Competition · AMC preparation · step 4 of 4
AMC 8 · 2007 · #4
Grade 4 countingPick an answer.
AMC 8 2007 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.
The trip splits into two independent jobs: first pick the entry window, then pick the exit window. Tool #7 (Identify Subproblems) handles that split — count each piece on its own, then combine by multiplication. Tool #2 (Systematic List) is the safety net: imagine writing the entry window in a column and the exit window beside it, with the rule that the two must differ. Listing a few rows confirms each entry has exactly 5 matching exits, so 6 × 5 gives the total.
Count the entry choices
Georgie can enter through any window, so the entry has 6 choices.
Pick one window from a set of 6 — that is 6 equal options for the first action.
3.OA.A.1Identify SubproblemsCount the exit choices
The exit must differ from the entry, so one window is gone, leaving 5 exit choices.
The "different window" rule removes exactly one window from the exit pool, leaving 5.
3.OA.A.1Identify SubproblemsMultiply the two counts
Each entry pairs with its 5 exits, so multiply: 6 × 5 = 30 trips.
A systematic list with 6 entry rows, each followed by its 5 allowed exits, has 6 × 5 = 30 rows total.
The total number of possible entry-exit trips equals the six entry windows multiplied by the five allowed exit windows for each entry.
▸ Why?
Every one of the six entry windows offers the same five allowed exits, so the trips split into six equal groups of five, and totaling equal groups is exactly a multiplication.
▸ Why?
Six equal groups of five means adding five six times over, and that repeated sum of equal amounts is what six times five stands for.
▸ Why?
Each entry uses one window, and the six windows split with no overlap into that one used window and the ones left for exit, so five remain.
Two actions in a row: count each, then multiply. Six entry windows times five remaining exit windows gives 30 — a Grade 4 multi-step multiplication problem.
- Count the entry choices
- Count the exit choices
- Multiply the two counts
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