Difference between revisions of "2020 AMC 10B Problems/Problem 19"
Arcticturn (talk | contribs) (→Solution 7) |
King.of.math (talk | contribs) (Added a new solution that I used to solve this and did not find in the other solutions listed above) |
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~Arcticturn | ~Arcticturn | ||
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+ | ==Solution 8 (Very time consuming)== | ||
+ | As stated in previous solutions, the number of ways to choose <math>10</math> from <math>52</math> is | ||
+ | <math>\binom{52}{10}=\frac{52\cdot51\cdot50\cdot49\cdot48\cdot47\cdot46\cdot45\cdot44\cdot43}{10\cdot9\cdot8\cdot7\cdot6\cdot5\cdot4\cdot3\cdot2\cdot1}</math> | ||
+ | |||
+ | Canceling out common factors (52/2=26, 51/3=17, 5*10=50, 49/7=7, 48=6*8, 45/9=5, 44/11=4) , you get this - | ||
+ | <math>\frac{\cancel{52}^{26}\cdot\cancel{51}^{17}\cdot\cancel{50}\cdot\cancel{49}^{7}\cdot\cancel{48}\cdot47\cdot46\cdot\cancel{45}^5\cdot\cancel{44}^{4}\cdot43}{\cancel{10}\cdot\cancel{9}\cdot\cancel{8}\cdot\cancel{7}\cdot\cancel{6}\cdot\cancel{5}\cdot\cancel{4}\cdot\cancel{3}\cdot\cancel{2}\cdot\cancel{1}}</math> | ||
+ | |||
+ | From this product, we can then determine that <math>A</math> is equal to <math>\boxed {\textbf{(A)2}}</math> | ||
+ | |||
+ | ~ KING.OF.MATH | ||
==Video Solutions== | ==Video Solutions== |
Revision as of 18:27, 8 November 2022
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
.
Solution 6
As in Solution 2, we see that
which contains no factors of Therefore, the sum of the digits must not be a multiple of
This sum is
It follows that cannot be a multiple of
ruling out choices
and
Therefore, our possibilities are
and
Now, notice that
is divisible by
Therefore, we can plug each possible value of
into
and test for divisibility by
Conveniently, we see that the first value,
works. Thus, the answer is
(To make our argument more rigorous, we can also test divisibility by
for
and
to show that these values do not work.)
--vaporwave
Solution 7
The total number of ways to choose from
is
Using divisibility rules, we have that A is not a multiple of . Then, divide this equation by 10. This implies that the new number
is divisible by
but not
. This means that
is either
or
. However,
is a multiple of
, meaning
has to be
~Arcticturn
Solution 8 (Very time consuming)
As stated in previous solutions, the number of ways to choose from
is
Canceling out common factors (52/2=26, 51/3=17, 5*10=50, 49/7=7, 48=6*8, 45/9=5, 44/11=4) , you get this -
From this product, we can then determine that is equal to
~ KING.OF.MATH
Video Solutions
Video Solution 1
~IceMatrix
Video Solution 2
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.