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Difference between revisions of "2020 AMC 10B Problems/Problem 19"

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<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>
 
<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>
  
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We need to get rid of the multiples of <math>3</math>, which will subsequently get rid of the multiples of <math>9</math>.
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We need to get rid of the multiples of <math>3</math>, which will subsequently get rid of the multiples of <math>9</math> (if we didn't, the zeroes would mess with the equation since you can't divide by 0)
  
 
<math>9\cdot5=45</math>, <math>8\cdot6=48</math>, <math>\frac{51}{3}</math> leaves us with 17.
 
<math>9\cdot5=45</math>, <math>8\cdot6=48</math>, <math>\frac{51}{3}</math> leaves us with 17.

Revision as of 15:19, 7 February 2020

Solution

$158A00A4AA0 \equiv 1+5+8+A+0+0+A+4+A+A+0 \equiv 4A \pmod{9}$

We're looking for the amount of ways we can get $10$ cards from a deck of $52$, which is represented by $\binom{52}{10}$.

$\binom{52}{10}=\frac{52\cdot51\cdot50\cdot49\cdot48\cdot47\cdot46\cdot45\cdot44\cdot43}{10\cdot9\cdot8\cdot7\cdot6\cdot5\cdot4\cdot3\cdot2\cdot1}$

We need to get rid of the multiples of $3$, which will subsequently get rid of the multiples of $9$ (if we didn't, the zeroes would mess with the equation since you can't divide by 0)

$9\cdot5=45$, $8\cdot6=48$, $\frac{51}{3}$ leaves us with 17.

$\frac{52\cdot\cancel{51}^{17}\cdot50\cdot49\cdot\cancel{48}\cdot47\cdot46\cdot\cancel{45}\cdot44\cdot43}{10\cdot\cancel{9}\cdot\cancel{8}\cdot7\cdot\cancel{6}\cdot\cancel{5}\cdot4\cdot\cancel{3}\cdot2\cdot1}$

Converting these into$\pmod{9}$, we have

$\binom{52}{10}\equiv \frac{(-2)\cdot(-1)\cdot(-4)\cdot4\cdot2\cdot1\cdot(-1)\cdot(-2)}{1\cdot(-2)\cdot4\cdot2\cdot1} \equiv (-1)\cdot(-4)\cdot(-1)\cdot(-2) \equiv 8 \pmod{9}$

$4A\equiv8\pmod{9} \implies A=\boxed{\textbf{(A) }2}$ ~quacker88

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