AMC 10 · 2020 · #4

Grade 4 number-theory
digit-constraintsdivisibility-rulessystematic-enumerationmultiplesparity caseworksystematic-enumeration ↑ Prerequisites: divisibility-rulesdigit-constraints
📏 Short solution 💡 2 insights
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Problem
Count the four-digit whole numbers from 1000 to 9999 whose every digit is even and that are also divisible by 5.

Pick an answer.

(A)
80
(B)
100
(C)
125
(D)
200
(E)
500
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

The two rules leave only zero.

units digit = 0 (1 choice)
2STEP 2

Count the leading digit

The leading digit cannot be zero.

thousands ∈ {2, 4, 6, 8} (4 choices)
3STEP 3

Count the hundreds digit

A middle digit has five choices.

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

Count the tens digit

The tens digit is the same.

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

Multiply the counts

Multiply the four slot counts.

4 × 5 × 5 × 1 = 100
6STEP 6

Match the choice

The result is 100.

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 500 / 5 = 100. The answer 100 fits in the middle of the choices and matches both counting paths.
💡Key takeaway

This AMC 12 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.