2020 AMC 10B Problem 19

Attempt Problem 19 of the 2020 AMC 10B below, then check your answer against the video solution and professionally curated solution from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2020 AMC 10B solutions, or check the answer key.

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19.

In a certain card game, a player is dealt a hand of 1010 cards from a deck of 5252 distinct cards. The number of distinct (unordered) hands that can be dealt to the player can be written as 158A00A4AA0.158A00A4AA0. What is the digit A?A?

22

33

44

66

77

Answer: A
Concepts:combinationsdigitsmodular arithmetic
Difficulty rating: 1620
Video solution:
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Written solution:

After canceling factors in the binomial coefficient, (5210)=101713747461143. \begin{aligned} \binom{52}{10} &=10\cdot17\cdot13\cdot7\\ &\quad\cdot47\cdot46\cdot11\cdot43. \end{aligned} Dividing 158A00A4AA0158A00A4AA0 by the final factor 10,10, the units digit of the remaining product is A.A.

Working modulo 10,10, A73776132(mod10). \begin{aligned} A&\equiv7\cdot3\cdot7\cdot7\cdot6\cdot1\cdot3\\ &\equiv2\pmod{10}. \end{aligned} Hence A=2.A=2. Thus, A is the correct answer.

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