Difference between revisions of "2003 AMC 10A Problems/Problem 23"
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<math> \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 </math> | <math> \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 </math> | ||
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− | + | == Solution 1== | |
There are <math>1+3+5+...+2003=1002^{2}=1004004</math> small equilateral triangles. | There are <math>1+3+5+...+2003=1002^{2}=1004004</math> small equilateral triangles. | ||
Revision as of 13:49, 6 December 2020
Contents
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 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 ~dolphin7
Solution 2
The first row of triangles has upward-facing triangle, the second row has upward-facing triangles, the third row has upward-facing triangles, and so on having upward-facing triangles in the row. The last row with small triangles has upward-facing triangles. By Gauss's formula, the number of the upward-facing triangles in the entire triangle are now , meaning that the number of toothpicks are , or .
~mathpro12345
Note
You don't have to calculate the value of , and you can use units digits to find the answer easily. The units digit of is , and has a unit digit of after being divided by . Then this is multiplied by , now the final number ending with a . This leaves only one answer choice possible, which is
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 |
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