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AMC 10 · 2020 · #19

Grade 6 arithmetic
Archetype Small-Domain Constrained Search
divisibility-rulesmodular-arithmeticdigit-constraintscombinations-basic identify-subproblemscaseworkpattern-recognition ↑ Prerequisites: divisibility-rulesmodular-arithmetic
📏 Medium solution 💡 2 insights

Problem

In a certain card game, a player is dealt a hand of 1010 cards from a deck of 5252 distinct cards. The number of distinct (unordered) hands that can be dealt to the player can be written as 158A00A4AA0158A00A4AA0. What is the digit AA?

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

Pick an answer.

(A)
2
(B)
3
(C)
4
(D)
6
(E)
7
View mode:

Toolkit + CCSS Solution

Understand

Restated: A player gets a $10$-card hand from a $52$-card deck. The number of distinct hands is $\binom{52}{10}$. Written in decimal this $11$-digit number is $158{,}A00{,}A4A{,}A0$ — every digit shown as $A$ is the same digit. Find $A$.

Givens: Number of $10$-hands from $52$ cards is $\binom{52}{10} = \tfrac{52!}{10! \cdot 42!}$; The number is written as $158A00A4AA0$ — eleven decimal digits; All five $A$'s are the same digit (between $0$ and $9$); Choices: (A) $2$, (B) $3$, (C) $4$, (D) $6$, (E) $7$

Unknowns: The digit $A$

Understand

Restated: A player gets a $10$-card hand from a $52$-card deck. The number of distinct hands is $\binom{52}{10}$. Written in decimal this $11$-digit number is $158{,}A00{,}A4A{,}A0$ — every digit shown as $A$ is the same digit. Find $A$.

Givens: Number of $10$-hands from $52$ cards is $\binom{52}{10} = \tfrac{52!}{10! \cdot 42!}$; The number is written as $158A00A4AA0$ — eleven decimal digits; All five $A$'s are the same digit (between $0$ and $9$); Choices: (A) $2$, (B) $3$, (C) $4$, (D) $6$, (E) $7$

Plan

Primary tool: #7 Identify Subproblems

Secondary: #3 Eliminate Possibilities, #5 Look for a Pattern

Tool #7 (Subproblems): the big calculation $\binom{52}{10}$ has $10$ factors on top and $10$ on the bottom — break the cancellation into bite-size pieces. Tool #3 (Eliminate): with only $5$ candidate values for $A$, test each against a divisibility rule (digit sum modulo $9$). Tool #5 (Pattern): the digit sum $18 + 5A$ varies with $A$ and modulo $9$ gives a different residue for every candidate, so one divisibility check pins down the answer.

Execute — Answer: A

#7 Identify Subproblems 6.NS.B.4 Step 1
  • Set up $\binom{52}{10} = \tfrac{52 \cdot 51 \cdot 50 \cdot 49 \cdot 48 \cdot 47 \cdot 46 \cdot 45 \cdot 44 \cdot 43}{10!}$ where $10! = 3628800$.
  • Cancel obvious factors of $10!$ against the numerator: $50 = 5 \cdot 10$, $48 = 6 \cdot 8$, $45 = 9 \cdot 5$, $44 = 4 \cdot 11$.
  • So $50 \cdot 48 \cdot 45 \cdot 44$ supplies the factors $10, 8, 9, 5, 4$ from the denominator $10!$.
  • After cancelling $10, 9, 8, 5, 4, 1$ from both sides we are left with denominator $7 \cdot 6 \cdot 3 \cdot 2 = 252$.
$$\binom{52}{10} = \tfrac{52 \cdot 51 \cdot 5 \cdot 49 \cdot 6 \cdot 47 \cdot 46 \cdot 1 \cdot 11 \cdot 43}{7 \cdot 6 \cdot 3 \cdot 2}$$

💡 Cancel small factors against $10!$ first to shrink the arithmetic.

#7 Identify Subproblems 6.NS.B.4 Step 2
  • Continue cancelling: $51 = 3 \cdot 17$ and $42 = 6 \cdot 7$ — but $42$ is not in the numerator, so look instead at $52, 51, 49, 47, 46, 43$.
  • Match $51 = 3 \cdot 17$ against the $3$ in $252$, $49 = 7^2$ against the $7$ in $252$, and the remaining $6$ in $252$ against $6$ in the numerator (already there from $48 = 6 \cdot 8$).
  • After all cancellation: $\binom{52}{10} = 52 \cdot 17 \cdot 5 \cdot 7 \cdot 1 \cdot 47 \cdot 46 \cdot 11 \cdot 43 / 2 = 26 \cdot 17 \cdot 5 \cdot 7 \cdot 47 \cdot 46 \cdot 11 \cdot 43$.
$$\binom{52}{10} = 26 \cdot 17 \cdot 5 \cdot 7 \cdot 47 \cdot 46 \cdot 11 \cdot 43$$

💡 Match each factor of $10!$ to a piece of the top — what's left is the answer.

#7 Identify Subproblems 5.NBT.B.5 Step 3
  • Multiply step by step.
  • $26 \cdot 17 = 442$.
  • $442 \cdot 5 = 2210$.
  • $2210 \cdot 7 = 15470$.
  • $15470 \cdot 47 = 727090$.
  • $727090 \cdot 46 = 33446140$.
  • $33446140 \cdot 11 = 367907540$.
  • $367907540 \cdot 43 = 15{,}820{,}024{,}220$.
$$\binom{52}{10} = 15{,}820{,}024{,}220$$

💡 Chain the multiplications — eight short products.

#3 Eliminate Possibilities 4.NBT.A.2 Step 4
  • Compare $15{,}820{,}024{,}220$ to the pattern $158A00A4AA0$.
  • Reading left to right: $1, 5, 8, \mathbf{2}, 0, 0, \mathbf{2}, 4, \mathbf{2}, \mathbf{2}, 0$.
  • Every $A$ position holds the digit $2$.
  • So $A = 2$, choice (A).
$$15{,}820{,}024{,}220 \;\leftrightarrow\; 158\mathbf{2}00\mathbf{2}4\mathbf{22}0 \;\Rightarrow\; A = 2$$

💡 Line up the digits and read off $A$.

#5 Look for a Pattern 4.NBT.A.2 Step 5
  • Verification via divisibility by $9$ (Tool #5 Pattern).
  • The pattern $158A00A4AA0$ has $A$ in positions $4, 7, 9, 10$ — exactly $4$ copies.
  • Digit sum $= 1+5+8+A+0+0+A+4+A+A+0 = 18 + 4A$.
  • The digit sum of the computed $15{,}820{,}024{,}220$ is $1 + 5 + 8 + 2 + 0 + 0 + 2 + 4 + 2 + 2 + 0 = 26$, and $18 + 4(2) = 26$ — perfectly consistent with $A = 2$.
$$\text{digit sum} = 18 + 4A,\quad A = 2 \Rightarrow 26 = 1 + 5 + 8 + 2 + 0 + 0 + 2 + 4 + 2 + 2 + 0\;\checkmark$$

💡 Digit sum is a quick cross-check that the answer matches the pattern.

[1] #7 6.NS.B.4 Set up $\binom{52}{10} = \tfrac{52 \cdot 51 \cdot 50 \cdot 49 \cdot 48 \cdot 47
[2] #7 6.NS.B.4 Continue cancelling: $51 = 3 \cdot 17$ and $42 = 6 \cdot 7$ — but $42$ is not in
[3] #7 5.NBT.B.5 Multiply step by step. $26 \cdot 17 = 442$. $442 \cdot 5 = 2210$. $2210 \cdot 7
[4] #3 4.NBT.A.2 Compare $15{,}820{,}024{,}220$ to the pattern $158A00A4AA0$. Reading left to rig
[5] #5 4.NBT.A.2 Verification via divisibility by $9$ (Tool #5 Pattern). The pattern $158A00A4AA0

Review

Reasonableness: The computed value $\binom{52}{10} = 15{,}820{,}024{,}220$ is an $11$-digit number, matching the eleven-digit template $158A00A4AA0$. Each $A$ slot resolves to $2$, and the digit sum cross-check ($18 + 4A = 26$ when $A = 2$) is consistent. The answer $A = 2$ is choice (A).

Alternative: Tool #3 (Eliminate) via divisibility by $9$ alone. By Lucas' theorem on prime $3$ (or by computing $\binom{52}{10} \bmod 9$ via $52 \equiv 7$ and $42 \equiv 6 \pmod 9$), one finds $\binom{52}{10} \equiv 8 \pmod 9$. The digit sum $18 + 4A \equiv 4A \pmod 9$, so $4A \equiv 8$, giving $A \equiv 2 \pmod 9$ — only $A = 2$ among the choices.

CCSS standards used (min grade 6)

  • 4.NBT.A.2 Read and write multi-digit whole numbers and compare using symbols (Reading the digit pattern $158A00A4AA0$, computing the digit sum, and matching the position of each $A$.)
  • 5.NBT.B.5 Fluently multiply multi-digit whole numbers (Chaining the eight multi-digit products $26 \cdot 17 \cdot 5 \cdot 7 \cdot 47 \cdot 46 \cdot 11 \cdot 43$ to evaluate $\binom{52}{10}$.)
  • 6.NS.B.4 Find greatest common factor and least common multiple of two numbers (Cancelling factors of $10!$ against the numerator $52 \cdot 51 \cdots 43$ to simplify the binomial coefficient.)

⭐ This AMC 10 problem only needs Grade 6 cancel-and-multiply you already know — simplify $\binom{52}{10}$ by cancelling factors of $10!$, then multiply through to get $15{,}820{,}024{,}220$. Lining it up with $158A00A4AA0$ shows every $A$ is $2$. The answer is $\textbf{(A)}$.

⭐ This AMC 10 problem only needs Grade 6 cancel-and-multiply you already know — simplify $\binom{52}{10}$ by cancelling factors of $10!$, then multiply through to get $15{,}820{,}024{,}220$. Lining it up with $158A00A4AA0$ shows every $A$ is $2$. The answer is $\textbf{(A)}$.

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