Difference between revisions of "2020 AMC 10B Problems/Problem 19"
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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 (mod 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 | + | It is obvious that <math>K \equiv 0 (mod 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> |
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==Video Solution== | ==Video Solution== |
Revision as of 11:34, 28 December 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 .
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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