Difference between revisions of "2020 AMC 10B Problems/Problem 19"
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~[[User:i_equal_tan_90|i_equal_tan_90]], revised by [[User:emerald_block|emerald_block]] | ~[[User:i_equal_tan_90|i_equal_tan_90]], revised by [[User:emerald_block|emerald_block]] | ||
− | ==Solution 5 | + | ==Solution 5== |
==Video Solution== | ==Video Solution== |
Revision as of 08:16, 22 May 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
Video Solution
~IceMatrix
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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