AMC12 2014 B
AMC12 2014 B · Q18
AMC12 2014 B · Q18. It mainly tests Basic counting (rules of product/sum), Casework.
The numbers 1, 2, 3, 4, 5 are to be arranged in a circle. An arrangement is bad if it is not true that for every $n$ from 1 to 15 one can find a subset of the numbers that appear consecutively on the circle that sum to $n$. Arrangements that differ only by a rotation or a reflection are considered the same. How many different bad arrangements are there?
将数字 1, 2, 3, 4, 5 排列成一个圆圈。如果对于从 1 到 15 的每个 $n$,不能找到圆圈上连续出现的数字子集使其和为 $n$,则该排列为坏排列。只考虑旋转或反射不同的排列为不同。坏排列有多少种?
(A)
1
1
(B)
2
2
(C)
3
3
(D)
4
4
(E)
5
5
Answer
Correct choice: (B)
正确答案:(B)
Solution
Answer (B): The circular arrangement 14352 is bad because the sum 6 cannot be achieved with consecutive numbers, and the circular arrangement 23154 is bad because the sum 7 cannot be so achieved. It remains to show that these are the only bad arrangements. Given a circular arrangement, sums 1 through 5 can be achieved with a single number, and if the sum $n$ can be achieved, then the sum $15-n$ can be achieved using the complementary subset. Therefore an arrangement is not bad as long as sums 6 and 7 can be achieved. Suppose 6 cannot be achieved. Then 1 and 5 cannot be adjacent, so by a suitable rotation and/or reflection, the arrangement is $1bc5e$. Furthermore, $\{b,c\}$ cannot equal $\{2,3\}$ because $1+2+3=6$; similarly $\{b,c\}$ cannot equal $\{2,4\}$. It follows that $e=2$, which then forces the arrangement to be 14352 in order to avoid consecutive 213. This arrangement is bad. Next suppose that 7 cannot be achieved. Then 2 and 5 cannot be adjacent, so again without loss of generality the arrangement is $2bc5e$. Reasoning as before, $\{b,c\}$ cannot equal $\{3,4\}$ or $\{1,4\}$, so $e=4$, and then $b=3$ and $c=1$, to avoid consecutive 421; therefore the arrangement is 23154, which is also bad. Thus there are only two bad arrangements up to rotation and reflection.
答案(B):环形排列 14352 是不好的,因为用相邻(连续)数字无法得到和为 6;环形排列 23154 也是不好的,因为同样无法得到和为 7。接下来需证明这两种是唯一的不良排列。给定一个环形排列,和为 1 到 5 都可以用单个数字得到;并且若和为 $n$ 可以得到,则利用补集子集也可得到和为 $15-n$。因此,只要和为 6 与 7 都能得到,该排列就不是不良排列。假设和为 6 不能得到,则 1 与 5 不能相邻,因此通过适当的旋转和/或翻转,不妨设排列为 $1bc5e$。此外,$\{b,c\}$ 不能等于 $\{2,3\}$,因为 $1+2+3=6$;同理,$\{b,c\}$ 也不能等于 $\{2,4\}$。于是得到 $e=2$,从而为避免出现连续的 213,排列被迫为 14352。该排列是不好的。再假设和为 7 不能得到,则 2 与 5 不能相邻,因此同样不失一般性可设排列为 $2bc5e$。同前推理,$\{b,c\}$ 不能等于 $\{3,4\}$ 或 $\{1,4\}$,所以 $e=4$,并且为避免出现连续的 421,有 $b=3$、$c=1$;因此排列为 23154,这也是不好的。故在旋转与翻转意义下,不良排列仅有两种。
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