2020 AMC 10B Problems/Problem 19

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 $9$ .

$9\cdot5=45$ , $8\cdot6=48$ , $\frac{51}{3}$ 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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2020 AMC 10B Problems/Problem 19

Solution