AMC 10 · 2003 · #23
Grade 6 countingA large equilateral triangle is constructed by using toothpicks to create rows of small equilateral triangles. For example, in the figure, we have 3 rows of small congruent equilateral triangles, with 5 small triangles in the base row. How many toothpicks would be needed to construct a large equilateral triangle if the base row of the triangle consists of 2003 small equilateral triangles?
Pick an answer.
AMC 10 2003 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 big equilateral triangle is tiled with rows of small equilateral triangles, built from toothpicks. When the bottom row holds $b$ small triangles, we want the total number of toothpicks used. The sample figure has $3$ rows with $5$ small triangles in the bottom row. Find the toothpick count when the bottom row has $2003$ small triangles.
Givens: The big triangle is made of small congruent equilateral triangles laid in rows; Sample case: $3$ rows correspond to $5$ small triangles in the bottom row; Target case: the bottom row has $2003$ small triangles; Answer choices: (A) 1,004,004 (B) 1,005,006 (C) 1,507,509 (D) 3,015,018 (E) 6,021,018
Unknowns: The total number of toothpicks needed to build the big triangle
Understand
Restated: A big equilateral triangle is tiled with rows of small equilateral triangles, built from toothpicks. When the bottom row holds $b$ small triangles, we want the total number of toothpicks used. The sample figure has $3$ rows with $5$ small triangles in the bottom row. Find the toothpick count when the bottom row has $2003$ small triangles.
Givens: The big triangle is made of small congruent equilateral triangles laid in rows; Sample case: $3$ rows correspond to $5$ small triangles in the bottom row; Target case: the bottom row has $2003$ small triangles; Answer choices: (A) 1,004,004 (B) 1,005,006 (C) 1,507,509 (D) 3,015,018 (E) 6,021,018
Plan
Primary tool: #16 Change Focus / Count the Complement
Secondary: #5 Look for a Pattern, #9 Solve an Easier Related Problem
Counting toothpicks directly is a trap, because most toothpicks are shared between an up-pointing and a down-pointing small triangle, so naive counting double-counts. Tool #16 (Change Focus) fixes this with one clean observation: look only at the up-pointing triangles. Every toothpick is a side of exactly one up-pointing triangle, and no toothpick is shared between two up-pointing triangles. So the total number of toothpicks is exactly $3$ times the number of up-pointing triangles — no overcounting to correct. Tool #5 (Look for a Pattern) then counts those up-pointing triangles row by row ($1, 2, 3, \dots$ per row), and Tool #9 (Solve an Easier Related Problem) lets us test the whole idea on the tiny given figure before trusting it on $2003$.
Execute — Answer: C
4.OA.C.5 Step 1 Count the rows from the bottom row
- Look at how many small triangles sit in each row.
- The top row has $1$, the next has $3$, then $5$, and so on — the odd numbers.
- The bottom row is the $r$-th odd number, which is $2r-1$.
- The sample checks out: $3$ rows give a bottom row of $2\cdot 3-1=5$.
- For our triangle the bottom row has $2003$ small triangles, so $2r-1=2003$, giving $r=1002$ rows.
💡 Each row down adds one triangle to each slanted side, so the counts climb through the odd numbers.
4.OA.C.5 Step 2 Look only at up-pointing triangles
- Here is the key move.
- Instead of counting toothpicks, count up-pointing small triangles.
- Every toothpick in the whole figure is one side of exactly one up-pointing triangle, and two different up-pointing triangles never share a toothpick (the shared sides always sit between an up triangle and a down triangle).
- So counting up-pointing triangles counts every toothpick once, with none left out and none doubled.
- That means the total toothpicks equal $3$ times the number of up-pointing triangles.
💡 Down-pointing triangles are built entirely from sides that already belong to up-pointing neighbors, so up-pointing triangles alone own every toothpick.
4.OA.A.3 Step 3 Test the idea on the sample figure
- Before trusting this on $2003$, try the given $3$-row figure.
- Its up-pointing triangles number $1+2+3=6$, so the rule predicts $3\times 6=18$ toothpicks.
- Check it another way: the whole figure has $3^2=9$ small triangles, using $3\times 9=27$ triangle-sides; the $3$ outer edges of the big triangle each hold $3$ toothpicks (that is $9$ single-count boundary toothpicks), and every other toothpick is shared by two triangles.
- Solving $27 = 2\cdot(\text{inside}) + 9$ gives $9$ inside toothpicks, for $9+9=18$ total.
- Both methods agree, so the up-pointing-triangle rule is sound.
💡 A rule you can confirm on a case you can see by hand is a rule you can push to a huge case with confidence.
6.EE.A.2 Step 4 Count all up-pointing triangles
- Row by row from the top, the up-pointing triangles come $1, 2, 3, \dots$ — the $k$-th row has exactly $k$ of them.
- With $1002$ rows, the total up-pointing count is $1+2+3+\cdots+1002$.
- Use the sum-of-first-$n$ rule $1+2+\cdots+n=\dfrac{n(n+1)}{2}$ with $n=1002$: this equals $\dfrac{1002\cdot 1003}{2}=501\cdot 1003=502{,}503$.
💡 Pairing the first term with the last, the second with the second-last, and so on turns a long addition into one multiplication.
5.NBT.B.5 Step 5 Multiply by three for the toothpick total
- Each up-pointing triangle owns $3$ toothpicks, and together they own every toothpick exactly once, so multiply the count by $3$: $3\times 502{,}503 = 1{,}507{,}509$.
- That matches answer choice (C).
💡 Once each toothpick is assigned to exactly one triangle, the grand total is just triangles times sides-per-triangle.
4.OA.C.5 Look at how many small triangles sit in each row. The top row has $1$, the next 4.OA.C.5 Here is the key move. Instead of counting toothpicks, count up-pointing small tr 4.OA.A.3 Before trusting this on $2003$, try the given $3$-row figure. Its up-pointing tr 6.EE.A.2 Row by row from the top, the up-pointing triangles come $1, 2, 3, \dots$ — the $ 5.NBT.B.5 Each up-pointing triangle owns $3$ toothpicks, and together they own every tooth Review
Reasonableness: The answer $1{,}507{,}509$ is choice (C). A sanity check on size: choice (A) $1{,}004{,}004=1002^2$ is exactly the number of small triangles, which is fewer than the toothpicks (every triangle carries $3$ sides, most shared, so toothpicks should be roughly $1.5$ times the triangle count) — and indeed $1{,}507{,}509 \approx 1.5\times 1{,}004{,}004$, which fits. A units-digit check also isolates (C): $1002\cdot 1003$ ends in $6$, halving gives a number ending in $3$, and tripling ends in $9$; only (C) ends in $9$.
Alternative: Count slanted and horizontal toothpicks separately. The bottom row needs $2\cdot 1002+1=2005$ slanted toothpicks, and each row up needs $2$ fewer, giving $2005+2003+\cdots+3$ slanted toothpicks. The horizontal toothpicks, counted as the base of each row, are $1002+1001+\cdots+1$. Adding the two totals gives the same $1{,}507{,}509$, but the up-pointing-triangle method reaches it in one multiplication instead of two long sums.
CCSS standards used (min grade 6)
4.OA.C.5Generate a number or shape pattern following a given rule (Seeing that row sizes follow the odd numbers ($2r-1$) to find $1002$ rows, and that each toothpick belongs to exactly one up-pointing triangle.)4.OA.A.3Solve multi-step word problems using four operations with whole numbers (Verifying the counting rule on the $3$-row sample two different ways ($3\times 6=18$ and boundary-plus-inside $=18$).)6.EE.A.2Write, read, and evaluate expressions in which letters stand for numbers (Summing $1+2+\cdots+1002$ with the formula $\frac{n(n+1)}{2}$ to get $502{,}503$ up-pointing triangles.)5.NBT.B.5Fluently multiply multi-digit whole numbers (Computing $1002\cdot 1003$ and the final $3\times 502{,}503 = 1{,}507{,}509$.)
⭐ When pieces share their edges, count the up-pointing pieces only — each edge belongs to exactly one of them, so the total edges are just three times that count.
⭐ When pieces share their edges, count the up-pointing pieces only — each edge belongs to exactly one of them, so the total edges are just three times that count.
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