Competition · AMC preparation · step 4 of 4

AMC 10 · 2020A · #6

Grade 4 number-theory

Archetype Small-Domain Constrained Search └ Sub-archetype Sum-Target Search

digit-constraintsdivisibility-rulessystematic-enumerationmultiplesparity caseworksystematic-enumeration ↑ Prerequisites: divisibility-rulesdigit-constraints

📏 Short solution 💡 2 insights

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Problem

Count the 4-digit whole numbers (from 1000 to 9999) whose every digit is even AND that are divisible by 5.

Pick an answer.

(A)

(B)

100

(C)

125

(D)

200

(E)

500

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

How to solve

Strategy Make a Systematic List

The four digits are independent once we pin down their allowed sets — Tool #7 splits the count into four subproblems (one per digit position). Tool #2 lists the allowed values for each position so we don't miss or double-count. Tool #3 eliminates illegal values: odd digits at every position, and 0 at the thousands place. Multiplying the four counts gives the answer in one line.

1STEP 1

Fix the units digit

Units must be even and a multiple of 5, so it can only be 0.

units digit = 0 (1 choice)

2STEP 2

Count the thousands choices

Thousands digit is even but can't be 0, so 2, 4, 6, 8 — 4 choices.

thousands ∈ {2, 4, 6, 8} (4 choices)

3STEP 3

Count the hundreds choices

Hundreds digit just has to be even — any of 0, 2, 4, 6, 8, so 5 choices.

hundreds ∈ {0, 2, 4, 6, 8} (5 choices)

4STEP 4

Count the tens choices

Tens digit follows the same even rule — again 5 choices.

tens ∈ {0, 2, 4, 6, 8} (5 choices)

5STEP 5

Multiply the four counts

Independent digit choices multiply: 4 × 5 × 5 × 1 = 100.

4 × 5 × 5 × 1 = 100

6STEP 6

Match against the choices

That count, 100, is choice (B).

N = 100 → (B)

Answer

100

Sanity check the size. Without any constraint there are 9 · 10 · 10 · 10 = 9000 four-digit numbers. Even-only digits already drop the count by a lot: 4 · 5 · 5 · 5 = 500 (choice E — but that ignored divisibility by 5). Adding the divisibility constraint then divides the units-digit options from 5 down to 1, giving

\frac{500}{5}

= 100. The answer 100 fits in the middle of the choices and matches both counting paths.

💡Key takeaway

This AMC 10 problem only needs Grade 4 'multiples of 5 end in 0 or 5' plus place value you already know — units must be 0, leading digit picks from {2, 4, 6, 8}, middle two pick from all five even digits, so 4 × 5 × 5 × 1 = 100.

Fix the units digit
Count the thousands choices
Count the hundreds choices
Count the tens choices
Multiply the four counts
Match against the choices

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