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Difference between revisions of "2017 AMC 10A Problems/Problem 23"

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==Solution 1==
 
==Solution 1==
 
There are a total of <math>\binom{25}{3}=2300</math> sets of three points. However, some of them form degenerate triangles (i.e., they have area of 0) if the three points are collinear. There are a total of 12 lines that go through 5 points (5 vertical, 5 horizontal, 2 diagonal), which contributes <math>\binom{5}{3} \cdot 12 = 120</math> degenerate triangles, 4 lines that go through exactly 4 points, which contributes <math>\binom{4}{3} \cdot 4 = 16</math> degenerate triangles, and 16 lines that go through exactly three points, which contributes <math>\binom{3}{3} \cdot 16 = 16</math> degenerate triangles. Subtracting these degenerate triangles, we get an answer of <math>2300-120-16-16=2300-152=\boxed{\textbf{(B) }2148}</math>.
 
There are a total of <math>\binom{25}{3}=2300</math> sets of three points. However, some of them form degenerate triangles (i.e., they have area of 0) if the three points are collinear. There are a total of 12 lines that go through 5 points (5 vertical, 5 horizontal, 2 diagonal), which contributes <math>\binom{5}{3} \cdot 12 = 120</math> degenerate triangles, 4 lines that go through exactly 4 points, which contributes <math>\binom{4}{3} \cdot 4 = 16</math> degenerate triangles, and 16 lines that go through exactly three points, which contributes <math>\binom{3}{3} \cdot 16 = 16</math> degenerate triangles. Subtracting these degenerate triangles, we get an answer of <math>2300-120-16-16=2300-152=\boxed{\textbf{(B) }2148}</math>.
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==Solution 2==
 
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We can find out that the least number of digits the number <math>N</math> is <math>142</math>, with <math>141</math> <math>9</math>'s and one <math>5</math>.
 
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By randomly mixing the digits up, we are likely to get: <math>9999</math>...<math>9995999</math>...<math>9999</math>.
 
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By adding 1 to this number, we get: <math>9999</math>...<math>9996000</math>...<math>0000</math>.
 
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We can subtract 6 from every available choice, and see if the number is divisible by 9 afterwards.
 
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After subtracting 6 from every number, we can conclude that <math>1233</math> (originally <math>1239</math>) is the only number divisible by 9.
 
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So our answer is <math>1239</math>
 
  
 
==See Also==
 
==See Also==
 
{{AMC10 box|year=2017|ab=A|num-b=22|num-a=24}}
 
{{AMC10 box|year=2017|ab=A|num-b=22|num-a=24}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 18:05, 9 February 2017

Problem

How many triangles with positive area have all their vertices at points $(i,j)$ in the coordinate plane, where $i$ and $j$ are integers between $1$ and $5$, inclusive?

$\textbf{(A)}\ 2128 \qquad\textbf{(B)}\ 2148 \qquad\textbf{(C)}\ 2160 \qquad\textbf{(D)}\ 2200 \qquad\textbf{(E)}\ 2300$

Solution 1

There are a total of $\binom{25}{3}=2300$ sets of three points. However, some of them form degenerate triangles (i.e., they have area of 0) if the three points are collinear. There are a total of 12 lines that go through 5 points (5 vertical, 5 horizontal, 2 diagonal), which contributes $\binom{5}{3} \cdot 12 = 120$ degenerate triangles, 4 lines that go through exactly 4 points, which contributes $\binom{4}{3} \cdot 4 = 16$ degenerate triangles, and 16 lines that go through exactly three points, which contributes $\binom{3}{3} \cdot 16 = 16$ degenerate triangles. Subtracting these degenerate triangles, we get an answer of $2300-120-16-16=2300-152=\boxed{\textbf{(B) }2148}$.

See Also

2017 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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