Question

在某种纸牌游戏中，一名玩家从 $52$ 张互不相同的牌组成的牌堆中拿到 $10$ 张牌。可以发给该玩家的不同无序手牌数可写成 $158A00A4AA0$ 。数字 $A$ 是多少？

In a certain card game, a player is dealt a hand of $10$ cards from a deck of $52$ distinct cards. The number of distinct (unordered) hands that can be dealt to the player can be written as $158A00A4AA0.$ What is the digit $A?$

$2$

$3$

$4$

$6$

$7$

答案：A

知识点：组合数字模运算

难度评级：1620

小提示：

手牌数为 $\binom{52}{10}$ 。

The number of hands is $\binom{52}{10}$

大提示：

提出末尾的一个 $10$ 后，只需要剩余乘积的个位数字。

After factoring out a final $10,$ only the units digit of the remaining product is needed

视频讲解：

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文字解答：

把二项式系数中的因子约分后，得到 $\begin{aligned} \binom{52}{10} &=10\cdot17\cdot13\cdot7\\ &\quad\cdot47\cdot46\cdot11\cdot43 \end{aligned}\text{。}$ 把 $158A00A4AA0$ 除以最后一个因子 $10$ ，剩下乘积的个位数字就是 $A$ 。

对 $10$ 取模计算，得 $\begin{aligned} A&\equiv7\cdot3\cdot7\cdot7\cdot6\cdot1\cdot3\\ &\equiv2\pmod{10} \end{aligned}\text{。}$ 因此 $A=2$ 。所以正确答案是 A。

After canceling factors in the binomial coefficient, $\begin{aligned} \binom{52}{10} &=10\cdot17\cdot13\cdot7\\ &\quad\cdot47\cdot46\cdot11\cdot43. \end{aligned}$ Dividing $158A00A4AA0$ by the final factor $10,$ the units digit of the remaining product is $A.$

Working modulo $10,$ $\begin{aligned} A&\equiv7\cdot3\cdot7\cdot7\cdot6\cdot1\cdot3\\ &\equiv2\pmod{10}. \end{aligned}$ Hence $A=2.$ Thus, A is the correct answer.

2020 AMC 10B 第 19 题