2020 AMC 10B Problem 19

Question

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

The number of hands is

\binom{52}{10}

After factoring out a final

10,

only the units digit of the remaining product is needed

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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.

Problem 19 in other years