Difference between revisions of "2018 AMC 8 Problems/Problem 23"

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How many positive three-digit integers have a remainder of 2 when divided by 6, a remainder of 5 when divided by 9, and a remainder of 7 when divided by 11?
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From a regular octagon, a triangle is formed by connecting three randomly chosen vertices of the octagon. What is the probability that at least one of the sides of the triangle is also a side of the octagon?
  
<math>\textbf{(A) }1\qquad\textbf{(B) }2\qquad\textbf{(C) }3\qquad\textbf{(D) }4\qquad \textbf{(E) }5</math>  
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<asy>
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size(3cm);
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pair A[];
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for (int i=0; i<9; ++i) {
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  A[i] = rotate(22.5+45*i)*(1,0);
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}
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filldraw(A[0]--A[1]--A[2]--A[3]--A[4]--A[5]--A[6]--A[7]--cycle,gray,black);
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for (int i=0; i<8; ++i) { dot(A[i]); }
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</asy>
  
==Solution==
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==Solution 1==
  
 
We will use constructive counting to solve this. There are <math>2</math> cases: Either all <math>3</math> points are adjacent, or exactly <math>2</math> points are adjacent.
 
We will use constructive counting to solve this. There are <math>2</math> cases: Either all <math>3</math> points are adjacent, or exactly <math>2</math> points are adjacent.
If all <math>3</math> points are adjacent, then we obviously have <math>8</math> choices. If we have exactly <math>2</math> adjacent points, then we will have <math>8</math> places to put the adjacent points and also <math>4</math> places to put the remaining point, so we have <math>8*4</math> choices. The total amount of choices is <math>{8 \choose 3} = 8*7</math>.
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If all <math>3</math> points are adjacent, then we have <math>8</math> choices. If we have exactly <math>2</math> adjacent points, then we will have <math>8</math> places to put the adjacent points and also <math>4</math> places to put the remaining point, so we have <math>8\cdot4</math> choices. The total amount of choices is <math>{8 \choose 3} = 8\cdot7</math>.
Thus our answer is <math>\frac{8+8*4}{8*7}= \frac{1+4}{7}=\boxed{\textbf{(D) } \frac 57}</math>
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Thus our answer is <math>\frac{8+8\cdot4}{8\cdot7}= \frac{1+4}{7}=\boxed{\textbf{(D) } \frac 57}</math>
  
==Solution 2 (Complementary)==
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==Solution 2 ==
  
We can decide <math>2</math> adjacent points with <math>8</math> choices. The remaining point will have <math>6</math> choices. However, we have counted the case with <math>3</math> adjacent points twice, so we need to subtract this case once. The case with the <math>3</math> adjacent points has <math>8</math> arrangements, so our answer is <math>\frac{8*6-8}{8*7}=\boxed{\textbf{(D) } \frac 57}</math>
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We can decide <math>2</math> adjacent points with <math>8</math> choices. The remaining point will have <math>6</math> choices. However, we have counted the case with <math>3</math> adjacent points twice, so we need to subtract this case once. The case with the <math>3</math> adjacent points has <math>8</math> arrangements, so our answer is <math>\frac{8\cdot6-8}{{8 \choose 3 }}</math><math>=\frac{8\cdot6-8}{8 \cdot 7 \cdot 6 \div 6}\Longrightarrow\boxed{\textbf{(D) } \frac 57}</math>
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==Solution 3==
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Let 1 point of the triangle be fixed at the top. Then, there are 7C2 = 21 ways to chose the other 2 points. There must be 3 spaces in the points and 3 points themselves. This leaves 2 extra points to be placed anywhere. By stars and bars, there are 3 things (triangle points) and 2 things (extra points) distributed so there are 4C2 = 6ways to arrange the bars and stars. Thus, the probability is (21 - 6)\21 = 5/7.
  
 
==See Also==
 
==See Also==

Revision as of 01:44, 12 November 2019

From a regular octagon, a triangle is formed by connecting three randomly chosen vertices of the octagon. What is the probability that at least one of the sides of the triangle is also a side of the octagon?

[asy] size(3cm); pair A[]; for (int i=0; i<9; ++i) {   A[i] = rotate(22.5+45*i)*(1,0); } filldraw(A[0]--A[1]--A[2]--A[3]--A[4]--A[5]--A[6]--A[7]--cycle,gray,black); for (int i=0; i<8; ++i) { dot(A[i]); } [/asy]

Solution 1

We will use constructive counting to solve this. There are $2$ cases: Either all $3$ points are adjacent, or exactly $2$ points are adjacent. If all $3$ points are adjacent, then we have $8$ choices. If we have exactly $2$ adjacent points, then we will have $8$ places to put the adjacent points and also $4$ places to put the remaining point, so we have $8\cdot4$ choices. The total amount of choices is ${8 \choose 3} = 8\cdot7$. Thus our answer is $\frac{8+8\cdot4}{8\cdot7}= \frac{1+4}{7}=\boxed{\textbf{(D) } \frac 57}$

Solution 2

We can decide $2$ adjacent points with $8$ choices. The remaining point will have $6$ choices. However, we have counted the case with $3$ adjacent points twice, so we need to subtract this case once. The case with the $3$ adjacent points has $8$ arrangements, so our answer is $\frac{8\cdot6-8}{{8 \choose 3 }}$$=\frac{8\cdot6-8}{8 \cdot 7 \cdot 6 \div 6}\Longrightarrow\boxed{\textbf{(D) } \frac 57}$

Solution 3

Let 1 point of the triangle be fixed at the top. Then, there are 7C2 = 21 ways to chose the other 2 points. There must be 3 spaces in the points and 3 points themselves. This leaves 2 extra points to be placed anywhere. By stars and bars, there are 3 things (triangle points) and 2 things (extra points) distributed so there are 4C2 = 6ways to arrange the bars and stars. Thus, the probability is (21 - 6)\21 = 5/7.

See Also

2018 AMC 8 (ProblemsAnswer KeyResources)
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 AJHSME/AMC 8 Problems and Solutions

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