Sample Question 19:

In a certain card game, a player is dealt a hand of \(10\) cards from a deck of \(52\) distinct cards. The number of distinct (unordered) hands that can be dealt to the player can be written as \(158A00A4AA0\). What is the digit \(A\)?

\(\textbf{(A) } 2 \qquad\textbf{(B) } 3 \qquad\textbf{(C) } 4 \qquad\textbf{(D) } 6 \qquad\textbf{(E) } 7\)

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Answer Keys

Question 19: A

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Solutions

Question 19
Step 1: Calculate the modulo from the number \(158A00A4AA0\). This is done by summing the digits and their residue when divided by 9. This gives us \(1+5+8+A+0+0+A+4+A+A+0 \equiv 4A \pmod{9}\).

Step 2: Determine the amount of ways to get 10 cards from a deck of 52, which is represented by \(\binom{52}{10}\). This is equal to \(\frac{52\cdot51\cdot50\cdot49\cdot48\cdot47\cdot46\cdot45\cdot44\cdot43}{10\cdot9\cdot8\cdot7\cdot6\cdot5\cdot4\cdot3\cdot2\cdot1}\).

Step 3: Simplify out the multiples of 3 from the equation to also get rid of multiples of 9. This results in \(9\cdot5=45\), \(8\cdot6=48\), and \(\frac{51}{3}\) leaving us with 17.

Step 4: Convert the binomial into mod 9, this results in \(\binom{52}{10}\equiv \frac{(-2)\cdot(-1)\cdot(-4)\cdot4\cdot2\cdot1\cdot(-1)\cdot(-2)}{1\cdot(-2)\cdot4\cdot2\cdot1} \equiv (-1)\cdot(-4)\cdot(-1)\cdot(-2) \equiv 8 \pmod{9}\).

Step 5: Equate the two modulus equations such as \(4A\equiv8\pmod{9}\) which implies \(A= 2\). This corresponds to answer \textbf{(A)} 2.