imc

$158A00A4AA0\equiv 1+5+8+A+0+0+A+4+A+A+0\equiv 4A(\mathrm{mod}9)$ We're looking for the amount of ways we can get $10$ cards from a deck of $52$, which is represented by $\left(\genfrac{}{}{0ex}{}{52}{10}\right)$. $\left(\genfrac{}{}{0ex}{}{52}{10}\right)=\frac{52\cdot 51\cdot 50\cdot 49\cdot 48\cdot 47\cdot 46\cdot 45\cdot 44\cdot 43}{10\cdot 9\cdot 8\cdot 7\cdot 6\cdot 5\cdot 4\cdot 3\cdot 2\cdot 1}$ 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) $9\cdot 5=45$, $8\cdot 6=48$, $\frac{51}{3}$ leaves us with 17. $\frac{52\cdot {\cancel{51}}^{17}\cdot 50\cdot 49\cdot \cancel{48}\cdot 47\cdot 46\cdot \cancel{45}\cdot 44\cdot 43}{10\cdot \cancel{9}\cdot \cancel{8}\cdot 7\cdot \cancel{6}\cdot \cancel{5}\cdot 4\cdot \cancel{3}\cdot 2\cdot 1}$ Converting these into$(\mathrm{mod}9)$, we have $\left(\genfrac{}{}{0ex}{}{52}{10}\right)\equiv \frac{(-2)\cdot (-1)\cdot (-4)\cdot 4\cdot 2\cdot 1\cdot (-1)\cdot (-2)}{1\cdot (-2)\cdot 4\cdot 2\cdot 1}\equiv (-1)\cdot (-4)\cdot (-1)\cdot (-2)\equiv 8(\mathrm{mod}9)$ $4A\equiv 8(\mathrm{mod}9)⟹A=\boxed{2}$