Competition · AMC preparation · step 4 of 4
AMC 8 · 2015 · #24
Grade 7 algebranumber-theoryPick an answer.
AMC 8 2015 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.
Tool #1 (Write an Equation) is the natural first move: count one team's opponents in each division and translate "76 games total" into a single linear equation in N and M. That equation alone has many solutions, so we then use Tool #12 (Use Cases) — combine the two inequalities (N ＞ 2M, M ＞ 4) with the requirement that N and M are integers to narrow M to a tiny list (M = 5, 6, 7) and test each case. Only one survives, giving the unique (N, M) pair that answers the question.
Set up the games equation
One team meets 3 rivals in its own division (3N games) and 4 teams across (4M), so 3N + 4M = 76.
Using letters N and M to stand for unknown game counts and writing one equation from the word problem is the Grade 6 "use variables to represent numbers and write expressions" standard.
6.EE.B.6Draw A DiagramSolve for N in terms of M
Solve for N so the condition N ＞ 2M becomes a bound on M alone: N = .
Isolating one variable in a two-variable equation is the algebraic move taught in Grade 7.
7.EE.B.4Draw A DiagramTurn the inequality into a bound
Substituting into N ＞ 2M gives M ＜ 7.6; with M ＞ 4 and M an integer, M ∈ {5, 6, 7}.
Solving a multi-step inequality and reading off the integer values inside the range is exactly the Grade 7 inequalities standard.
7.EE.B.4Draw A Venn DiagramTest each candidate M
Only M = 7 makes 76 - 4M divisible by 3, so M = 5 and 6 fail and N = 16.
Checking divisibility by 3 for three small numbers is the Grade 6 factors-and-multiples skill at work.
Among the surviving candidates M = 5, 6, 7, only M = 7 makes N = (76 - 4M)/3 a whole number, so the schedule is forced to M = 7 and N = 16.
▸ Why?
Only M = 7 turns 76 - 4M into a whole multiple of 3 (namely 48 = 3 × 16), so it is the one candidate that gives a whole-number N.
▸ Why?
N counts how many times two teams actually play, and a count of games is never a fraction, so N must be a whole number.
▸ Why?
For a whole N, the top 76 - 4M must be exactly 3 times a whole number, because N = (76 - 4M)/3 means 3N rebuilds 76 - 4M, and dividing by 3 simply undoes multiplying by 3.
▸ Why?
Testing the three values, only M = 7 passes: 76 - 4(7) = 48 = 3 × 16 is three equal whole groups, while 76 - 4(5) = 56 and 76 - 4(6) = 52 cannot be split into three equal whole groups.
Count the intra-division games
Check 16 ＞ 2 × 7 holds, then the same-division total is 3N = 3 × 16 = 48.
Substituting the solved value back into the expression 3N is the standard "evaluate at a specific value" move from Grade 6.
6.EE.B.6Draw A DiagramWrite the count as one equation, then let the inequalities and the "games must be whole numbers" rule shrink the choices until just one (N, M) pair is left.
- Set up the games equation
- Solve for N in terms of M
- Turn the inequality into a bound
- Test each candidate M
- Count the intra-division games
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