2003 AMC 10A Problems/Problem 23

Problem

A 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?

$\mathrm{(A) \ } 1,004,004 \qquad \mathrm{(B) \ } 1,005,006 \qquad \mathrm{(C) \ } 1,507,509 \qquad \mathrm{(D) \ } 3,015,018 \qquad \mathrm{(E) \ } 6,021,018$

Solution

Solution 1

There are $1+3+5+...+2003=1002^{2}=1004004$ small equilateral triangles.

Each small equilateral triangle needs $3$ toothpicks to make it.

But, each toothpick that isn't one of the $1002\cdot3=3006$ toothpicks on the outside of the large equilateral triangle is a side for $2$ small equilateral triangles.

So, the number of toothpicks on the inside of the large equilateral triangle is $\frac{10040004\cdot3-3006}{2}=1504503$

Therefore the total number of toothpicks is $1504503+3006=\boxed{\mathrm{(C)}\ 1,507,509}$

Solution 2

We see that the bottom row of $2003$ small triangles are formed from $1002$ upward-facing triangles and $1001$ downward-facing triangles. Since each upward-facing triangle uses three distinct toothpicks, and since the total number of upward-facing triangles is $1002+1001+...+1=\frac{1003\cdot1002}{2}=502503$ , we have that the total number of toothpicks is $3\cdot 502503=\boxed{\mathrm{(C)}\ 1,507,509}$

Solution 3

Experimenting a bit we find that the number of toothpicks needs a triangle with $1$ , $2$ and $3$ rows is $1\cdot{3}$ , $3\cdot{3}$ and $6\cdot{3}$ respectively. Since $1$ , $3$ and $6$ are triangular numbers we know that depending on how many rows there are in the triangle, the number we multiply by $3$ to find total no.toothpicks is the corresponding triangular number. Since the triangle in question has $2n-1=2003\implies{n=1002}$ rows, we can use $\frac{n(n+1)}{2}$ to find the triangular number for that row and multiply by $3$ , hence finding the total no.toothpicks; this is just $\frac{3\cdot{1002}\cdot{1003}}{2}=3\cdot{501}\cdot{1003}=\boxed{\mathrm{(C)}\ 1,507,509}$ .

Note

In the final step of the problem, we know that the units digit of the answer must $9$ , so the only answer choice applicable must be $\boxed{\mathrm{(C)}\ 1,507,509}$ . This saves you the time it takes to compute $3\cdot{501}\cdot{1003}$ .

2003 AMC 10A (Problems • Answer Key • Resources)
Preceded by Problem 22	Followed by Problem 24
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
All AMC 10 Problems and Solutions

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