AMC 8 · 2000 · #5
Easy mode Grade 3Problem
Imagine a school called Lincoln High School. Every principal there stays for exactly 3 years, then a new principal takes over.
Now picture an 8-year stretch at this school. During those 8 years, several principals may come and go. The first one might be near the end of their term, and the last one might be just starting.
What is the largest number of different principals the school could have during those 8 years?
Pick an answer.
AMC 8 2000 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: Every principal of Lincoln High School serves exactly one $3$-year term. Over an $8$-year stretch, what is the largest possible number of different principals?
Givens: Each principal serves exactly $3$ consecutive years; We watch the school for an $8$-year window; Answer choices: (A) $2$, (B) $3$, (C) $4$, (D) $5$, (E) $8$
Unknowns: The maximum number of different principals whose terms touch the $8$-year window
Understand
Restated: Every principal of Lincoln High School serves exactly one $3$-year term. Over an $8$-year stretch, what is the largest possible number of different principals?
Givens: Each principal serves exactly $3$ consecutive years; We watch the school for an $8$-year window; Answer choices: (A) $2$, (B) $3$, (C) $4$, (D) $5$, (E) $8$
Plan
Primary tool: #1 Draw a Diagram
Secondary: #6 Guess and Check
The $8$-year window is small enough to draw as a strip of $8$ boxes, and a $3$-year term is just $3$ shaded boxes. Tool #1 (Draw a Diagram) turns the word problem into a visible timeline. To find the maximum we use Tool #6 (Guess and Check): try $3$ principals, then $4$, then $5$, and see which arrangement still fits. The picture also reveals the trick — terms can poke past the window at each end, so the first and last principal only need to cover $1$ year inside.
Execute — Answer: C
3.MD.A.1 Step 1 - Draw the $8$-year window as a row of $8$ boxes labeled Year $1$ through Year $8$.
- Each principal's $3$-year term will be a group of $3$ consecutive boxes — but the group can start before Year $1$ or end after Year $8$, as long as at least one box of the group sits inside the window.
💡 A picture of the $8$ years lets us slide $3$-year blocks across it and see what fits — exactly the Grade 3 "measure and represent time intervals" idea.
3.OA.D.8 Step 2 - Guess and check: try fitting $4$ principals.
- Let Principal A's term end on Year $1$ (years $-1, 0, 1$).
- Principal B serves Years $2{-}4$.
- Principal C serves Years $5{-}7$.
- Principal D's term starts on Year $8$ (years $8, 9, 10$).
- Inside the window the years used are $1 + 3 + 3 + 1 = 8$ — a perfect fit, so $4$ principals is possible.
💡 $1 + 3 + 3 + 1 = 8$. The first and last principals donate just $1$ year each to the window, leaving room for two full $3$-year terms in the middle.
3.OA.D.8 Step 3 - Now check whether $5$ principals could fit.
- The first and last principals each cover at least $1$ year of the window.
- The other $3$ principals are stuck entirely inside, so they each cover $3$ full years.
- Minimum total years needed: $1 + 3 + 3 + 3 + 1 = 11$.
- But the window is only $8$ years long, so $5$ principals will not fit.
💡 Squeezing harder doesn't help — once you have a middle principal, their whole $3$ years live in the window. Three middle terms alone already cost $9 > 8$ years.
3.OA.D.8 Step 4 - We constructed a working schedule for $4$ principals and proved $5$ cannot fit.
- So the maximum is $4$, matching choice $\textbf{(C)}$.
💡 The diagram shows it: $4$ principals exactly tile the $8$ years when the two end terms each chip in just $1$ year.
3.MD.A.1 Draw the $8$-year window as a row of $8$ boxes labeled Year $1$ through Year $8$ 3.OA.D.8 Guess and check: try fitting $4$ principals. Let Principal A's term end on Year 3.OA.D.8 Now check whether $5$ principals could fit. The first and last principals each c 3.OA.D.8 We constructed a working schedule for $4$ principals and proved $5$ cannot fit. Review
Reasonableness: Check the construction directly. A on Year $1$, B on Years $2{-}4$, C on Years $5{-}7$, D on Year $8$ — every year is covered by exactly one principal, no overlaps, every term is still a full $3$ years (parts of A's and D's terms simply lie outside the window). Four different people. And the bound $1 + 3(n-2) + 1 \le 8$ gives $n \le 4$, so we cannot do better. Answer (C) $= 4$ checks out from both sides.
Alternative: Tool #13 (Convert to Algebra): let $n$ be the number of principals. The first and last serve at least $1$ year of the window, the $n-2$ middle principals serve their full $3$ years. So $1 + 3(n-2) + 1 \le 8 \Rightarrow 3n \le 12 \Rightarrow n \le 4$. The diagram-and-check approach gets the same answer with no algebra, which is why we picked it first.
CCSS standards used (min grade 3)
3.MD.A.1Tell and write time and measure time intervals; solve word problems involving addition and subtraction of time intervals (Drawing the $8$-year window as a strip of years and representing each $3$-year term as a contiguous block of time on that strip.)3.OA.D.8Solve two-step word problems using the four operations, including assessing the reasonableness of answers (Checking the candidate arrangement with $1 + 3 + 3 + 1 = 8$ and ruling out $5$ principals with $1 + 3 + 3 + 3 + 1 = 11 > 8$.)
⭐ Draw the $8$ years as a strip, slide in $3$-year blocks, and let the first and last terms poke past the edges — that little trick fits $4$ principals into an $8$-year window.
⭐ Draw the $8$ years as a strip, slide in $3$-year blocks, and let the first and last terms poke past the edges — that little trick fits $4$ principals into an $8$-year window.
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