AMC 8 · 2000 · #5
Grade 3 arithmeticProblem
Each principal of Lincoln High School serves exactly one 3-year term. What is the maximum number of principals this school could have during an 8-year period?
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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