Question

有多少个非负整数可以写成其中对 $0\le i \le 7$ ，都有 $a_i\in \{-1,0,1\}$ ？ $a_7\cdot3^7+a_6\cdot3^6+a_5\cdot3^5$ $+a_4\cdot3^4+a_3\cdot3^3+a_2\cdot3^2$ $+a_1\cdot3^1+a_0\cdot3^0,$

How many nonnegative integers can be written in the form $a_7\cdot3^7+a_6\cdot3^6+a_5\cdot3^5$ $+a_4\cdot3^4+a_3\cdot3^3+a_2\cdot3^2$ $+a_1\cdot3^1+a_0\cdot3^0,$ where $a_i\in \{-1,0,1\}$ for $0\le i \le 7?$

$512$

$729$

$1094$

$3281$

$59,048$

答案：D

知识点：进制对称性配对与分组

难度评级：1770

解答：

每一组系数得到的和必为正数、负数或零。

由于对称性，正数与负数的数量相等：把所有 $1$ 换成 $-1$ ，同时把所有 $-1$ 换成 $1$ 即可一一配对。

只有当所有系数都为 $0$ 时，和才等于 $0$ 。

系数的组合共有 $3^8 = 6561$ 种。若两组系数不同，则最高次的不同项的绝对值至少为相应的 $3$ 的幂，而所有低次项的差的绝对值之和比它小，所以两组系数不可能得到同一个整数。 $3^k$ ， $2(1+3+\cdots+3^{k-1})=3^k-1$ 。

因此，不同的非负整数共有。 $\dfrac{6561 - 1}{2} + 1 = 3281$

因此正确答案是 D。

Note that every number formed by this sum is either positive, negative, or zero.

The number of positive numbers equals the number of negative numbers due to symmetry (flip the $1$ s to $-1$ s and $-1$ s to $1$ s).

The only way for the sum to be $0$ is if all the coefficients are $0.$

The total number of numbers is $3^8 = 6561.$ Because each power of $3$ is larger than the sum of all previous powers of three, each combination of coefficients yields a different value. More explicitly, at the highest place where two combinations differ, the difference has magnitude at least $3^k,$ while all lower places together can cancel at most $2(1+3+\cdots+3^{k-1})=3^k-1.$

Therefore, there are $\dfrac{6561 - 1}{2} + 1 = 3281$ distinct nonnegative integers.

Thus, D is the correct answer.

2018 AMC 10A 第 18 题

答案：D

难度评级：1770

解答：

其他年份的第 18 题