Competition · AMC preparation · step 4 of 4
AMC 8 · 2012 · #10
Grade 7 countingPick an answer.
AMC 8 2012 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.
There are only a handful of arrangements of {0, 1, 2, 2}, so Tool #4 (Make a Systematic List) is the safest way to count without missing or double-counting — especially since the two 2s are identical. Tool #7 (Identify Subproblems) splits the list cleanly by the leading digit: either the number starts with 1 or it starts with 2 (it cannot start with 0). Counting each case separately and adding is much easier than juggling the whole thing at once.
Decide which digits can lead
A 4-digit number over 1000 needs a nonzero leading digit, and only 1 and 2 qualify — so split into two cases.
Knowing that the thousands place controls how big the number is — and that a 0 there shrinks it to 3 digits — is Grade 4 place-value reasoning.
4.NBT.A.2Identify SubproblemsList the numbers starting with 1
Case 1 — lead with 1, then arrange {0, 2, 2} in the last three slots; the lone 0 has three spots, giving 3 numbers.
Writing out the three positions for the lone 0 is exactly the Grade 7 "organized list" method for counting outcomes.
7.SP.C.8Introduce A VariableList the numbers starting with 2
Case 2 — lead with 2, then arrange the distinct digits {0, 1, 2}; three different digits order 6 ways, giving 6 numbers.
Three distinct items have 3 × 2 × 1 = 6 orderings, and an organized list confirms every one.
7.SP.C.8Introduce A VariableAdd the two cases
The two cases can't overlap, so just add them: 3 + 6 = 9 numbers, which is (D).
Combining disjoint case counts by addition is the Grade 4 multi-step word-problem move.
The count of these numbers is the count that start with 1 added to the count that start with 2.
▸ Why?
Every valid number begins with either 1 or 2 and never with 0, so sorting them by their first digit drops each number into exactly one group with nothing left over and no number in two groups.
▸ Why?
To be a full four-digit number above 1000 the thousands place must hold a nonzero digit, and among 0, 1, 2, 2 the only nonzero choices are 1 and 2.
▸ Why?
When groups do not overlap and together use up every case, their separate sizes add back to the size of the whole.
▸ Why?
Fixing 1 in front leaves 0, 2, 2 for the last three places, and because the two 2s are identical the whole arrangement is decided only by where the single 0 sits, so there are as many numbers as there are slots for that 0.
▸ Why?
Each of the three possible slots for the lone 0 matches exactly one finished number, so counting the numbers is the same as counting the slots.
▸ Why?
Fixing 2 in front leaves 0, 1, 2 — three different digits — for the last three places, and filling those places one at a time offers 3 choices, then 2, then 1.
▸ Why?
Each position is an independent choice from the digits not yet used, and the number of options at each stage is fixed at 3, then 2, then 1 no matter what was picked before; because successive independent choices combine by multiplying, the total is 3 × 2 × 1.
An organized list — split by what the leading digit can be — turns this AMC 8 counting problem into a quick Grade 7 sample-space exercise.
- Decide which digits can lead
- List the numbers starting with 1
- List the numbers starting with 2
- Add the two cases
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