Difference between revisions of "2020 AMC 10B Problems/Problem 19"
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It is obvious that <math>K \cong 0 (mod 4)</math>. Also (although this not so obvious), <math>K \cong (13)(17)(7)(47)(46)(5)(22)(43) \cong (13)(-8)(7)(-3)(-4)(5)(-3)(-7) \cong (13)(-96)(21)(35) \cong (13)(4)(-4)(10) \cong (13)(-16)(10) \cong (13)(9)(10) \cong (117)(10) \cong (-8)(10) \cong 20 (mod 25)</math>. Therefore, <math>K \cong 20 (mod 100)</math>. Thus <math>K=20</math>, implying that <math>A=2</math>. (A) | It is obvious that <math>K \cong 0 (mod 4)</math>. Also (although this not so obvious), <math>K \cong (13)(17)(7)(47)(46)(5)(22)(43) \cong (13)(-8)(7)(-3)(-4)(5)(-3)(-7) \cong (13)(-96)(21)(35) \cong (13)(4)(-4)(10) \cong (13)(-16)(10) \cong (13)(9)(10) \cong (117)(10) \cong (-8)(10) \cong 20 (mod 25)</math>. Therefore, <math>K \cong 20 (mod 100)</math>. Thus <math>K=20</math>, implying that <math>A=2</math>. (A) | ||
− | + | (I need help with finding the modulo congruence symbol. Thanks!) | |
==Video Solution== | ==Video Solution== |
Revision as of 07:27, 22 May 2020
Contents
[hide]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
. (A)
(I need help with finding the modulo congruence symbol. Thanks!)
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.