Difference between revisions of "2020 AMC 10B Problems/Problem 19"
(Created page with "==Solution== <math>158A00A4AA0 \equiv 1+5+8+A+0+0+A+4+A+A+0 \equiv 4A \pmod{9}</math> We're looking for the amount of ways we can get <math>10</math> cards from a deck of <m...") |
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<math>\binom{52}{10}\equiv \frac{(-2)\cdot(-1)\cdot(-4)\cdot4\cdot2\cdot1\cdot(-1)\cdot(-2)}{1\cdot(-2)\cdot4\cdot2\cdot1} \equiv (-1)\cdot(-4)\cdot(-1)\cdot(-2) \equiv 8 \pmod{9} </math> | <math>\binom{52}{10}\equiv \frac{(-2)\cdot(-1)\cdot(-4)\cdot4\cdot2\cdot1\cdot(-1)\cdot(-2)}{1\cdot(-2)\cdot4\cdot2\cdot1} \equiv (-1)\cdot(-4)\cdot(-1)\cdot(-2) \equiv 8 \pmod{9} </math> | ||
| − | <math>4A\equiv8\pmod{9} \implies A=\boxed{\textbf{(A) }2}</math> | + | <math>4A\equiv8\pmod{9} \implies A=\boxed{\textbf{(A) }2}</math> ~quacker88 |
Revision as of 15:18, 7 February 2020
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 
.
, 
, 
leaves us with 17.
Converting these into
, we have
~quacker88
