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Difference between revisions of "2014 AMC 10B Problems/Problem 24"

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==Problem==
 
==Problem==
−
The numbers 1, 2, 3, 4, 5 are to be arranged in a circle. An arrangement is <math>bad</math> if it is not true that for every <math>n</math> from <math>1</math> to <math>15</math> one can find a subset of the numbers that appear consecutively on the circle that sum to <math>n</math>. Arrangements that differ only by a rotation or a reflection are considered the same. How many different bad arrangements are there?
+
The numbers <math>1, 2, 3, 4, 5</math> are to be arranged in a circle. An arrangement is <math>\textit{bad}</math> if it is not true that for every <math>n</math> from <math>1</math> to <math>15</math> one can find a subset of the numbers that appear consecutively on the circle that sum to <math>n</math>. Arrangements that differ only by a rotation or a reflection are considered the same. How many different bad arrangements are there?
  
 
<math> \textbf {(A) } 1 \qquad \textbf {(B) } 2 \qquad \textbf {(C) } 3 \qquad \textbf {(D) } 4 \qquad \textbf {(E) } 5 </math>
 
<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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==Solution==
 
==Solution==
  
−
We see that there are <math>5!</math> total ways to arrange the numbers. However, we can always rotate these numbers so that, for example, the number 1 is always at the top of the circle. Thus, there are only <math>4!</math> ways under rotation, which is not difficult to list out. We systematically list out all <math>24</math> cases.  
+
We see that there are <math>5!</math> total ways to arrange the numbers. However, we can always rotate these numbers so that, for example, the number <math>1</math> is always at the top of the circle. Thus, there are only <math>4!</math> ways under rotation. Every case has exactly <math>1</math> reflection, so that gives us only <math>4!/2</math>, or <math>12</math> cases, which is not difficult to list out. We systematically list out all <math>12</math> cases.  
  
−
Now, we must examine if they satisfy the conditions. We can see that by choosing one number at a time, we can always obtain subsets with sums 1, 2, 3, 4, and 5. By choosing the full circle, we can obtain 15. By choosing everything except for 1, 2, 3, 4, and 5, we can obtain subsets with sums of 10, 11, 12, 13, and 14.  
+
Now, we must examine if they satisfy the conditions. We can see that by choosing one number at a time, we can always obtain subsets with sums <math>1, 2, 3, 4,</math> and <math>5</math>. By choosing the full circle, we can obtain <math>15</math>. By choosing everything except for <math>1, 2, 3, 4,</math> and <math>5</math>, we can obtain subsets with sums of <math>10, 11, 12, 13,</math> and <math>14</math>.  
  
−
This means that we now only need to check for 6, 7, 8, and 9. However, once we have found a set summing to 6, we can choose everything else and obtain a set summing to 9, and similarly for 7 and 8. Thus, we only need to check each case for whether or not we can obtain 6 or 7.  
+
This means that we now only need to check for <math>6, 7, 8,</math> and <math>9</math>. However, once we have found a set summing to <math>6</math>, we can choose everything else and obtain a set summing to <math>9</math>, and similarly for <math>7</math> and <math>8</math>. Thus, we only need to check each case for whether or not we can obtain <math>6</math> or <math>7</math>.  
  
−
We find that there are only 4 arrangements that satisfy these conditions. However, each of these is a reflection of another. We divide by 2 for these reflections to obtain a final answer of <math>\boxed{\textbf {(B) }2}</math>.
+
We can make <math>6</math> by having <math>4, 2</math>, or <math>3, 2, 1</math>, or <math>5, 1</math>. We can start with the group of three. To separate <math>3, 2, 1</math> from each other, they must be grouped two together and one separate, like this.
 +
 
 +
<asy>
 +
draw(circle((0, 0), 5));
 +
pair O, A, B, C, D, E;
 +
O=origin;
 +
A=(0, 5);
 +
B=rotate(72)*A;
 +
C=rotate(144)*A;
 +
D=rotate(216)*A;
 +
E=rotate(288)*A;
 +
label("$x$", A, N);
 +
label("$y$", C, SW);
 +
label("$z$", D, SE);
 +
</asy>
 +
 
 +
Now, we note that <math>x</math> is next to both blank spots, so we can't have a number from one of the pairs. So since we can't have <math>1</math>, because it is part of the <math>5, 1</math> pair, and we can't have <math>2</math> there, because it's part of the <math>4, 2</math> pair, we must have <math>3</math> inserted into the <math>x</math> spot. We can insert <math>1</math> and <math>2</math> in <math>y</math> and <math>z</math> interchangeably, since reflections are considered the same.
 +
 
 +
<asy>
 +
draw(circle((0, 0), 5));
 +
pair O, A, B, C, D, E;
 +
O=origin;
 +
A=(0, 5);
 +
B=rotate(72)*A;
 +
C=rotate(144)*A;
 +
D=rotate(216)*A;
 +
E=rotate(288)*A;
 +
label("$3$", A, N);
 +
label("$2$", C, SW);
 +
label("$1$", D, SE);
 +
</asy>
 +
 
 +
We have <math>4</math> and <math>5</math> left to insert. We can't place the <math>4</math> next to the <math>2</math> or the <math>5</math> next to the <math>1</math>, so we must place <math>4</math> next to the <math>1</math> and <math>5</math> next to the <math>2</math>.
 +
 
 +
<asy>
 +
draw(circle((0, 0), 5));
 +
pair O, A, B, C, D, E;
 +
O=origin;
 +
A=(0, 5);
 +
B=rotate(72)*A;
 +
C=rotate(144)*A;
 +
D=rotate(216)*A;
 +
E=rotate(288)*A;
 +
label("$3$", A, N);
 +
label("$5$", B, NW);
 +
label("$2$", C, SW);
 +
label("$1$", D, SE);
 +
label("$4$", E, NE);
 +
</asy>
 +
 
 +
This is the only solution to make <math>6</math> "bad."
 +
 
 +
Next we move on to <math>7</math>, which can be made by <math>3, 4</math>, or <math>5, 2</math>, or <math>4, 2, 1</math>. We do this the same way as before. We start with the three group. Since we can't have <math>4</math> or <math>2</math> in the top slot, we must have one there, and <math>4</math> and <math>2</math> are next to each other on the bottom. When we have <math>3</math> and <math>5</math> left to insert, we place them such that we don't have the two pairs adjacent.
 +
 
 +
<asy>
 +
draw(circle((0, 0), 5));
 +
pair O, A, B, C, D, E;
 +
O=origin;
 +
A=(0, 5);
 +
B=rotate(72)*A;
 +
C=rotate(144)*A;
 +
D=rotate(216)*A;
 +
E=rotate(288)*A;
 +
label("$1$", A, N);
 +
label("$3$", B, NW);
 +
label("$2$", C, SW);
 +
label("$4$", D, SE);
 +
label("$5$", E, NE);
 +
</asy>
 +
 
 +
This is the only solution to make <math>7</math> "bad."
 +
 
 +
We've covered all needed cases, and the two examples we found are distinct, therefore the answer is <math>\boxed{\textbf {(B) }2}</math>.
 +
 
 +
Note by Williamgolly:
 +
Notice how we took care of the case of 3 numbers first. That is because that case is the most severe and we usually take care of the most severe cases first in counting problems.
 +
 
 +
==Video Solution==
 +
https://youtu.be/45TQEV8OjRk
  
 
==See Also==
 
==See Also==

Revision as of 20:43, 5 October 2020

The following problem is from both the 2014 AMC 12B #18 and 2014 AMC 10B #24, so both problems redirect to this page.

Problem

The numbers $1, 2, 3, 4, 5$ are to be arranged in a circle. An arrangement is $\textit{bad}$ if it is not true that for every $n$ from $1$ to $15$ one can find a subset of the numbers that appear consecutively on the circle that sum to $n$. Arrangements that differ only by a rotation or a reflection are considered the same. How many different bad arrangements are there?

$\textbf {(A) } 1 \qquad \textbf {(B) } 2 \qquad \textbf {(C) } 3 \qquad \textbf {(D) } 4 \qquad \textbf {(E) } 5$

Solution

We see that there are $5!$ total ways to arrange the numbers. However, we can always rotate these numbers so that, for example, the number $1$ is always at the top of the circle. Thus, there are only $4!$ ways under rotation. Every case has exactly $1$ reflection, so that gives us only $4!/2$, or $12$ cases, which is not difficult to list out. We systematically list out all $12$ cases.

Now, we must examine if they satisfy the conditions. We can see that by choosing one number at a time, we can always obtain subsets with sums $1, 2, 3, 4,$ and $5$. By choosing the full circle, we can obtain $15$. By choosing everything except for $1, 2, 3, 4,$ and $5$, we can obtain subsets with sums of $10, 11, 12, 13,$ and $14$.

This means that we now only need to check for $6, 7, 8,$ and $9$. However, once we have found a set summing to $6$, we can choose everything else and obtain a set summing to $9$, and similarly for $7$ and $8$. Thus, we only need to check each case for whether or not we can obtain $6$ or $7$.

We can make $6$ by having $4, 2$, or $3, 2, 1$, or $5, 1$. We can start with the group of three. To separate $3, 2, 1$ from each other, they must be grouped two together and one separate, like this.

[asy] draw(circle((0, 0), 5)); pair O, A, B, C, D, E; O=origin; A=(0, 5); B=rotate(72)*A; C=rotate(144)*A; D=rotate(216)*A; E=rotate(288)*A; label("$x$", A, N); label("$y$", C, SW); label("$z$", D, SE); [/asy]

Now, we note that $x$ is next to both blank spots, so we can't have a number from one of the pairs. So since we can't have $1$, because it is part of the $5, 1$ pair, and we can't have $2$ there, because it's part of the $4, 2$ pair, we must have $3$ inserted into the $x$ spot. We can insert $1$ and $2$ in $y$ and $z$ interchangeably, since reflections are considered the same.

[asy] draw(circle((0, 0), 5)); pair O, A, B, C, D, E; O=origin; A=(0, 5); B=rotate(72)*A; C=rotate(144)*A; D=rotate(216)*A; E=rotate(288)*A; label("$3$", A, N); label("$2$", C, SW); label("$1$", D, SE); [/asy]

We have $4$ and $5$ left to insert. We can't place the $4$ next to the $2$ or the $5$ next to the $1$, so we must place $4$ next to the $1$ and $5$ next to the $2$.

[asy] draw(circle((0, 0), 5)); pair O, A, B, C, D, E; O=origin; A=(0, 5); B=rotate(72)*A; C=rotate(144)*A; D=rotate(216)*A; E=rotate(288)*A; label("$3$", A, N); label("$5$", B, NW); label("$2$", C, SW); label("$1$", D, SE); label("$4$", E, NE); [/asy]

This is the only solution to make $6$ "bad."

Next we move on to $7$, which can be made by $3, 4$, or $5, 2$, or $4, 2, 1$. We do this the same way as before. We start with the three group. Since we can't have $4$ or $2$ in the top slot, we must have one there, and $4$ and $2$ are next to each other on the bottom. When we have $3$ and $5$ left to insert, we place them such that we don't have the two pairs adjacent.

[asy] draw(circle((0, 0), 5)); pair O, A, B, C, D, E; O=origin; A=(0, 5); B=rotate(72)*A; C=rotate(144)*A; D=rotate(216)*A; E=rotate(288)*A; label("$1$", A, N); label("$3$", B, NW); label("$2$", C, SW); label("$4$", D, SE); label("$5$", E, NE); [/asy]

This is the only solution to make $7$ "bad."

We've covered all needed cases, and the two examples we found are distinct, therefore the answer is $\boxed{\textbf {(B) }2}$.

Note by Williamgolly: Notice how we took care of the case of 3 numbers first. That is because that case is the most severe and we usually take care of the most severe cases first in counting problems.

Video Solution

https://youtu.be/45TQEV8OjRk

See Also

2014 AMC 10B (Problems • Answer Key • Resources)
Preceded by
Problem 23 		Followed by
Problem 25
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
2014 AMC 12B (Problems • Answer Key • Resources)
Preceded by
Problem 17 		Followed by
Problem 19
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 12 Problems and Solutions

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