2014 AMC 10B Problems/Problem 24

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 1

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 $\boxed{\textbf {(B) }2}$.

Solution 2

Note that any ordering satisfies the following numbers:

$1$ through $5:$ choose the number

$10$ through $14:$ choose all numbers excluding a specific one (such as $1, 2, 3, 4$ in some order for $10$)

$15:$ choose all the numbers.

Now, the pairs $7, 8$ and $6, 9$ are the same cases, since if a sequence satisfies a number, we can choose all the remaining numbers to make the other number. ($1, 2, 3$ for $6$, then $4, 5$ for $9$.)

Thus, we have two cases, whether a sequence doesn't make $6$ or whether a sequence doesn't make $7.$

$6$ can be made by $(1,2,3), (1, 5), (2, 4).$ We can put $1, 2, 3$ around the circle. $4$ and $5$ now need to go in $2$ of the $3$ spots in between $1, 2, 3.$ Also keeping in mind the other two ways to make $6,$ $4$ has to go in the spot opposite of $2$ and $5$ has to go in the spot opposite of $1.$ Thus the only ordering that works is $1, 4, 3, 5, 2$ (ignore rotations and reflections).

Similarly, for the case with $7,$ the only ordering that works is $1, 5, 4, 3, 2$ with gives the answer of $\mathbf{(B) }2.$

Solution 3 (Minimal Casework)

Note that $\{1,2,3,4,5,10,11,12,13,14,15\}$ will always be there. Thus, we need to prevent either the $(6,9)$ or the $(7,8)$ pair. We consider the neighbors of 5.

Case 1: No 7s. Then we have $1-5-4$, and the neighbor of 4 cannot be 3 so the full config must be $3-1-5-4-2$.

Case 2: No 6s. Then we have $2-5-3$, and the neighbor of 2 cannot be 4, so the full config must be $1-2-5-3-4$.

Both of these are bad, and they're the only bads, so the answer is $\textbf{(B)} \text{ }2$.

Video Solution by icematrix

https://youtu.be/45TQEV8OjRk

See Also

2014 AMC 10B (ProblemsAnswer KeyResources)
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 (ProblemsAnswer KeyResources)
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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