AMC 8 · 2016 · #17
Grade 4 countingPick an answer.
AMC 8 2016 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.
Counting allowed passwords directly is messy — we would have to handle every first-digit case carefully. The forbidden set is much smaller, so Tool #16 (Complement) wins: count ALL 4-digit strings, count just the forbidden ones starting with 9,1,1, and subtract. Tool #2 (Systematic List) is the natural way to confirm the forbidden count — list 9110, 9111, …, 9119 and check we have exactly 10 items.
Count the whole universe first: 4 positions, 10 choices each, so 10,000 passwords with no rules.
Multiplying choices across independent positions is the basic counting move from Grade 4 multiplicative reasoning.
4.OA.A.1Count The ComplementCount the forbidden ones: first three digits pinned to 9, 1, 1 and only the 4th is free, giving 10 banned passwords.
Fixing positions reduces the count: a forced slot contributes a factor of 1, an open slot contributes its full count.
4.OA.A.1Count The ComplementList them in order to be sure: 9110, 9111, …, 9119 — exactly 10 items, only the last digit varies.
Listing in order is the safest way to confirm a small count and avoid off-by-one errors.
3.OA.A.1Make A Systematic ListSubtract forbidden from total — the complement step: 10,000 - 10 = 9,990 allowed passwords (D).
"Allowed = total minus forbidden" is the complement principle in one line.
4.NBT.B.4Count The ComplementWhen the "not allowed" cases are few, count those instead and subtract — that's the complement trick, and it only needs Grade 4 multiplication and subtraction.