Difference between revisions of "2020 AMC 10B Problems/Problem 19"
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| + | ==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>? | ||
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| + | <math>\textbf{(A) } 2 \qquad\textbf{(B) } 3 \qquad\textbf{(C) } 4 \qquad\textbf{(D) } 6 \qquad\textbf{(E) } 7</math> | ||
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==Solution== | ==Solution== | ||
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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 | ||
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| + | ==See Also== | ||
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| + | {{AMC10 box|year=2020|ab=B|num-b=18|num-a=20}} | ||
| + | {{MAA Notice}} | ||
Revision as of 16:25, 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
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
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. 
