Difference between revisions of "2018 AIME II Problems/Problem 15"

Revision as of 14:11, 27 March 2018

Problem

Find the number of functions $f$ from $\{0, 1, 2, 3, 4, 5, 6\}$ to the integers such that $f(0) = 0$ , $f(6) = 12$ , and

$|x - y|$ $\leq$ $|f(x) - f(y)|$ $\leq$ $3|x - y|$

for all $x$ and $y$ in $\{0, 1, 2, 3, 4, 5, 6\}$ .

Solution

First, suppose $x = y + 1$ . Then, the inequality becomes $1 \leq |f(y + 1) - f(y)| \leq 3$ . In other words, the (positive) difference between consecutive function values is $1$ , $2$ , or $3$ . Let $d_k := f(k) - f(k - 1)$ . Note that

$f(x) - f(y) = \sum_{k=y+1}^{x} d_k.$

Thus, $\sum_{k=1}^{6} d_k = f(6) - f(0) = 12$ . Note that at most one value of $d_k$ in $\left\{d_1, d_2, d_3, d_4, d_5, d_6\right\}$ can be negative. This is because the maximum value of $d_1 + d_2 + d_3 + d_4 + d_5 + d_6$ would be $3 + 3 + 3 + 3 - 1 - 1 = 10$ if more than one value of $d_k$ is negative. Plugging $x = y + 2$ into the original inequality yields $2 \leq |f(y + 2) - f(y)| \leq 6$ , which becomes $2 \leq \left|d_{y+2} + d_{y+1}\right| \leq 6$ . The only way for $d_{y+2} + d_{y+1}$ to be negative while satisfying this inequality is for $\left(d_{y+2}, d_{y+1}\right)$ to equal $(1, -3)$ or $(-3, 1)$ . However, this forces $d_1 + d_2 + d_3 + d_4 + d_5 + d_6 \leq 3 + 3 + 3 + 3 + 1 - 3 = 10 < 12$ , which is disallowed. Hence, we conclude that the following stronger inequality,

$2 \leq d_{y + 2} + d_{y + 1} \leq 6,$

is always true. We now have two cases of functions to count. For future reference let $D$ be the (ordered) sequence $\left\{d_1, d_2, d_3, d_4, d_5, d_6\right\}$ .

Case 1: There is exactly one instance of $d_k = -1$ .

By the "stronger" inequality above, $d_{k-1} = 3$ if $k > 1$ , and $d_{k+1} = 3$ if $k < 6$ . If $2 \leq k \leq 5$ , then $D$ contains the subsequence $\left\{3, -1, 3\right\}$ , and the other three $d$ -values sum to $7$ , so they are either $3$ , $3$ , and $1$ in some order, or they are $2$ , $2$ , and $3$ in some order. Thus, each $k \in \{2, 3, 4, 5\}$ for which $d_k = -1$ produces $6$ sequences $D$ . If $k = 1$ or $k = 6$ , then $D$ begins with $\{-1, 3\}$ or ends with $\{3, -1\}$ , respectively. Either way, the remaining four $d$ -values sum to $10$ , so they can be any permutation of $\{3, 3, 2, 2\}$ (six permutations) or $\{3, 3, 3, 1\}$ (four permutations). Each of these vaues of $k$ yields $6 + 4 = 10$ sequences, so our total count for Case 1 is $4 \cdot 6 + 2 \cdot 10 = 44$ .

Case 2: All values of $d$ are positive.

Then, $D$ is a permutation of $\{3, 3, 3, 1, 1, 1\}$ , $\{3, 3, 2, 2, 1, 1\}$ , $\{3, 2, 2, 2, 2, 1\}$ , or $\{2, 2, 2, 2, 2, 2\}$ . The number of ways to permute three $3$ s and three $1$ s is $\frac{6!}{3! \cdot 3!} = 20$ . The number of ways to permute two $3$ s, two $2$ s, and two $1$ s is $\frac{6!}{2! \cdot 2! \cdot 2!} = 90$ . The number of ways to permute one $3$ , four $2$ s, and one $1$ is $\frac{6!}{1! \cdot 4! \cdot 1!} = 30$ . Finally, there is obviously only $1$ way to permute six $2$ s. Our total count for Case 2 is $20 + 90 + 30 + 1 = 141$ .

To complete the justification that all of these $44 + 141 = \boxed{185}$ cases satisfy the original inequality, we leverage the fact that $\{f(x)\}_{x=0}^{6}$ is either monotonically increasing (Case 2) or the union of two monotonically increasing subsequences (Case 1). Consider any monotonically increasing subsequence that starts at $x = a$ and ends at $x = b$ . For $a < k \leq b$ , $d_k$ will be positive, allowing us to remove the absolute value bars from the original inequality:

$x - y \leq f(x) - f(y) \leq 3(x - y).$

Now, the inequality is transitive; supposing $a \leq x_3 < x_2 < x_1 \leq b$ , if the inequality is satisfied at $(x, y) = \left(x_1, x_2\right)$ and at $(x, y) = \left(x_2, x_3\right)$ , then it is also satisfied at $(x, y) = \left(x_1, x_3\right)$ . If we ever have a decreasing part where $f(x + 1) < f(x)$ , then we can use some variant of the inequality $2 \leq f(x + 1) - f(x - 1) \leq 6$ , which we derived earlier. This is a specific case of $x - y \leq f(x) - f(y) \leq 3(x - y)$ , so we can finish off the argument by invoking transitivity.

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@@ Line 6: / Line 6: @@
 for all <math>x</math> and <math>y</math> in <math>\{0, 1, 2, 3, 4, 5, 6\}</math>.
+==Solution==
+First, suppose <math>x = y + 1</math>. Then, the inequality becomes <math>1 \leq |f(y + 1) - f(y)| \leq 3</math>. In other words, the (positive) difference between consecutive function values is <math>1</math>, <math>2</math>, or <math>3</math>. Let <math>d_k := f(k) - f(k - 1)</math>. Note that
+<math>f(x) - f(y) = \sum_{k=y+1}^{x} d_k.</math>
+Thus, <math>\sum_{k=1}^{6} d_k = f(6) - f(0) = 12</math>. Note that at most one value of <math>d_k</math> in <math>\left\{d_1, d_2, d_3, d_4, d_5, d_6\right\}</math> can be negative. This is because the maximum value of <math>d_1 + d_2 + d_3 + d_4 + d_5 + d_6</math> would be <math>3 + 3 + 3 + 3 - 1 - 1 = 10</math> if more than one value of <math>d_k</math> is negative. Plugging <math>x = y + 2</math> into the original inequality yields <math>2 \leq |f(y + 2) - f(y)| \leq 6</math>, which becomes <math>2 \leq \left|d_{y+2} + d_{y+1}\right| \leq 6</math>. The only way for <math>d_{y+2} + d_{y+1}</math> to be negative while satisfying this inequality is for <math>\left(d_{y+2}, d_{y+1}\right)</math> to equal <math>(1, -3)</math> or <math>(-3, 1)</math>. However, this forces <math>d_1 + d_2 + d_3 + d_4 + d_5 + d_6 \leq 3 + 3 + 3 + 3 + 1 - 3 = 10 < 12</math>, which is disallowed. Hence, we conclude that the following stronger inequality,
+<math>2 \leq d_{y + 2} + d_{y + 1} \leq 6,</math>
+is always true. We now have two cases of functions to count. For future reference let <math>D</math> be the (ordered) sequence <math>\left\{d_1, d_2, d_3, d_4, d_5, d_6\right\}</math>.
+Case 1: There is exactly one instance of <math>d_k = -1</math>.
+By the "stronger" inequality above, <math>d_{k-1} = 3</math> if <math>k > 1</math>, and <math>d_{k+1} = 3</math> if <math>k < 6</math>. If <math>2 \leq k \leq 5</math>, then <math>D</math> contains the subsequence <math>\left\{3, -1, 3\right\}</math>, and the other three <math>d</math>-values sum to <math>7</math>, so they are either <math>3</math>, <math>3</math>, and <math>1</math> in some order, or they are <math>2</math>, <math>2</math>, and <math>3</math> in some order. Thus, each <math>k \in \{2, 3, 4, 5\}</math> for which <math>d_k = -1</math> produces <math>6</math> sequences <math>D</math>. If <math>k = 1</math> or <math>k = 6</math>, then <math>D</math> begins with <math>\{-1, 3\}</math> or ends with <math>\{3, -1\}</math>, respectively. Either way, the remaining four <math>d</math>-values sum to <math>10</math>, so they can be any permutation of <math>\{3, 3, 2, 2\}</math> (six permutations) or <math>\{3, 3, 3, 1\}</math> (four permutations). Each of these vaues of <math>k</math> yields <math>6 + 4 = 10</math> sequences, so our total count for Case 1 is <math>4 \cdot 6 + 2 \cdot 10 = 44</math>.
+Case 2: All values of <math>d</math> are positive.
+Then, <math>D</math> is a permutation of <math>\{3, 3, 3, 1, 1, 1\}</math>, <math>\{3, 3, 2, 2, 1, 1\}</math>, <math>\{3, 2, 2, 2, 2, 1\}</math>, or <math>\{2, 2, 2, 2, 2, 2\}</math>.
+The number of ways to permute three <math>3</math>s and three <math>1</math>s is <math>\frac{6!}{3! \cdot 3!} = 20</math>.
+The number of ways to permute two <math>3</math>s, two <math>2</math>s, and two <math>1</math>s is <math>\frac{6!}{2! \cdot 2! \cdot 2!} = 90</math>.
+The number of ways to permute one <math>3</math>, four <math>2</math>s, and one <math>1</math> is <math>\frac{6!}{1! \cdot 4! \cdot 1!} = 30</math>.
+Finally, there is obviously only <math>1</math> way to permute six <math>2</math>s. Our total count for Case 2 is <math>20 + 90 + 30 + 1 = 141</math>.
+To complete the justification that all of these <math>44 + 141 = \boxed{185}</math> cases satisfy the original inequality, we leverage the fact that <math>\{f(x)\}_{x=0}^{6}</math> is either monotonically increasing (Case 2) or the union of two monotonically increasing subsequences (Case 1). Consider any monotonically increasing subsequence that starts at <math>x = a</math> and ends at <math>x = b</math>. For <math>a < k \leq b</math>, <math>d_k</math> will be positive, allowing us to remove the absolute value bars from the original inequality:
+<math>x - y \leq f(x) - f(y) \leq 3(x - y).</math>
+Now, the inequality is transitive; supposing <math>a \leq x_3 < x_2 < x_1 \leq b</math>, if the inequality is satisfied at <math>(x, y) = \left(x_1, x_2\right)</math> and at <math>(x, y) = \left(x_2, x_3\right)</math>, then it is also satisfied at <math>(x, y) = \left(x_1, x_3\right)</math>. If we ever have a decreasing part where <math>f(x + 1) < f(x)</math>, then we can use some variant of the inequality <math>2 \leq f(x + 1) - f(x - 1) \leq 6</math>, which we derived earlier. This is a specific case of <math>x - y \leq f(x) - f(y) \leq 3(x - y)</math>, so we can finish off the argument by invoking transitivity.
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Difference between revisions of "2018 AIME II Problems/Problem 15"

Revision as of 14:11, 27 March 2018

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