AMC10 2020 B

AMC10 2020 B · Q19

AMC10 2020 B · Q19. It mainly tests Combinations, Remainders & modular arithmetic.

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 ?

在某种纸牌游戏中，玩家从 52 张不同牌的牌堆中分到 10 张牌。可以分给玩家的不同（无序）牌型数量可以写成 158A00A4AA0。A 是多少？

(A) 2 2

(B) 3 3

(D) 6 6

(E) 7 7

Answer

Correct choice: (A)

正确答案：（A）

Solution

The number of ways of dealing 10 cards from 52 is $\binom{52}{10}$. Expanding and simplifying this quantity gives $$\frac{52 \cdot 51 \cdot 50 \cdot 49 \cdot 48 \cdot 47 \cdot 46 \cdot 45 \cdot 44 \cdot 43}{10 \cdot 9 \cdot 8 \cdot 7 \cdot 6 \cdot 5 \cdot 4 \cdot 3 \cdot 2} = 2^2 \cdot 5 \cdot 7 \cdot 11 \cdot 13 \cdot 17 \cdot 23 \cdot 43 \cdot 47.$$ Dividing this quantity by 10 gives $$2 \cdot 7 \cdot 11 \cdot 13 \cdot 17 \cdot 23 \cdot 43 \cdot 47.$$ The value of A can then be found by computing this quantity modulo 10: $$2 \cdot ((-3) \cdot 1 \cdot 3) \cdot ((-3) \cdot 3) \cdot (3 \cdot (-3)) \equiv 2 \cdot 1 \cdot 1 \cdot 1 = 2.$$

从 52 张牌中分 10 张的方式数为 $\binom{52}{10}$。展开并简化此量得 $$\frac{52 \cdot 51 \cdot 50 \cdot 49 \cdot 48 \cdot 47 \cdot 46 \cdot 45 \cdot 44 \cdot 43}{10 \cdot 9 \cdot 8 \cdot 7 \cdot 6 \cdot 5 \cdot 4 \cdot 3 \cdot 2} = 2^2 \cdot 5 \cdot 7 \cdot 11 \cdot 13 \cdot 17 \cdot 23 \cdot 43 \cdot 47。$$ 除以 10 得 $$2 \cdot 7 \cdot 11 \cdot 13 \cdot 17 \cdot 23 \cdot 43 \cdot 47。$$ 通过计算此量模 10 可得 A 的值： $$2 \cdot ((-3) \cdot 1 \cdot 3) \cdot ((-3) \cdot 3) \cdot (3 \cdot (-3)) \equiv 2 \cdot 1 \cdot 1 \cdot 1 = 2。$$

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