AMC 8 · 2000 · #12
Grade 4 arithmeticgeometry-2dProblem
A block wall 100 feet long and 7 feet high will be constructed using blocks that are 1 foot high and either 2 feet long or 1 foot long (no blocks may be cut). The vertical joins in the blocks must be staggered as shown, and the wall must be even on the ends. What is the smallest number of blocks needed to build this wall?
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: 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.
Givens: Wall is $100$ ft long and $7$ ft high; Each block is $1$ ft tall; lengths are $1$ ft or $2$ ft; No blocks may be cut; Vertical joints between consecutive rows must be staggered, as in the picture; Each row must be exactly $100$ ft long, with flat ends; Answer choices: (A) $344$, (B) $347$, (C) $350$, (D) $353$, (E) $356$
Unknowns: The minimum total number of blocks needed
Understand
Restated: 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.
Givens: Wall is $100$ ft long and $7$ ft high; Each block is $1$ ft tall; lengths are $1$ ft or $2$ ft; No blocks may be cut; Vertical joints between consecutive rows must be staggered, as in the picture; Each row must be exactly $100$ ft long, with flat ends; Answer choices: (A) $344$, (B) $347$, (C) $350$, (D) $353$, (E) $356$
Plan
Primary tool: #7 Break Into Subproblems
Secondary: #1 Draw a Diagram
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.
Execute — Answer: D
4.OA.A.3 Step 1 - Set up the rows.
- Each block is $1$ ft tall and the wall is $7$ ft tall, so the wall has $7$ stacked rows.
- To use the fewest blocks per row, we want as many $2$-ft blocks as possible.
- The bottom row in the figure does exactly that: it is $50$ blocks of length $2$, all lined up, for a total of $100$ ft.
- Call this the "bottom-style" row.
💡 Dividing the $100$-ft length by the longest block ($2$ ft) gives $50$ blocks — the fewest a single row can use.
4.OA.A.3 Step 2 - Now look at the row above it.
- A bottom-style row's joints land at every even foot ($2, 4, 6, \dots, 98$).
- The stagger rule says the row above must not place a joint at any of those positions, so it cannot be another all-$2$-ft row.
- The simplest fix, shown in the picture, is to start and end the upper row with a $1$-ft block and fill the middle with $2$-ft blocks.
- The middle length is $100 - 1 - 1 = 98$ ft, which uses $98 \div 2 = 49$ two-foot blocks.
- Call this the "top-style" row.
💡 Adding two $1$-ft "end caps" shifts every middle joint by $1$ ft, so the upper joints fall at the odd feet $3, 5, 7, \dots, 97$ — never above a bottom-row joint.
4.OA.A.3 Step 3 - Check that alternating these two styles really is the cheapest staggered pattern.
- Any row must total $100$ ft from $1$s and $2$s, so its block count equals $100 - (\text{number of }2\text{-ft blocks})$.
- Fewer $1$-ft blocks means fewer total blocks.
- A bottom-style row uses zero $1$-ft blocks ($50$ total).
- A row staggered against it has joints forbidden at every even foot, which forces at least two $1$-ft blocks (one on each end — the wall ends must stay flat at $0$ and $100$, and the next inside joint must shift by an odd amount), giving at least $51$ total.
- So alternating $50$-block and $51$-block rows is optimal.
💡 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.
4.OA.A.3 Step 4 - Stack $7$ rows alternating bottom-style, top-style, bottom-style, $\dots$ Rows $1, 3, 5, 7$ are bottom-style ($4$ rows of $50$ blocks).
- Rows $2, 4, 6$ are top-style ($3$ rows of $51$ blocks).
- Add the totals.
💡 Four cheap rows plus three slightly more expensive rows — the multiplication-then-addition wraps up the three subproblems.
4.OA.A.3 Set up the rows. Each block is $1$ ft tall and the wall is $7$ ft tall, so the w 4.OA.A.3 Now look at the row above it. A bottom-style row's joints land at every even foo 4.OA.A.3 Check that alternating these two styles really is the cheapest staggered pattern 4.OA.A.3 Stack $7$ rows alternating bottom-style, top-style, bottom-style, $\dots$ Rows $ Review
Reasonableness: An all-$2$-ft fantasy wall (ignoring the stagger rule) would use $7 \times 50 = 350$ blocks, which is choice (C). The stagger rule forces three of those seven rows to swap in two $1$-ft end caps, adding $3 \times 1 = 3$ 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.
Alternative: Tool #9 (Solve an Easier Problem): replace the $100$-ft wall with a $6$-ft wall, still $7$ rows tall. Bottom-style rows take $3$ blocks ($3 \times 2$); top-style rows take $1 + 2 \times 2 + 1 = 4$ blocks. Four bottom-style rows and three top-style rows give $4 \times 3 + 3 \times 4 = 24$ blocks — and the figure in the problem (which shows a $6 \times 2$ slice with $3$ blocks below and $4$ above) matches exactly. Scaling the row counts from $3 \to 50$ and $4 \to 51$ recovers $4 \times 50 + 3 \times 51 = 353$.
CCSS standards used (min grade 4)
4.OA.A.3Solve multistep word problems with the four operations (Counting blocks per row by division ($100 \div 2 = 50$ and $98 \div 2 = 49$), then combining row counts with multiplication and addition to get $4 \times 50 + 3 \times 51 = 353$.)3.MD.C.7Relate area to the operations of multiplication and addition, including decomposing rectangles into non-overlapping parts (Treating each $100$-ft row as a length tiled by $1$-ft and $2$-ft segments that add to $100$, the same decomposition idea used to break a rectangle into non-overlapping smaller pieces.)
⭐ 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 $4 \times 50 + 3 \times 51 = 353$ blocks, choice (D).
⭐ 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 $4 \times 50 + 3 \times 51 = 353$ blocks, choice (D).
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