AMC 8 · 2000 · #12

Grade 4 arithmeticgeometry-2d

Archetype Grid Tiling / Packing with Divisibility └ Sub-archetype Rectangular Tiling — Minimize Filler Count

spatial-visualizationmulti-digit-arithmeticpattern-recognitionarea-rectangles identify-subproblemsoptimization-countingpattern-recognition ↑ Prerequisites: multi-digit-arithmeticarea-rectangles

📏 Short solution 💡 3 insights 📊 Diagram

📘 View easy version →

Problem

A wall is 100 feet long and 7 feet high. It is built from blocks that are 1 foot tall and either 2 feet or 1 foot long. No block may be cut, the wall must have flat (100-foot) ends in every row, and vertical joints between adjacent rows must be staggered (no joint sits directly above the joint below it). Find the smallest possible total number of blocks.

Pick an answer.

(A)

344

(B)

347

(C)

350

(D)

353

(E)

356

AMC 8 2000 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve

Strategy Break Into Subproblems

The picture in the problem already does most of the planning. Tool #1 (Draw a Diagram) lets us read off the two row "styles" we are allowed to alternate: a bottom-style row of all 2-ft blocks, and a top-style row that starts and ends with a 1-ft block. Tool #7 (Break Into Subproblems) then splits the count into three short pieces: (a) count blocks in one bottom-style row, (b) count blocks in one top-style row, (c) figure out how many rows of each type appear in 7 alternating rows and add. Fewer blocks per row means longer blocks, so each row separately wants as many 2-ft blocks as it can use — and the stagger rule forces the top-style row to add exactly two 1-ft blocks. No algebra is needed; each subproblem is one quick multiplication or addition.

1STEP 1

Blocks are 1 ft tall, so a 7-ft wall has 7 rows. Filling a row with 2-ft blocks is cheapest: 100 ÷ 2 gives a 50-block bottom-style row.

bottom-style row : 50 blocks of length 2 → 50 × 2 = 100 ft

2STEP 2

The row above can't be all 2-ft (joints align). Cap both ends with 1-ft blocks, fill 98 ft with 49 two-ft blocks: a 51-block top row.

top-style row : 1 + 49 × 2 + 1 = 100 ft, blocks = 1 + 49 + 1 = 51

3STEP 3

A row's block count is 100 minus its 2-ft blocks, so a staggered row's forced end caps cost at least 51. Alternating 50 and 51 is optimal.

minimum per row = 50 & if row below is top-style ; 51 & if row below is bottom-style

4STEP 4

Stack 7 rows alternating: rows 1, 3, 5, 7 bottom-style (4 × 50), rows 2, 4, 6 top-style (3 × 51). Total 200 + 153 = 353.

4 × 50 + 3 × 51 = 200 + 153 = 353 → (D)

Answer

353

An all-2-ft fantasy wall (ignoring the stagger rule) would use 7 × 50 = 350 blocks, which is choice (C). The stagger rule forces three of those seven rows to swap in two 1-ft end caps, adding 3 × 1 = 3 extra blocks (51 instead of 50 on each of those rows). So 350 + 3 = 353, matching (D). Choice (E) 356 would be "50 everywhere plus 6 extras", which would happen only if every staggered row needed 4 end caps instead of 2. Choices (A) 344 and (B) 347 are below the 350 floor and are impossible.

💡Key takeaway

Only two row styles can appear in a staggered wall: the cheap 50-block row (all 2-ft blocks) and the slightly pricier 51-block row (2-ft blocks with 1-ft end caps). Alternating them for 7 rows gives 4 × 50 + 3 × 51 = 353 blocks, choice (D).

More like this