AMC 8 · 2015 · #24

Grade 7 algebranumber-theory
linear-diophantinedivisibility-rulescombinations-basic convert-to-algebrabound-inequality-then-enumeratecasework ↑ Prerequisites: linear-equations-one-vardivisibility-rules
📏 Medium solution 💡 3 insights

Problem

A baseball league consists of two four-team divisions. Each team plays every other team in its division NN games. Each team plays every team in the other division MM games with N>2MN>2M and M>4M>4. Each team plays a 7676 game schedule. How many games does a team play within its own division?

(A) 36(B) 48(C) 54(D) 60(E) 72\textbf{(A) } 36 \qquad \textbf{(B) } 48 \qquad \textbf{(C) } 54 \qquad \textbf{(D) } 60 \qquad \textbf{(E) } 72

Pick an answer.

(A)
36
(B)
48
(C)
54
(D)
60
(E)
72
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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

#1 Write an Equation 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 $76$.
$$3N + 4M = 76$$

💡 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.

#1 Write an Equation 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.

$$3N = 76 - 4M \;\Rightarrow\; N = \dfrac{76 - 4M}{3}$$

💡 Isolating one variable in a two-variable equation is the algebraic move taught in Grade 7.

#12 Use Cases 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.
$\dfrac{76 - 4M}{3} > 2M \;\Rightarrow\; 76 - 4M > 6M \;\Rightarrow\; 76 > 10M \;\Rightarrow\; M < 7.6$. With $M > 4$ and $M$ an integer: $M \in \{5, 6, 7\}$.

💡 Solving a multi-step inequality and reading off the integer values inside the range is exactly the Grade 7 inequalities standard.

#12 Use Cases 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 $3$.
$M = 5$: $76 - 20 = 56$, and $56 / 3$ is not an integer. $\;$ $M = 6$: $76 - 24 = 52$, and $52 / 3$ is not an integer. $\;$ $M = 7$: $76 - 28 = 48$, and $48 / 3 = 16$. $\checkmark$

💡 Checking divisibility by $3$ for three small numbers is the Grade 6 factors-and-multiples skill at work.

#1 Write an Equation 6.EE.B.6 Step 5

Confirm the surviving case satisfies the original $N > 2M$, then plug $N = 16$ into the intra-division count $3N$.

$N = 16$, $M = 7$: $N > 2M \Leftrightarrow 16 > 14 \;\checkmark$. Intra-division games $= 3N = 3 \times 16 = 48 \;\Rightarrow\; \textbf{(B)}$.

💡 Substituting the solved value back into the expression $3N$ is the standard "evaluate at a specific value" move from Grade 6.

[1] #1 6.EE.B.6 Count opponents and write the total-games equation. Inside a team's own division
[2] #1 7.EE.B.4 Solve the equation for $N$ in terms of $M$ so the inequality $N > 2M$ becomes a
[3] #12 7.EE.B.4 Substitute that expression into $N > 2M$ and simplify to an upper bound on $M$.
[4] #12 6.NS.B.4 Test each candidate $M$ to see which one makes $N = (76 - 4M)/3$ a positive inte
[5] #1 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.6 Use 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.4 Find 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.4 Use 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.