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| − | {{duplicate|[[2014 AMC 12B Problems|2014 AMC 12B #18]] and [[2014 AMC 10B Problems|2014 AMC 10B #24]]}}
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| − | ==Problem==
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| − | 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?
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| − | <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 1==
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| − | 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.
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| − | 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>.
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| − | 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>.
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| − | 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.
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| − | <asy>
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| − | draw(circle((0, 0), 5));
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| − | pair O, A, B, C, D, E;
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| − | O=origin;
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| − | A=(0, 5);
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| − | B=rotate(72)*A;
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| − | C=rotate(144)*A;
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| − | D=rotate(216)*A;
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| − | E=rotate(288)*A;
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| − | label("$x$", A, N);
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| − | label("$y$", C, SW);
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| − | label("$z$", D, SE);
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| − | </asy>
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| − | 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.
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| − | <asy>
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| − | draw(circle((0, 0), 5));
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| − | pair O, A, B, C, D, E;
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| − | O=origin;
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| − | A=(0, 5);
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| − | B=rotate(72)*A;
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| − | C=rotate(144)*A;
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| − | D=rotate(216)*A;
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| − | E=rotate(288)*A;
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| − | label("$3$", A, N);
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| − | label("$2$", C, SW);
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| − | label("$1$", D, SE);
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| − | </asy>
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| − | 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>.
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| − | <asy>
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| − | draw(circle((0, 0), 5));
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| − | pair O, A, B, C, D, E;
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| − | O=origin;
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| − | A=(0, 5);
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| − | B=rotate(72)*A;
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| − | C=rotate(144)*A;
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| − | D=rotate(216)*A;
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| − | E=rotate(288)*A;
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| − | label("$3$", A, N);
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| − | label("$5$", B, NW);
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| − | label("$2$", C, SW);
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| − | label("$1$", D, SE);
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| − | label("$4$", E, NE);
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| − | </asy>
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| − | This is the only solution to make <math>6</math> "bad."
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| − | 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.
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| − |
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| − | <asy>
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| − | draw(circle((0, 0), 5));
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| − | pair O, A, B, C, D, E;
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| − | O=origin;
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| − | A=(0, 5);
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| − | B=rotate(72)*A;
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| − | C=rotate(144)*A;
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| − | D=rotate(216)*A;
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| − | E=rotate(288)*A;
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| − | label("$1$", A, N);
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| − | label("$3$", B, NW);
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| − | label("$2$", C, SW);
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| − | label("$4$", D, SE);
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| − | label("$5$", E, NE);
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| − | </asy>
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| − | This is the only solution to make <math>7</math> "bad."
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| − | We've covered all needed cases, and the two examples we found are distinct, therefore the answer is <math>\boxed{\textbf {(B) }2}</math>.
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| − | ==Solution 2==
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| − | Note that any ordering satisfies the following numbers:
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| − | <math>1</math> through <math>5:</math> choose the number
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| − | <math>10</math> through <math>14:</math> choose all numbers excluding a specific one (such as <math>1, 2, 3, 4</math> in some order for <math>10</math>)
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| − | <math>15:</math> choose all the numbers.
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| − | Now, the pairs <math>7, 8</math> and <math>6, 9</math> are the same cases, since if a sequence satisfies a number, we can choose all the remaining numbers to make the other number. (<math>1, 2, 3</math> for <math>6</math>, then <math>4, 5</math> for <math>9</math>.)
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| − | Thus, we have two cases, whether a sequence doesn't make <math>6</math> or whether a sequence doesn't make <math>7.</math>
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| − | <math>6</math> can be made by <math>(1,2,3), (1, 5), (2, 4).</math> We can put <math>1, 2, 3</math> around the circle. <math>4</math> and <math>5</math> now need to go in <math>2</math> of the <math>3</math> spots in between <math>1, 2, 3.</math> Also keeping in mind the other two ways to make <math>6,</math> <math>4</math> has to go in the spot opposite of <math>2</math> and <math>5</math> has to go in the spot opposite of <math>1.</math> Thus the only ordering that works is <math>1, 4, 3, 5, 2</math> (ignore rotations and reflections).
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| − | Similarly, for the case with <math>7,</math> the only ordering that works is <math>1, 5, 4, 3, 2</math> with gives the answer of <math>\mathbf{(B) }2.</math>
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| − | ==Video Solution by icematrix==
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| − | https://youtu.be/45TQEV8OjRk
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| − | ==See Also==
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| − | {{AMC10 box|year=2014|ab=B|num-b=23|num-a=25}}
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| − | {{AMC12 box|year=2014|ab=B|num-b=17|num-a=19}}
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| − | {{MAA Notice}}
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