Difference between revisions of "2020 AMC 10B Problems/Problem 19"

Revision as of 15:33, 10 April 2022

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

Solution 5 (Very Factor Bashy CRT)

We note that: $\frac{(52)(51)(50)(49)(48)(47)(46)(45)(44)(43)}{(10)(9)(8)(7)(6)(5)(4)(3)(2)(1)} = (13)(17)(7)(47)(46)(5)(22)(43).$ Let $K=(13)(17)(7)(47)(46)(5)(22)(43)$ . This will help us find the last two digits modulo $4$ and modulo $25$ . It is obvious that $K \equiv 0 \pmod{4}$ . Also (although this not so obvious), $K \equiv (13)(17)(7)(47)(46)(5)(22)(43)$ $\equiv (13)(-8)(7)(-3)(-4)(5)(-3)(-7)$ $\equiv (13)(-96)(21)(35)$ $\equiv (13)(4)(-4)(10)$ $\equiv (13)(-16)(10)$ $\equiv (13)(9)(10)$ $\equiv (117)(10)$ $\equiv (-8)(10)$ $\equiv 20 \pmod{25}.$ Therefore, $K \equiv 20 \mod 100$ . Thus $K=20$ , implying that $\boxed{\textbf{(A) }2}$ .

Solution 6

As in Solution 2, we see that

$\binom{52}{10}=26\cdot17\cdot5\cdot7\cdot47\cdot46\cdot11\cdot43,$

which contains no factors of $3.$ Therefore, the sum of the digits must not be a multiple of $3.$ This sum is

$1+5+8+A+0+0+A+4+A+A+0=18+4A.$

It follows that $4A$ cannot be a multiple of $3,$ ruling out choices $(B)$ and $(D).$ Therefore, our possibilities are $A=2,4,$ and $7.$ Now, notice that $\binom{52}{10}$ is divisible by $7.$ Therefore, we can plug each possible value of $A$ into $158A00A4AA0$ and test for divisibility by $7.$ Conveniently, we see that the first value, $A=2,$ works. Thus, the answer is $\boxed{(A) 2}.$ (To make our argument more rigorous, we can also test divisibility by $7$ for $A=4$ and $7$ to show that these values do not work.)

--vaporwave

Video Solutions

Video Solution 1

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

~IceMatrix

Video Solution 2

https://www.youtube.com/watch?v=ApqZFuuQJ18&list=PLLCzevlMcsWNcTZEaxHe8VaccrhubDOlQ&index=6 ~ MathEx

@@ Line 52: / Line 52: @@
 Let <math>K=(13)(17)(7)(47)(46)(5)(22)(43)</math>. This will help us find the last two digits modulo <math>4</math> and modulo <math>25</math>.
 It is obvious that <math>K \equiv 0 \pmod{4}</math>. Also (although this not so obvious), <cmath>K \equiv (13)(17)(7)(47)(46)(5)(22)(43)</cmath> <cmath>\equiv (13)(-8)(7)(-3)(-4)(5)(-3)(-7)</cmath> <cmath>\equiv (13)(-96)(21)(35)</cmath> <cmath>\equiv (13)(4)(-4)(10)</cmath> <cmath>\equiv (13)(-16)(10)</cmath> <cmath>\equiv (13)(9)(10)</cmath> <cmath>\equiv (117)(10)</cmath> <cmath>\equiv (-8)(10)</cmath> <cmath>\equiv 20 \pmod{25}.</cmath> Therefore, <math>K \equiv 20 \mod 100</math>. Thus <math>K=20</math>, implying that <math>\boxed{\textbf{(A) }2}</math>.
+==Solution 6==
+As in Solution 2, we see that
+<cmath>\binom{52}{10}=26\cdot17\cdot5\cdot7\cdot47\cdot46\cdot11\cdot43,</cmath>
+which contains no factors of <math>3.</math> Therefore, the sum of the digits must not be a multiple of <math>3.</math> This sum is
+<cmath>1+5+8+A+0+0+A+4+A+A+0=18+4A.</cmath>
+It follows that <math>4A</math> cannot be a multiple of <math>3,</math> ruling out choices <math>(B)</math> and <math>(D).</math> Therefore, our possibilities are <math>A=2,4,</math> and <math>7.</math> Now, notice that <math>\binom{52}{10}</math> is divisible by <math>7.</math> Therefore, we can plug each possible value of <math>A</math> into <math>158A00A4AA0</math> and test for divisibility by <math>7.</math> Conveniently, we see that the first value, <math>A=2,</math> works. Thus, the answer is <math>\boxed{(A) 2}.</math> (To make our argument more rigorous, we can also test divisibility by <math>7</math> for <math>A=4</math> and <math>7</math> to show that these values do not work.)
+--vaporwave
 ==Video Solutions==

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