Seventh-Grade “100 Lockers” Problem — Comprehensive Study Notes

Classroom Context & Curriculum Integration

• Seventh-grade scope up to this point
  • Sequences, order-of-operations review
  • “Algebra block”: basic symbolic manipulation, then a short rest period before light review work
  • Extensive practice with signed numbers (adding, subtracting, multiplying, dividing integers)
  • Second half of year: Pythagorean Theorem (where algebra re-appears)
• Teacher comment: the locker task fits any time of year because it requires no explicit algebra

Problem Narrative ("100 Lockers" / “Locker-Toggling”)

• Physical setting: hallway with lockers numbered $1\text{–}100$
• Initial condition: all lockers start open
• Sequence of student actions (each “toggles” certain locker doors)
  • Chris: closes every even-numbered locker (multiples of $2$)
  • Mary: toggles lockers that are multiples of $3$
  • Peter: toggles lockers that are multiples of $4$
  • Process continues with successive students $n=5,6,7,\dots,100$
  • Final student only touches locker $100$
• Questions posed to students
  1. “Which lockers are open at the end?” (for $100$ lockers)
  2. “If the hallway continued to $1000$ lockers, which numbers stay open? Explain in one sentence.”

Suggested Problem-Solving Strategy

• Teacher’s #1 heuristic reminder: “Make it simpler.”
  • Try a reduced model (e.g.
  • $30$ lockers instead of $100$) to look for patterns
• Group-work time: ≈ 15 minutes

Mathematical Analysis

• Key idea: A locker’s door changes state once for every divisor of its number
  • Example: Locker $12$ has divisors $1,2,3,4,6,12$ ⇒ toggled $6$ times
• Parity principle
  • Even number of divisors ⇒ ends closed (starts open)
  • Odd number of divisors ⇒ ends open
• Fact: Only perfect squares have an odd divisor count
  • Reason: divisors pair symmetrically $a \times b = n$ except when $a=b=\sqrt n$, leaving one unpaired divisor
  • Divisor-count formula (prime factorization $n=p<em>1^{e</em>1}\dots p<em>k^{e</em>k}$)
    $d(n)=\prod<em>{i=1}^{k}(e</em>i+1)$
    • $d(n)$ is odd ⇔ each $e_i$ is even ⇔ $n$ is a perfect square

Solution for $100$ Lockers

• Perfect squares ≤ $100$: $1^2,2^2,3^2,\dots,10^2$
  • Numerical list: $1,4,9,16,25,36,49,64,81,100$
• These $10$ lockers remain open; all others end closed

Extension to $1000$ Lockers

• Highest square ≤ $1000$: $31^2=961$ (since 32^2=1024>1000)
• Open lockers: perfect squares $1^2 \text{–} 31^2$ ⇒ $31$ lockers stay open
• Concise sentence: “In a hallway of $1000$ lockers, the lockers numbered by perfect squares up to $31^2$—that is, $1,4,9,\dots,961$—will be open because only perfect squares have an odd number of factors.”

Worked Micro-Example (30 Locker Simplification)

• Perfect squares ≤ $30$: $1,4,9,16,25$
• After full toggling cycle, lockers $1,4,9,16,25$ are open; others closed
• Enables students to spot the square-number pattern quickly

Curriculum & Real-World Connections

• Bridges arithmetic (factors/divisors) and algebraic reasoning (pattern identification)
• Prepares groundwork for later number-theory ideas (divisor parity, square numbers) and geometry (square roots in Pythagorean Theorem)
• Embodies computational thinking: state toggling resembles on/off bits or switch circuits

Pedagogical Notes & Classroom Logistics

• Teacher struggled with chat/recording tech but read problem aloud for clarity
• Encourages collaborative exploration before formal explanation
• Example showcases integrating heuristics (“simplify”) into everyday problem solving

Ethical / Philosophical Angle

• Implicit lesson: value of pattern recognition over brute-force checking
• Encourages persistence and creative restructuring of complex tasks

Quick Reference: Numerical Facts & Formulas

• Divisor count: $d(n)=\prod<em>{i}(e</em>i+1)$
• Perfect-square test: $n$ is a perfect square ⇔ $\sqrt n$ is integer ⇔ all prime exponents even
• Open-locker rule: locker $k$ open ⇔ $k$ perfect square
• Counts: $100$ lockers → $10$ open; $1000$ lockers → $31$ open