Competition · AMC preparation · step 4 of 4

AMC 10 · 2020B · #19

Grade 6 arithmetic
Archetype Small-Domain Constrained Search └ Sub-archetype Factor / Divisibility Constraint
divisibility-rulesmodular-arithmeticdigit-constraintscombinations-basic identify-subproblemscaseworkpattern-recognition ↑ Prerequisites: divisibility-rulesmodular-arithmetic
📏 Medium solution 💡 2 insights
Problem
A player gets a 10-card hand from a 52-card deck. The number of distinct hands is C(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.

Pick an answer.

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

AMC 10 2020 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve
Strategy Identify Subproblems

Tool #7 (Subproblems): the big calculation C(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.

1STEP 1

Set up the binomial coefficient

Write C(52, 10) as ten falling factors over 10!; cancel 10!'s small factors against the top, leaving denominator 252.

C(52, 10) = (52 · 51 · 5 · 49 · 6 · 47 · 46 · 1 · 11 · 43)/(7 · 6 · 3 · 2)
2STEP 2

Cancel the common factors

Cancel 252 against the top (51=3·17, 49=7·7, the 6, 52=2·26), leaving C(52, 10) = 26 · 17 · 5 · 7 · 47 · 46 · 11 · 43.

C(52, 10) = 26 · 17 · 5 · 7 · 47 · 46 · 11 · 43
3STEP 3

Multiply it out

Chain the eight products (26·17=442, ·5, ·7, ·47, ·46, ·11, ·43) to reach C(52, 10) = 15,820,024,220.

C(52, 10) = 15,820,024,220
4STEP 4

Match the digit pattern

Line up 15,820,024,220 with 158A00A4AA0 digit by digit; every A slot holds the same value, giving A = 2, choice (A).

15,820,024,220 ⇔ 15820024220 → A = 2
5STEP 5

Check with divisibility by 9

Cross-check by 9s: the template's digit sum is 18 + 4A, and 15,820,024,220 sums to 26 = 18 + 4·2, consistent with A = 2.

digit sum = 18 + 4A, A = 2 → 26 = 1 + 5 + 8 + 2 + 0 + 0 + 2 + 4 + 2 + 2 + 0 ✓
Answer
2
The computed value C(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).
💡Key takeaway

This AMC 10 problem only needs Grade 6 cancel-and-multiply you already know — simplify C(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 (A).

  • Set up the binomial coefficient
  • Cancel the common factors
  • Multiply it out
  • Match the digit pattern
  • Check with divisibility by 9

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