makes the next row entirely 11s as well. Hence, by induction, f(i,2)=1f(i,2)=1 for every i≥5.i\ge5.
Since 2015≥5,2015 \ge 5, we get f(2015,2)=1.f(2015, 2) = 1.
Thus, the correct answer is B.
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