Question

随机选择一个七位正整数。已知它的各位数字之和为 $61$ ，求它能被 $11$ 整除的概率。

A seven-digit positive integer is chosen at random. What is the probability that the number is divisible by $11,$ given that the sum of its digits is $61?$

$\dfrac{3}{14}$

$\dfrac{3}{11}$

$\dfrac{2}{7}$

$\dfrac{4}{11}$

$\dfrac{3}{7}$

答案：A

知识点：条件概率整除性隔板法

难度评级：2040

解答：

数字和 $61 = 63 - 2$ ，表示七个数字全为 $9$ 后总亏缺为 $2$ ，这样的数有 $\binom{2 + 6}{6} = 28$ 个。要能被 $11$ 整除，需要 $O - E \equiv 0 \pmod{11}$ ，其中 $O$ 是 $4$ 个奇数位数字之和， $E$ 是 $3$ 个偶数位数字之和。设亏缺为 $d_O + d_E = 2$ 。则 $O - E = 9 - d_O$ $+ d_E = 11 - 2 d_O$ ，只有当 $d_O = 0$ 时才是 $11$ 的倍数。因此全部亏缺 $2$ 都在 $3$ 个偶数位上，有 $\binom{2 + 2}{2} = 6$ 种。概率为 $\tfrac{6}{28} = \tfrac{3}{14}$ 。因此正确答案是 A。

A digit sum of $61 = 63 - 2$ means all seven digits are $9$ except for a total deficit of $2,$ which gives $\binom{2 + 6}{6} = 28$ numbers. For divisibility by $11$ we need $O - E \equiv 0 \pmod{11},$ where $O$ sums the $4$ odd-position digits and $E$ the $3$ even ones. Write the deficits as $d_O + d_E = 2.$ Then $O - E = 9 - d_O$ $+ d_E = 11 - 2 d_O,$ a multiple of $11$ only when $d_O = 0.$ So all of the deficit $2$ falls on the $3$ even positions, giving $\binom{2 + 2}{2} = 6$ ways. The probability is $\tfrac{6}{28} = \tfrac{3}{14}.$ Therefore, the answer is A.

2025 AMC 10B 第 22 题

答案：A

难度评级：2040

解答：

其他年份的第 22 题