AMC 8 · 2015 · #24
Grade 7 algebranumber-theoryProblem
A baseball league consists of two four-team divisions. Each team plays every other team in its division N games. Each team plays every team in the other division M games with N>2M and M>4. Each team plays a 76 game schedule. How many games does a team play within its own division?
Pick an answer.
AMC 8 2015 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.
Try it yourself first — the explanation is most useful after you’ve attempted it.
Toolkit + CCSS Solution
Understand
Restated: A baseball league has two divisions of $4$ teams each. Inside a division, every pair of teams plays $N$ games against each other; across divisions, every pair plays $M$ games. We're told $N > 2M$, $M > 4$, and each team plays a total of $76$ games. How many of those $76$ games are against teams in the same division?
Givens: $2$ divisions, each with $4$ teams; Intra-division: each opponent is played $N$ times; Inter-division: each opponent is played $M$ times; Constraints: $N > 2M$ and $M > 4$; Each team's full season $= 76$ games; Answer choices: (A) $36$, (B) $48$, (C) $54$, (D) $60$, (E) $72$
Unknowns: The number of intra-division games on one team's schedule
Understand
Restated: A baseball league has two divisions of $4$ teams each. Inside a division, every pair of teams plays $N$ games against each other; across divisions, every pair plays $M$ games. We're told $N > 2M$, $M > 4$, and each team plays a total of $76$ games. How many of those $76$ games are against teams in the same division?
Givens: $2$ divisions, each with $4$ teams; Intra-division: each opponent is played $N$ times; Inter-division: each opponent is played $M$ times; Constraints: $N > 2M$ and $M > 4$; Each team's full season $= 76$ games; Answer choices: (A) $36$, (B) $48$, (C) $54$, (D) $60$, (E) $72$
Plan
Primary tool: #1 Write an Equation
Secondary: #12 Use Cases
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.
Execute — Answer: B
6.EE.B.6 Step 1 - Count opponents and write the total-games equation.
- Inside a team's own division there are $4 - 1 = 3$ other teams, each played $N$ times, contributing $3N$ games.
- The other division has $4$ teams, each played $M$ times, contributing $4M$ games.
- The season total is $76$.
💡 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.
7.EE.B.4 Step 2 Solve the equation for $N$ in terms of $M$ so the inequality $N > 2M$ becomes a condition on $M$ alone.
💡 Isolating one variable in a two-variable equation is the algebraic move taught in Grade 7.
7.EE.B.4 Step 3 - Substitute that expression into $N > 2M$ and simplify to an upper bound on $M$.
- Combined with $M > 4$, this pins $M$ down to a short list of integers.
💡 Solving a multi-step inequality and reading off the integer values inside the range is exactly the Grade 7 inequalities standard.
6.NS.B.4 Step 4 - Test each candidate $M$ to see which one makes $N = (76 - 4M)/3$ a positive integer.
- The numerator must be divisible by $3$.
💡 Checking divisibility by $3$ for three small numbers is the Grade 6 factors-and-multiples skill at work.
6.EE.B.6 Step 5 Confirm the surviving case satisfies the original $N > 2M$, then plug $N = 16$ into the intra-division count $3N$.
💡 Substituting the solved value back into the expression $3N$ is the standard "evaluate at a specific value" move from Grade 6.
6.EE.B.6 Count opponents and write the total-games equation. Inside a team's own division 7.EE.B.4 Solve the equation for $N$ in terms of $M$ so the inequality $N > 2M$ becomes a 7.EE.B.4 Substitute that expression into $N > 2M$ and simplify to an upper bound on $M$. 6.NS.B.4 Test each candidate $M$ to see which one makes $N = (76 - 4M)/3$ a positive inte 6.EE.B.6 Confirm the surviving case satisfies the original $N > 2M$, then plug $N = 16$ i Review
Reasonableness: Plug $N = 16$, $M = 7$ back into the full season: $3(16) + 4(7) = 48 + 28 = 76$. $\checkmark$ Both inequalities hold: $16 > 2 \cdot 7 = 14$ and $7 > 4$. The split $48$ intra $+ 28$ inter $= 76$ matches the answer choice (B), and the intra-division share is more than half the schedule, which fits "$N > 2M$" (you play same-division opponents more often than cross-division ones).
Alternative: Tool #6 (Guess and Check) directly on the answer choices: if intra-division games $= 3N$ equals one of the choices, then $N$ must be choice$/3$, an integer. (A) $36 \Rightarrow N = 12$, $4M = 40$, $M = 10$, but $N > 2M$ fails ($12 \not> 20$). (B) $48 \Rightarrow N = 16$, $4M = 28$, $M = 7$: $16 > 14$ $\checkmark$ and $7 > 4$ $\checkmark$. (C) $54 \Rightarrow N = 18$, $4M = 22$, $M = 5.5$, not an integer. (D) $60 \Rightarrow N = 20$, $4M = 16$, $M = 4$, but $M > 4$ fails. (E) $72 \Rightarrow N = 24$, $4M = 4$, $M = 1$, but $M > 4$ fails. Only (B) survives.
CCSS standards used (min grade 7)
6.EE.B.6Use variables to represent numbers and write expressions when solving real-world problems (Letting $N$ and $M$ stand for the unknown intra- and inter-division game counts and writing the schedule equation $3N + 4M = 76$, then evaluating $3N$ at $N = 16$.)6.NS.B.4Find common factors and multiples; apply divisibility reasoning (Deciding which of $M = 5, 6, 7$ makes $76 - 4M$ divisible by $3$ so that $N$ is a whole number of games.)7.EE.B.4Use variables to represent quantities in real-world problems and construct simple equations and inequalities to solve them (Solving the equation $3N + 4M = 76$ for $N$ and then solving the multi-step inequality $(76 - 4M)/3 > 2M$ down to $M < 7.6$.)
⭐ Write 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.
⭐ Write 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.
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