Difference between revisions of "2020 AMC 10B Problems/Problem 19"
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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 | 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 | ||
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| + | ==Solution 3== | ||
==Video Solution== | ==Video Solution== | ||
Revision as of 19:20, 7 February 2020
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
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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
