Competition · AMC preparation · step 4 of 4

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

Count blocks in a bottom row

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

Count blocks in a top row

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

Check the pattern is cheapest

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

Each 1-ft block we swap in costs one extra block, so we want to swap in as few as possible while still breaking the joint alignment.

A row that must stagger against an all-2-foot row uses at least 51 blocks, one more than the 50 blocks of an all-2-foot row, so the cheapest legal wall alternates 50-block rows with 51-block rows.

▸ Why?

An all-2-foot row is the cheapest a single row can be, at 50 blocks, because a row holds fewer blocks the more of its 100 feet it covers with 2-foot blocks instead of 1-foot ones.

▸ Why?

The blocks in one row fill its 100 feet with no gaps and no overlaps, so their lengths add to exactly 100 feet and any change must keep that total.

Whole is the sum of its partsBreak a thing into pieces with no gaps or overlaps and the pieces add back to the whole.

▸ Why?

Covering the full 100 feet with 2-foot blocks takes 50 of them, since 50 groups of 2 feet make 100 feet, and swapping any 2-foot block for two 1-foot blocks keeps the length but adds one block.

Multiplication as equal groupsa times b means a equal groups of b — it is just repeated adding of the same amount.

▸ Why?

The row directly above an all-2-foot row cannot be all 2-foot blocks and needs at least two 1-foot blocks, so it holds at least 51 blocks.

▸ Why?

An all-2-foot row has a joint at every even foot from 2 to 98, so a row above made only of 2-foot blocks would repeat those same even joints and break the staggering rule; it must place at least one 1-foot block to shift its joints off the even feet.

▸ Why?

Laying only 2-foot blocks from the flat end at foot 0, the joint after any number of blocks sits at twice that many feet, and twice any whole number is even, so every joint lands on an even foot.

Even and odd (parity)A whole number is even when it splits into two equal whole groups (a factor of 2), else odd; odd×odd stays odd and even×anything is even.

▸ Why?

One 1-foot block alone cannot work, because it leaves 99 feet for the 2-foot blocks and any run of 2-foot blocks covers an even number of feet while 99 is odd; two 1-foot blocks leave an even 98 feet, which splits into exactly 49 of them.

Even and odd (parity)A whole number is even when it splits into two equal whole groups (a factor of 2), else odd; odd×odd stays odd and even×anything is even.

4.OA.A.3Identify Subproblems

4STEP 4

Stack the seven rows

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).

Count blocks in a bottom row
Count blocks in a top row
Check the pattern is cheapest
Stack the seven rows

A parent dashboard for the family lives at sensimlab.com.

Count blocks in a bottom row

Count blocks in a top row

Check the pattern is cheapest

Stack the seven rows

More like this