2020 AMC 10B Problems/Problem 19

Problem

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 $A$ ?

$\textbf{(A) } 2 \qquad\textbf{(B) } 3 \qquad\textbf{(C) } 4 \qquad\textbf{(D) } 6 \qquad\textbf{(E) } 7$

Solution 1

$158A00A4AA0 \equiv 1+5+8+A+0+0+A+4+A+A+0 \equiv 4A \pmod{9}$

We're looking for the amount of ways we can get $10$ cards from a deck of $52$ , which is represented by $\binom{52}{10}$ .

$\binom{52}{10}=\frac{52\cdot51\cdot50\cdot49\cdot48\cdot47\cdot46\cdot45\cdot44\cdot43}{10\cdot9\cdot8\cdot7\cdot6\cdot5\cdot4\cdot3\cdot2\cdot1}$

We need to get rid of the multiples of $3$ , which will subsequently get rid of the multiples of $9$ (if we didn't, the zeroes would mess with the equation since you can't divide by 0)

$9\cdot5=45$ , $8\cdot6=48$ , $\frac{51}{3}$ leaves us with 17.

$\frac{52\cdot\cancel{51}^{17}\cdot50\cdot49\cdot\cancel{48}\cdot47\cdot46\cdot\cancel{45}\cdot44\cdot43}{10\cdot\cancel{9}\cdot\cancel{8}\cdot7\cdot\cancel{6}\cdot\cancel{5}\cdot4\cdot\cancel{3}\cdot2\cdot1}$

Converting these into $\pmod{9}$ , we have

$\binom{52}{10}\equiv \frac{(-2)\cdot(-1)\cdot(-4)\cdot4\cdot2\cdot1\cdot(-1)\cdot(-2)}{1\cdot(-2)\cdot4\cdot2\cdot1} \equiv (-1)\cdot(-4)\cdot(-1)\cdot(-2) \equiv 8 \pmod{9}$

$4A\equiv8\pmod{9} \implies A=\boxed{\textbf{(A) }2}$ ~quacker88

Solution 2

$\binom{52}{10}=\frac{52\cdot51\cdot50\cdot49\cdot48\cdot47\cdot46\cdot45\cdot44\cdot43}{10\cdot9\cdot8\cdot7\cdot6\cdot5\cdot4\cdot3\cdot2\cdot1}=26\cdot17\cdot5\cdot7\cdot47\cdot46\cdot11\cdot43$

Since this number is divisible by $4$ but not $8$ , the last $2$ digits must be divisible by $4$ but the last $3$ digits cannot be divisible by $8$ . This narrows the options down to $2$ and $6$ .

Also, the number cannot be divisible by $3$ . Adding up the digits, we get $18+4A$ . If $A=6$ , then the expression equals $42$ , a multiple of $3$ . This would mean that the entire number would be divisible by $3$ , which is not what we want. Therefore, the only option is $\boxed{\textbf{(A) }2}$ -PCChess

Solution 3

It is not hard to check that $13$ divides the number, $\binom{52}{10}=\frac{52\cdot51\cdot50\cdot49\cdot48\cdot47\cdot46\cdot45\cdot44\cdot43}{10\cdot9\cdot8\cdot7\cdot6\cdot5\cdot4\cdot3\cdot2\cdot1}=26\cdot17\cdot5\cdot7\cdot47\cdot46\cdot11\cdot43.$ As $10^3\equiv-1\pmod{13}$ , using $\pmod{13}$ we have $13|\overline{AA0}-\overline{0A4}+\overline{8A0}-\overline{15}=110A+781$ . Thus $6A+1\equiv0\pmod{13}$ , implying $A\equiv2\pmod{13}$ so the answer is $\boxed{\textbf{(A) }2}$ .

$\textbf{- Emathmaster}$

Solution 4

As mentioned above,
$\binom{52}{10}=\frac{52 \cdot 51 \cdot 50 \cdot 49 \cdot 48 \cdot 47 \cdot 46 \cdot 45 \cdot 44 \cdot 43}{10 \cdot 9 \cdot 8 \cdot 7 \cdot 6 \cdot 5 \cdot 4 \cdot 3 \cdot 2 \cdot 1} = {10 \cdot 17 \cdot 13 \cdot 7 \cdot 47 \cdot 46 \cdot 11 \cdot 43} = 158A00A4AA0.$ We can divide both sides of $10 \cdot 17 \cdot 13 \cdot 7 \cdot 47 \cdot 46 \cdot 11 \cdot 43 = 158A00A4AA0$ by 10 to obtain $17 \cdot 13 \cdot 7 \cdot 47 \cdot 46 \cdot 11 \cdot 43 = 158A00A4AA,$ which means $A$ is simply the units digit of the left-hand side. This value is $7 \cdot 3 \cdot 7 \cdot 7 \cdot 6 \cdot 1 \cdot 3 \equiv \boxed{\textbf{(A) }2} \pmod{10}.$ ~i_equal_tan_90, revised by emerald_block

Video Solution

https://youtu.be/3BvJeZU3T-M

~IceMatrix

2020 AMC 10B (Problems • Answer Key • Resources)
Preceded by Problem 18	Followed by Problem 20
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