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

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(→‎Solution 5 (Very Factor Bashy CRT))
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<cmath> \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). </cmath>
 
<cmath> \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). </cmath>
 
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>.
 
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), <math>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 (mod 25)</math>. Therefore, <math>K \equiv 20 \mod 100</math>. Thus <math>K=20</math>, implying that <math>A=2</math>. <math>\textbf{A}</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>A=2</math>. <math>\textbf{A}</math>
  
 
==Video Solution==
 
==Video Solution==

Revision as of 20:06, 29 December 2020

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 $A=2$. $\textbf{A}$

Video Solution

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

~IceMatrix

Video Solution

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


See Also

2020 AMC 10B (Problems • Answer Key • Resources)
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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