2020 AMC 10B Problems/Problem 19

Revision as of 20:31, 7 February 2020 by Emathmaster (talk | contribs) (Solution 3)


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 $\binom{52}{10}$.


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.


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


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 $\boxed{\textbf{(A) }2}$.

$\textbf{- Emathmaster}$

Video Solution



See Also

2020 AMC 10B (ProblemsAnswer KeyResources)
Preceded by
Problem 18
Followed by
Problem 20
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25
All AMC 10 Problems and Solutions

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