AMC 8 · 2000 · #5
Grade 3 arithmeticPick an answer.
AMC 8 2000 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.
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.
Draw the 8-year window as a row of 8 boxes; each 3-year term is a block of 3 that may poke past either end.
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.MD.A.1Draw A DiagramFit 4: A ends Year 1, B Years 2-4, C Years 5-7, D starts Year 8. Inside, 1 + 3 + 3 + 1 = 8 fits exactly, so 4 works.
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.8Guess And CheckTry 5: two end terms take 1 year, three middle terms take 3 — 1 + 3 + 3 + 3 + 1 = 11 years, over the 8, so 5 cannot 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.8Guess And CheckWe built 4 and ruled out 5, so the maximum is 4 — choice (C).
The diagram shows it: 4 principals exactly tile the 8 years when the two end terms each chip in just 1 year.
3.OA.D.8Draw A DiagramDraw 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.