Difference between revisions of "2020 AMC 10B Problems/Problem 19"
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It is not hard to check that <math>13</math> divides the number, | 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>. | <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>. | ||
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<math>\textbf{- Emathmaster}</math> | <math>\textbf{- Emathmaster}</math> | ||
Revision as of 20:31, 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
It is not hard to check that divides the number,
As
, using
we have
. Thus
, implying
so the answer is
.
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.