Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

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

m (→‎Solution)
m (→‎Solution 10 (Solution 8 but better))
 
(54 intermediate revisions by 20 users not shown)
Line 1: Line 1:
−
==Solution==
+
==Problem==
 +
 
 +
In a certain card game, a player is dealt a hand of <math>10</math> cards from a deck of <math>52</math> distinct cards. The number of distinct (unordered) hands that can be dealt to the player can be written as <math>158A00A4AA0</math>. What is the digit <math>A</math>?
 +
 
 +
<math>\textbf{(A) } 2 \qquad\textbf{(B) } 3 \qquad\textbf{(C) } 4 \qquad\textbf{(D) } 6 \qquad\textbf{(E) } 7</math>
 +
 
 +
==Solution 1==
  
 
<math>158A00A4AA0 \equiv 1+5+8+A+0+0+A+4+A+A+0 \equiv 4A \pmod{9}</math>
 
<math>158A00A4AA0 \equiv 1+5+8+A+0+0+A+4+A+A+0 \equiv 4A \pmod{9}</math>
Line 7: Line 13:
 
<math>\binom{52}{10}=\frac{52\cdot51\cdot50\cdot49\cdot48\cdot47\cdot46\cdot45\cdot44\cdot43}{10\cdot9\cdot8\cdot7\cdot6\cdot5\cdot4\cdot3\cdot2\cdot1}</math>
 
<math>\binom{52}{10}=\frac{52\cdot51\cdot50\cdot49\cdot48\cdot47\cdot46\cdot45\cdot44\cdot43}{10\cdot9\cdot8\cdot7\cdot6\cdot5\cdot4\cdot3\cdot2\cdot1}</math>
  
−
We need to get rid of the multiples of <math>3</math>, which will subsequently get rid of the multiples of <math>9</math>.
+
We need to get rid of the multiples of <math>3</math>, which will subsequently get rid of the multiples of <math>9</math> (if we didn't, the zeroes would mess with the equation since you can't divide by 0)
  
 
<math>9\cdot5=45</math>, <math>8\cdot6=48</math>, <math>\frac{51}{3}</math> leaves us with 17.
 
<math>9\cdot5=45</math>, <math>8\cdot6=48</math>, <math>\frac{51}{3}</math> leaves us with 17.
Line 18: Line 24:
  
 
<math>4A\equiv8\pmod{9} \implies A=\boxed{\textbf{(A) }2}</math> ~quacker88
 
<math>4A\equiv8\pmod{9} \implies A=\boxed{\textbf{(A) }2}</math> ~quacker88
 +
 +
==Solution 1 but easier==
 +
We're looking for the amount of ways we can get <math>10</math> cards from a deck of <math>52</math>, which is represented by <math>\binom{52}{10}</math>.
 +
 +
<math>\binom{52}{10}=\frac{52\cdot51\cdot50\cdot49\cdot48\cdot47\cdot46\cdot45\cdot44\cdot43}{10\cdot9\cdot8\cdot7\cdot6\cdot5\cdot4\cdot3\cdot2\cdot1}</math>
 +
 +
And after simplifying, we get <math>26\cdot17\cdot7\cdot47\cdot46\cdot5\cdot11\cdot43</math>.
 +
Now, if we examine the number <math>158A00A4AA0</math>, we can notice that it is equal to some number <math>n</math> times 10.
 +
Therefore, we can divide 10 from the aforementioned expression and find the units digit, which will be <math>A</math>.
 +
 +
Now, after dividing ten, we will have <math>26\cdot17\cdot7\cdot47\cdot23\cdot11\cdot43</math>.
 +
We can then use modulo 10 and find that the unit digit of the expression is <math>\boxed{\textbf{(A) }2}</math>
 +
~lucaswujc
 +
 +
==Solution 2==
 +
<math>\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</math>
 +
 +
Since this number is divisible by <math>4</math> but not <math>8</math>, the last <math>2</math> digits must be divisible by <math>4</math> but the last <math>3</math> digits cannot be divisible by <math>8</math>. This narrows the options down to <math>2</math> and <math>6</math>.
 +
 +
Also, the number cannot be divisible by <math>3</math>. Adding up the digits, we get <math>18+4A</math>. If <math>A=6</math>, then the expression equals <math>42</math>, a multiple of <math>3</math>. This would mean that the entire number would be divisible by <math>3</math>, which is not what we want. Therefore, the only option is <math>\boxed{\textbf{(A) }2}</math>-PCChess
 +
 +
==Solution 3==
 +
It is not hard to check that <math>13</math> divides the number,
 +
<cmath>\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.</cmath> As <math>10^3\equiv-1\pmod{13}</math>, using <math>\pmod{13}</math> we have <math>13|\overline{AA0}-\overline{0A4}+\overline{8A0}-\overline{15}=110A+781</math>. Thus <math>6A+1\equiv0\pmod{13}</math>, implying <math>A\equiv2\pmod{13}</math> so the answer is <math>\boxed{\textbf{(A) }2}</math>.
 +
 +
<math>\textbf{- Emathmaster}</math>
 +
 +
==Solution 4==
 +
As mentioned above, <br>
 +
<cmath>\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.</cmath>
 +
We can divide both sides of <math>10 \cdot 17 \cdot 13 \cdot 7 \cdot 47 \cdot 46 \cdot 11 \cdot 43 = 158A00A4AA0</math> by 10 to obtain
 +
<cmath>17 \cdot 13 \cdot 7 \cdot 47 \cdot 46 \cdot 11 \cdot 43 = 158A00A4AA,</cmath>
 +
which means <math>A</math> is simply the units digit of the left-hand side. This value is
 +
<cmath>7 \cdot 3 \cdot 7 \cdot 7 \cdot 6 \cdot 1 \cdot 3 \equiv \boxed{\textbf{(A) }2} \pmod{10}.</cmath>
 +
~[[User:i_equal_tan_90|i_equal_tan_90]], revised by [[User:emerald_block|emerald_block]]
 +
 +
==Solution 5 (Very Factor Bashy CRT)==
 +
We note that:
 +
<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>.
 +
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{\bold{(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
 +
 +
==Solution 7==
 +
The total number of ways to choose <math>10</math> from <math>52</math> is
 +
<math>\binom{52}{10}=\frac{52\cdot51\cdot50\cdot49\cdot48\cdot47\cdot46\cdot45\cdot44\cdot43}{10\cdot9\cdot8\cdot7\cdot6\cdot5\cdot4\cdot3\cdot2\cdot1}</math>
 +
 +
Using divisibility rules, we have that A is not a multiple of <math>3</math>. Then, divide this equation by 10. This implies that the new number <math>158A00A4AA0</math> is divisible by <math>2</math> but not <math>4</math>. This means that <math>A</math> is either <math>2</math> or <math>6</math>. However, <math>6</math> is a multiple of <math>3</math>, meaning <math>A</math> has to be <math>\boxed {\textbf{(A)2}}</math>
 +
 +
~Arcticturn
 +
 +
==Solution 8 (Very time consuming)==
 +
As stated in previous solutions, the number of ways to choose <math>10</math> from <math>52</math> is
 +
<math>\binom{52}{10}=\frac{52\cdot51\cdot50\cdot49\cdot48\cdot47\cdot46\cdot45\cdot44\cdot43}{10\cdot9\cdot8\cdot7\cdot6\cdot5\cdot4\cdot3\cdot2\cdot1}</math>
 +
 +
Canceling out common factors <math>(\dfrac{52}{2} = 26 \text{, } \dfrac{51}{3} = 17 \text{, } 5 \cdot 10 = 50 \text{, } \dfrac{49}{7} = 7 \text{, } 48 = 6 \cdot 8 \text{, } \dfrac{45}{9} = 5 \text{, } \dfrac{44}{4} = 11)</math>, you get this -
 +
<math>\frac{\cancel{52}^{26}\cdot\cancel{51}^{17}\cdot\cancel{50}\cdot\cancel{49}^{7}\cdot\cancel{48}\cdot47\cdot46\cdot\cancel{45}^5\cdot\cancel{44}^{11}\cdot43}{\cancel{10}\cdot\cancel{9}\cdot\cancel{8}\cdot\cancel{7}\cdot\cancel{6}\cdot\cancel{5}\cdot\cancel{4}\cdot\cancel{3}\cdot\cancel{2}\cdot\cancel{1}}</math>
 +
 +
When you multiply the remaining numbers, you get the product as <math>15820024220</math>. From this product, we can then determine that <math>A</math> is equal to <math>\boxed {\textbf{(A)2}}</math>
 +
 +
~ KING.OF.MATH
 +
 +
~[https://artofproblemsolving.com/wiki/index.php/User: Puck_0] (Minor LaTeX)
 +
 +
==Solution 9==
 +
Compute <math>\frac{52!}{10!42!} = 15820024220.</math> Therefore our answer is <math>\boxed {\textbf{(A)2}}.</math>
 +
 +
~ Sliced_Bread
 +
 +
==Solution 10 (Solution 8 but better)==
 +
As in solution 8, you get <math>26\cdot17\cdot7\cdot47\cdot46\cdot5\cdot11\cdot43</math>
 +
This ends in a 0, so let's divide by 10.
 +
<math>13\cdot17\cdot7\cdot47\cdot46\cdot11\cdot43</math>
 +
We want the last digit of this:
 +
 +
<math>3\cdot7\cdot7\cdot7\cdot6\cdot1\cdot3</math>
 +
 +
<math>\equiv2\cdot(3\cdot2\cdot2\cdot2\cdot3\cdot1\cdot3)</math>
 +
 +
<math>\equiv2\cdot(1\cdot2\cdot2\cdot3\cdot3)</math>
 +
 +
<math>\equiv2\cdot(2\cdot2\cdot3\cdot3)</math>
 +
 +
<math>\equiv2\cdot(4\cdot3\cdot3)</math>
 +
 +
<math>\equiv2\cdot(2\cdot3)</math>
 +
 +
<math>\equiv2\cdot1</math>
 +
 +
<math>\equiv2</math>
 +
 +
so A=2
 +
 +
==Video Solutions==
 +
==Video Solution (HOW TO THINK CREATIVELY!!!)==
 +
https://youtu.be/QNUzrwXWQ2A
 +
 +
~Education, the Study of Everything
 +
 +
 +
 +
 +
 +
 +
===Video Solution===
 +
https://youtu.be/3BvJeZU3T-M
 +
 +
===Video Solution 2===
 +
https://www.youtube.com/watch?v=ApqZFuuQJ18&list=PLLCzevlMcsWNcTZEaxHe8VaccrhubDOlQ&index=6 ~ MathEx
 +
 +
==See Also==
 +
 +
{{AMC10 box|year=2020|ab=B|num-b=18|num-a=20}}
 +
{{MAA Notice}}

Latest revision as of 17:57, 20 October 2024

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 1 but easier

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

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

And after simplifying, we get $26\cdot17\cdot7\cdot47\cdot46\cdot5\cdot11\cdot43$. Now, if we examine the number $158A00A4AA0$, we can notice that it is equal to some number $n$ times 10. Therefore, we can divide 10 from the aforementioned expression and find the units digit, which will be $A$.

Now, after dividing ten, we will have $26\cdot17\cdot7\cdot47\cdot23\cdot11\cdot43$. We can then use modulo 10 and find that the unit digit of the expression is $\boxed{\textbf{(A) }2}$ ~lucaswujc

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

Solution 7

The total number of ways to choose $10$ from $52$ is $\binom{52}{10}=\frac{52\cdot51\cdot50\cdot49\cdot48\cdot47\cdot46\cdot45\cdot44\cdot43}{10\cdot9\cdot8\cdot7\cdot6\cdot5\cdot4\cdot3\cdot2\cdot1}$

Using divisibility rules, we have that A is not a multiple of $3$. Then, divide this equation by 10. This implies that the new number $158A00A4AA0$ is divisible by $2$ but not $4$. This means that $A$ is either $2$ or $6$. However, $6$ is a multiple of $3$, meaning $A$ has to be $\boxed {\textbf{(A)2}}$

~Arcticturn

Solution 8 (Very time consuming)

As stated in previous solutions, the number of ways to choose $10$ from $52$ is $\binom{52}{10}=\frac{52\cdot51\cdot50\cdot49\cdot48\cdot47\cdot46\cdot45\cdot44\cdot43}{10\cdot9\cdot8\cdot7\cdot6\cdot5\cdot4\cdot3\cdot2\cdot1}$

Canceling out common factors $(\dfrac{52}{2} = 26 \text{, } \dfrac{51}{3} = 17 \text{, } 5 \cdot 10 = 50 \text{, } \dfrac{49}{7} = 7 \text{, } 48 = 6 \cdot 8 \text{, } \dfrac{45}{9} = 5 \text{, } \dfrac{44}{4} = 11)$, you get this - $\frac{\cancel{52}^{26}\cdot\cancel{51}^{17}\cdot\cancel{50}\cdot\cancel{49}^{7}\cdot\cancel{48}\cdot47\cdot46\cdot\cancel{45}^5\cdot\cancel{44}^{11}\cdot43}{\cancel{10}\cdot\cancel{9}\cdot\cancel{8}\cdot\cancel{7}\cdot\cancel{6}\cdot\cancel{5}\cdot\cancel{4}\cdot\cancel{3}\cdot\cancel{2}\cdot\cancel{1}}$

When you multiply the remaining numbers, you get the product as $15820024220$. From this product, we can then determine that $A$ is equal to $\boxed {\textbf{(A)2}}$

~ KING.OF.MATH

~Puck_0 (Minor LaTeX)

Solution 9

Compute $\frac{52!}{10!42!} = 15820024220.$ Therefore our answer is $\boxed {\textbf{(A)2}}.$

~ Sliced_Bread

Solution 10 (Solution 8 but better)

As in solution 8, you get $26\cdot17\cdot7\cdot47\cdot46\cdot5\cdot11\cdot43$ This ends in a 0, so let's divide by 10. $13\cdot17\cdot7\cdot47\cdot46\cdot11\cdot43$ We want the last digit of this:

$3\cdot7\cdot7\cdot7\cdot6\cdot1\cdot3$

$\equiv2\cdot(3\cdot2\cdot2\cdot2\cdot3\cdot1\cdot3)$

$\equiv2\cdot(1\cdot2\cdot2\cdot3\cdot3)$

$\equiv2\cdot(2\cdot2\cdot3\cdot3)$

$\equiv2\cdot(4\cdot3\cdot3)$

$\equiv2\cdot(2\cdot3)$

$\equiv2\cdot1$

$\equiv2$

so A=2

Video Solutions

Video Solution (HOW TO THINK CREATIVELY!!!)

https://youtu.be/QNUzrwXWQ2A

~Education, the Study of Everything




Video Solution

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

Video Solution 2

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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2020_AMC_10B_Problems/Problem_19&oldid=230203"