Difference between revisions of "2003 AMC 10A Problems/Problem 23"
(Switched "upwards" and "downwards" in solution 2. There are always more upwards-facing triangles than downwards-facing triangles in each row.) |
|||
Line 1: | Line 1: | ||
== Problem == | == Problem == | ||
− | A large [[equilateral triangle]] is constructed by using toothpicks to create rows of small equilateral triangles. For example, in the figure we have <math>3</math> rows of small congruent equilateral triangles, with <math>5</math> 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 <math>2003</math> small equilateral triangles? | + | A large [[equilateral triangle]] is constructed by using toothpicks to create rows of small equilateral triangles. For example, in the figure, we have <math>3</math> rows of small congruent equilateral triangles, with <math>5</math> 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 <math>2003</math> small equilateral triangles? |
[[Image:2003amc10a23.gif]] | [[Image:2003amc10a23.gif]] | ||
Line 19: | Line 19: | ||
===Solution 2=== | ===Solution 2=== | ||
− | We see that the bottom row of <math>2003</math> small triangles | + | We see that the bottom row of <math>2003</math> small triangles are formed from <math>1002</math> upward-facing triangles and <math>1001</math> downward-facing triangles. Since each upward-facing triangle uses three distinct toothpicks, and since the total number of upward-facing triangles is <math>1002+1001+...+1=\frac{1003\cdot1002}{2}=502503</math>, we have that the total number of toothpicks is <math>3\cdot 502503=\boxed{\mathrm{(C)}\ 1,507,509}</math> |
===Solution 3=== | ===Solution 3=== | ||
− | Experimenting a bit we find that the number of toothpicks needs | + | Experimenting a bit we find that the number of toothpicks needs a triangle with <math>1</math>, <math>2</math> and <math>3</math> rows is <math>1\cdot{3}</math>, <math>3\cdot{3}</math> and <math>6\cdot{3}</math> respectively. Since <math>1</math>, <math>3</math> and <math>6</math> are triangular numbers we know that depending on how many rows there are in the triangle, the number we multiply by <math>3</math> to find total no.toothpicks is the corresponding triangular number. Since the triangle in question has <math>2n-1=2003\implies{n=1002}</math> rows, we can use <math>\frac{n(n+1)}{2}</math> to find the triangular number for that row and multiply by <math>3</math>, hence finding the total no.toothpicks; this is just <math>\frac{3\cdot{1002}\cdot{1003}}{2}=3\cdot{501}\cdot{1003}=\boxed{\mathrm{(C)}\ 1,507,509}</math>. |
+ | ===Note=== | ||
+ | In the final step of the problem, we know that the units digit of the answer must <math>9</math>, so the only answer choice applicable must be <math>\boxed{\mathrm{(C)}\ 1,507,509}</math>. This saves you the time it takes to compute <math>3\cdot{501}\cdot{1003}</math>. | ||
== See Also == | == See Also == |
Revision as of 23:54, 29 December 2017
Problem
A large equilateral triangle is constructed by using toothpicks to create rows of small equilateral triangles. For example, in the figure, we have rows of small congruent equilateral triangles, with 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 small equilateral triangles?
Solution
Solution 1
There are small equilateral triangles.
Each small equilateral triangle needs toothpicks to make it.
But, each toothpick that isn't one of the toothpicks on the outside of the large equilateral triangle is a side for small equilateral triangles.
So, the number of toothpicks on the inside of the large equilateral triangle is
Therefore the total number of toothpicks is
Solution 2
We see that the bottom row of small triangles are formed from upward-facing triangles and downward-facing triangles. Since each upward-facing triangle uses three distinct toothpicks, and since the total number of upward-facing triangles is , we have that the total number of toothpicks is
Solution 3
Experimenting a bit we find that the number of toothpicks needs a triangle with , and rows is , and respectively. Since , and are triangular numbers we know that depending on how many rows there are in the triangle, the number we multiply by to find total no.toothpicks is the corresponding triangular number. Since the triangle in question has rows, we can use to find the triangular number for that row and multiply by , hence finding the total no.toothpicks; this is just .
Note
In the final step of the problem, we know that the units digit of the answer must , so the only answer choice applicable must be . This saves you the time it takes to compute .
See Also
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.