AMC 10 · 2020 · #6
Grade 4 number-theoryPick an answer.
AMC 10 2020 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.
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.
Units must be even and a multiple of 5, so it can only be 0.
Two constraints overlap to leave a single legal digit.
4.OA.B.4Eliminate PossibilitiesThousands digit is even but can't be 0, so 2, 4, 6, 8 — 4 choices.
A 4-digit number must start with a non-zero digit.
2.NBT.A.1Make A Systematic ListHundreds digit just has to be even — any of 0, 2, 4, 6, 8, so 5 choices.
Any even digit works in a non-leading slot.
2.NBT.A.1Make A Systematic ListTens digit follows the same even rule — again 5 choices.
Tens place follows the same even rule, no leading constraint.
2.NBT.A.1Make A Systematic ListIndependent digit choices multiply: 4 × 5 × 5 × 1 = 100.
Independent choices multiply — like a 4-slot decision tree.
3.OA.A.1Identify SubproblemsThat count, 100, is choice (B).
Read the matching answer choice.
3.OA.A.3Eliminate PossibilitiesThis 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.