Difference between revisions of "2020 AMC 10B Problems/Problem 19"
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− | ==Solution== | + | ==Problem== |
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+ | 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> | ||
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<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 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 \cong 0 (mod 4)</math>. Also (although this not so obvious), <math>K \cong (13)(17)(7)(47)(46)(5)(22)(43) \cong (13)(-8)(7)(-3)(-4)(5)(-3)(-7) \cong (13)(-96)(21)(35) \cong (13)(4)(-4)(10) \cong (13)(-16)(10) \cong (13)(9)(10) \cong (117)(10) \cong (-8)(10) \cong 20 (mod 25)</math>. Therefore, <math>K \cong 20 (mod 100)</math>. Thus <math>K=20</math>, implying that <math>A=2</math>. (A) | ||
+ | |||
+ | (I need help with finding the modulo congruence symbol. Thanks!) | ||
+ | |||
+ | ==Video Solution== | ||
+ | https://youtu.be/3BvJeZU3T-M | ||
+ | |||
+ | ~IceMatrix | ||
+ | |||
+ | ==Video Solution== | ||
+ | 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}} |
Revision as of 13:08, 28 June 2020
Contents
Problem
In a certain card game, a player is dealt a hand of cards from a deck of distinct cards. The number of distinct (unordered) hands that can be dealt to the player can be written as . What is the digit ?
Solution 1
We're looking for the amount of ways we can get cards from a deck of , which is represented by .
We need to get rid of the multiples of , which will subsequently get rid of the multiples of (if we didn't, the zeroes would mess with the equation since you can't divide by 0)
, , leaves us with 17.
Converting these into, we have
~quacker88
Solution 2
Since this number is divisible by but not , the last digits must be divisible by but the last digits cannot be divisible by . This narrows the options down to and .
Also, the number cannot be divisible by . Adding up the digits, we get . If , then the expression equals , a multiple of . This would mean that the entire number would be divisible by , which is not what we want. Therefore, the only option is -PCChess
Solution 3
It is not hard to check that divides the number, As , using we have . Thus , implying so the answer is .
Solution 4
As mentioned above,
We can divide both sides of by 10 to obtain
which means is simply the units digit of the left-hand side. This value is
~i_equal_tan_90, revised by emerald_block
Solution 5 (Very Factor Bashy CRT)
We note that: Let . This will help us find the last two digits modulo and modulo . It is obvious that . Also (although this not so obvious), . Therefore, . Thus , implying that . (A)
(I need help with finding the modulo congruence symbol. Thanks!)
Video Solution
~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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