AMC 8 · 2011 · #23

Grade 5 counting

Archetype Casework by Position / Vertex Type └ Sub-archetype Digit Place Position

systematic-enumerationcombinations-basicdigit-constraintsdivisibility-rules caseworksystematic-enumeration ↑ Prerequisites: systematic-enumerationdivisibility-rules

📏 Long solution 💡 4 insights

Problem

Count the 4-digit positive integers ABCD whose four digits are all different, with leading digit A ≠ 0, the whole number divisible by 5, and the digit 5 being the largest digit used.

Pick an answer.

(A)

(B)

(C)

(D)

(E)

108

AMC 8 2011 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve

Strategy Identify Subproblems

The divisibility-by-5 rule forces D = 0 or D = 5, so Tool #7 (Identify Subproblems) splits the count into two clean cases on the last digit. Inside each case, Tool #2 (Make a Systematic List) counts the choices position by position using the multiplication principle: pick A, then B, then C from the digits that remain in {0,1,2,3,4,5}. Tool #3 (Eliminate Possibilities) handles the two side-conditions — A ≠ 0 removes one option from the thousands slot, and "5 must appear somewhere" forces a placement choice in the D = 0 case where 5 isn't already used as the last digit.

1STEP 1

Since 5 is the largest digit, every digit lies in {0,1,2,3,4,5}; divisibility by 5 forces D ∈ {0, 5}, splitting into two disjoint cases.

Case 1: D = 5 | Case 2: D = 0

2STEP 2

Case 1 (D = 5): 5 is already used, so fill A (≠ 0, 4 ways), B (4 ways), C (3 ways) from {0,1,2,3,4} → 48.

4 × 4 × 3 = 48 numbers

3STEP 3

Case 2 (D = 0): 5 must sit in A, B, or C — place it (3 ways), then fill two slots from {1,2,3,4} (4 × 3) → 36.

3 × (4 × 3) = 3 × 12 = 36 numbers

4STEP 4

The two cases are disjoint (D = 5 vs D = 0), so add them: 48 + 36 = 84.

48 + 36 = 84 ⟹ (D)

Answer

Sanity-check the sizes. The total count of 4-digit numbers with distinct digits drawn from {0,1,2,3,4,5} and A ≠ 0 is 5 × 5 × 4 × 3 = 300. Of those, the fraction with D ∈ {0, 5} should be roughly

\frac{2}{6}

\frac{1}{3}

, giving about 100. Our 48 + 36 = 84 sits just below that estimate — lower because requiring the digit 5 to actually appear trims out the all-{0,1,2,3,4} numbers. 84 matches choice (D) and is small enough to be plausible, large enough to be more than the trivial counts in (A)-(B).

💡Key takeaway

This AMC 8 problem only needs the Grade 5 habit of splitting into cases from a rule plus slot-by-slot multiplication you learned in Grade 3!

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