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Difference between revisions of "2015 AIME I Problems/Problem 9"

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Let <math>S</math> be the set of all ordered triple of integers <math>(a_1,a_2,a_3)</math> with <math>1 \le a_1,a_2,a_3 \le 10</math>. Each ordered triple in <math>S</math> generates a sequence according to the rule <math>a_n=a_{n-1}\cdot | a_{n-2}-a_{n-3} |</math> for all <math>n\ge 4</math>. Find the number of such sequences for which <math>a_n=0</math> for some <math>n</math>.
 
Let <math>S</math> be the set of all ordered triple of integers <math>(a_1,a_2,a_3)</math> with <math>1 \le a_1,a_2,a_3 \le 10</math>. Each ordered triple in <math>S</math> generates a sequence according to the rule <math>a_n=a_{n-1}\cdot | a_{n-2}-a_{n-3} |</math> for all <math>n\ge 4</math>. Find the number of such sequences for which <math>a_n=0</math> for some <math>n</math>.
  
−
==Solution==
+
==Solution 1==
−
Let <math>a_1=x, a_2=y, a_3=z</math>. First note that if any absolute value equals 0, then <math>a_n</math>=0.
+
Let <math>a_1=x, a_2=y, a_3=z</math>. First note that if any absolute value equals 0, then <math>a_n=0</math>.
 
Also note that if at any position, <math>a_n=a_{n-1}</math>, then <math>a_{n+2}=0</math>.
 
Also note that if at any position, <math>a_n=a_{n-1}</math>, then <math>a_{n+2}=0</math>.
−
Then, if any absolute value equals 1, then <math>a_n</math>=0.
+
Then, if any absolute value equals 1, then <math>a_n=0</math>.
 
Therefore, if either <math>|y-x|</math> or <math>|z-y|</math> is less than or equal to 1, then that ordered triple meets the criteria.
 
Therefore, if either <math>|y-x|</math> or <math>|z-y|</math> is less than or equal to 1, then that ordered triple meets the criteria.
 
Assume that to be the only way the criteria is met.
 
Assume that to be the only way the criteria is met.
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However, since the minimum values of <math>a_5</math> and <math>a_6</math> are equal, there must be a scenario where the criteria was met that does not meet our earlier scenarios. Calculation shows that to be <math>z=1</math>, <math>|y-x|=2</math>. Again assume that any other scenario will not meet criteria.
 
However, since the minimum values of <math>a_5</math> and <math>a_6</math> are equal, there must be a scenario where the criteria was met that does not meet our earlier scenarios. Calculation shows that to be <math>z=1</math>, <math>|y-x|=2</math>. Again assume that any other scenario will not meet criteria.
 
To prove, divide the other scenarios into two cases: <math>z>1</math>, <math>|y-x|>1</math>, and <math>|z-y|>1</math>; and <math>z=1</math>, <math>|y-x|>2</math>, and <math>|z-y|>1</math>.
 
To prove, divide the other scenarios into two cases: <math>z>1</math>, <math>|y-x|>1</math>, and <math>|z-y|>1</math>; and <math>z=1</math>, <math>|y-x|>2</math>, and <math>|z-y|>1</math>.
−
For the first one, <math>a_4</math> \ge 2z, <math>a_5</math> \ge 4z, <math>a_6</math> \ge 8z, and <math>a_7</math> \ge 16z, by which point we see that this function diverges.
+
For the first one, <math>a_4 \ge 2z</math>, <math>a_5 \ge 4z</math>, <math>a_6 \ge 8z</math>, and <math>a_7 \ge 16z</math>, by which point we see that this function diverges.
 
For the second one, <math>a_4 \ge 3</math>, <math>a_5 \ge 6</math>, <math>a_6 \ge 18</math>, and <math>a_7 \ge 54</math>, by which point we see that this function diverges.
 
For the second one, <math>a_4 \ge 3</math>, <math>a_5 \ge 6</math>, <math>a_6 \ge 18</math>, and <math>a_7 \ge 54</math>, by which point we see that this function diverges.
−
Therefore, the only scenarios where <math>a_n</math>=0 is when any of the following are met:
+
 
−
<math>|y-x|</math><2 (280 options)
+
 
−
<math>|z-y|</math><2 (280 options, 80 of which coincide with option 1)
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Therefore, the only scenarios where <math>a_n=0</math> is when any of the following are met:
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z=1, <math>|y-x|</math>=2. (16 options, 2 of which coincide with either option 1 or option 2)
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<math>|y-x|<2</math> (280 options)
−
Adding the total number of such ordered triples yields <math>280+280-80+16-2=494</math>.
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<math>|z-y|<2</math> (280 options, 80 of which coincide with option 1)
 +
<math>z=1</math>, <math>|y-x|=2</math>. (16 options, 2 of which coincide with either option 1 or option 2)
 +
Adding the total number of such ordered triples yields <math>280+280-80+16-2=\boxed{494}</math>.
 +
 
 +
*Note to author:
 +
 
 +
Because <math>a_4 \ge 2z</math>, <math>a_5 \ge 4z</math>, <math>a_6 \ge 8z</math>, and <math>a_7 \ge 16z</math> doesn't mean the function diverges. What if <math>z = 7</math>, <math>a_4 = 60</math>, and <math>a_5 = 30</math> too?
 +
 
 +
*Note to the Note to author:
 +
 
 +
This isn't possible because the difference between is either 0, 1, or some number greater than 1. If <math>a_4 = 63</math> (since it must be a multiple of <math>z</math>), then a_5 is either 0, 63, or some number greater than 63.
 +
 
 +
==Solution 2==
 +
Note that the only way for a <math>0</math> to be produced at <math>a_n</math> is if either <math>a_{n-1} = 0</math> or <math>a_{n-2} = a_{n-3}</math>. Since the first one will eventually get to the first three assuming that there is no <math>a_{n-2} = a_{n-3}</math> for any <math>n</math>, that is not possible because <math>a_1 , a_2 , a_3 >= 1</math>. Therefore, we need <math>a_{n - 2} = a_{n - 3}</math>.
 +
 
 +
If <math>2</math> consecutive numbers out of <math>a_1 , a_2 , a_3</math> are equal, then those cases work(<math>a_1</math> and <math>a_2</math> or <math>a_2</math> and <math>a_3</math>  <math>\textbf{NOT}</math> <math>a_1</math> and <math>a_3</math>). This is simply <math>10 \cdot 10 + 10 \cdot 10 - 10 = 190</math> by PIE.
 +
 
 +
Now, note that if any of the first three numbers have difference of <math>1</math>, we have another working case. First, we calculate how many there are given exactly one of <math>a_1,a_2</math> or <math>a_2,a_3</math> have difference <math>1</math>. Given <math>3</math> numbers such that the first <math>2</math> have difference <math>1</math>, exactly <math>4</math> permutations work(assuming the numbers are <math>x,y,</math> and <math>z</math> such that <math>|x-y| = 1</math>): <math>x,y,z</math>; <math>y,x,z</math>; <math>z,x,y</math>; and <math>z,y,x</math>. If the two consecutive numbers are <math>1</math> and <math>2</math>, then the last number has <math>7</math> possiblities: <math>4,5,6,\cdots , 10</math>. This is symmetric for <math>9</math> and <math>10</math>. If the consecutive numbers are <math>(2,3),\cdots , (8,9)</math>, there are <math>6</math> possibilities(<math>10</math> minus the numbers themselves and the numbers directly above and below). Note that we are not counting any cases already counted in the first case. Therefore, this case gives you <math>4(7 + 6 * 7 + 7) = 224</math>. Now we consider the case that there both adjacent <math>a</math>s have increments of <math>1</math>. <cmath>+1 , +1 \rightarrow 8</cmath> <cmath>-1 , -1 \rightarrow 8</cmath> <cmath>+1 , -1 \rightarrow 9</cmath> <cmath>-1 , +1 \rightarrow 9</cmath> Therefore this gives <math>224 + 8 + 8 + 9 + 9 = 258</math>. However, note that we have to add the case where you have <math>3</math> consecutive numbers in arrangement such that only <math>2</math> consecutive numbers have difference <math>1</math>. For example, <math>1,3,2</math> is one such triple. There are <math>8</math> triples of consecutive numbers and <math>4</math> ways to arrange each one(e.x: <math>(1,3,2);(3,1,2);(2,1,3);(2,3,1)</math>). This adds on 32 working cases, so this case gives <math>258 + 32 = 290</math>.
 +
 
 +
Note that there is an ultraspecial(Yes, I know that's not a word) case where we generate a pair of <math>a_i</math> that have difference one. This can only happen if <math>a_3 = 1</math> and <math>a_1</math> and <math>a_2</math> have difference <math>2</math>. This contributes <math>14</math> cases(<math>2 * 8</math> and then subtract <math>2</math> because of the cases <math>3,1,1</math> and <math>4,2,1</math>).
 +
 
 +
Therefore, our answer is <math>190 + 290 + 14 = \boxed{494}</math>.
 +
Solution by hyxue
 +
 
 +
 
 +
==Video Solution==
 +
https://youtu.be/nxieVbkk0jY
 +
 
 +
~MathProblemSolvingSkills.com
 +
 
  
 
==See Also==
 
==See Also==
 
{{AIME box|year=2015|n=I|num-b=8|num-a=10}}
 
{{AIME box|year=2015|n=I|num-b=8|num-a=10}}
 
{{MAA Notice}}
 
{{MAA Notice}}
 +
 +
[[Category: Intermediate Combinatorics Problems]]

Latest revision as of 19:35, 28 July 2023

Problem

Let $S$ be the set of all ordered triple of integers $(a_1,a_2,a_3)$ with $1 \le a_1,a_2,a_3 \le 10$. Each ordered triple in $S$ generates a sequence according to the rule $a_n=a_{n-1}\cdot | a_{n-2}-a_{n-3} |$ for all $n\ge 4$. Find the number of such sequences for which $a_n=0$ for some $n$.

Solution 1

Let $a_1=x, a_2=y, a_3=z$. First note that if any absolute value equals 0, then $a_n=0$. Also note that if at any position, $a_n=a_{n-1}$, then $a_{n+2}=0$. Then, if any absolute value equals 1, then $a_n=0$. Therefore, if either $|y-x|$ or $|z-y|$ is less than or equal to 1, then that ordered triple meets the criteria. Assume that to be the only way the criteria is met. To prove, let $|y-x|>1$, and $|z-y|>1$. Then, $a_4 \ge 2z$, $a_5 \ge 4z$, and $a_6 \ge 4z$. However, since the minimum values of $a_5$ and $a_6$ are equal, there must be a scenario where the criteria was met that does not meet our earlier scenarios. Calculation shows that to be $z=1$, $|y-x|=2$. Again assume that any other scenario will not meet criteria. To prove, divide the other scenarios into two cases: $z>1$, $|y-x|>1$, and $|z-y|>1$; and $z=1$, $|y-x|>2$, and $|z-y|>1$. For the first one, $a_4 \ge 2z$, $a_5 \ge 4z$, $a_6 \ge 8z$, and $a_7 \ge 16z$, by which point we see that this function diverges. For the second one, $a_4 \ge 3$, $a_5 \ge 6$, $a_6 \ge 18$, and $a_7 \ge 54$, by which point we see that this function diverges.


Therefore, the only scenarios where $a_n=0$ is when any of the following are met: $|y-x|<2$ (280 options) $|z-y|<2$ (280 options, 80 of which coincide with option 1) $z=1$, $|y-x|=2$. (16 options, 2 of which coincide with either option 1 or option 2) Adding the total number of such ordered triples yields $280+280-80+16-2=\boxed{494}$.

  • Note to author:

Because $a_4 \ge 2z$, $a_5 \ge 4z$, $a_6 \ge 8z$, and $a_7 \ge 16z$ doesn't mean the function diverges. What if $z = 7$, $a_4 = 60$, and $a_5 = 30$ too?

  • Note to the Note to author:

This isn't possible because the difference between is either 0, 1, or some number greater than 1. If $a_4 = 63$ (since it must be a multiple of $z$), then a_5 is either 0, 63, or some number greater than 63.

Solution 2

Note that the only way for a $0$ to be produced at $a_n$ is if either $a_{n-1} = 0$ or $a_{n-2} = a_{n-3}$. Since the first one will eventually get to the first three assuming that there is no $a_{n-2} = a_{n-3}$ for any $n$, that is not possible because $a_1 , a_2 , a_3 >= 1$. Therefore, we need $a_{n - 2} = a_{n - 3}$.

If $2$ consecutive numbers out of $a_1 , a_2 , a_3$ are equal, then those cases work($a_1$ and $a_2$ or $a_2$ and $a_3$ $\textbf{NOT}$ $a_1$ and $a_3$). This is simply $10 \cdot 10 + 10 \cdot 10 - 10 = 190$ by PIE.

Now, note that if any of the first three numbers have difference of $1$, we have another working case. First, we calculate how many there are given exactly one of $a_1,a_2$ or $a_2,a_3$ have difference $1$. Given $3$ numbers such that the first $2$ have difference $1$, exactly $4$ permutations work(assuming the numbers are $x,y,$ and $z$ such that $|x-y| = 1$): $x,y,z$; $y,x,z$; $z,x,y$; and $z,y,x$. If the two consecutive numbers are $1$ and $2$, then the last number has $7$ possiblities: $4,5,6,\cdots , 10$. This is symmetric for $9$ and $10$. If the consecutive numbers are $(2,3),\cdots , (8,9)$, there are $6$ possibilities($10$ minus the numbers themselves and the numbers directly above and below). Note that we are not counting any cases already counted in the first case. Therefore, this case gives you $4(7 + 6 * 7 + 7) = 224$. Now we consider the case that there both adjacent $a$s have increments of $1$. \[+1 , +1 \rightarrow 8\] \[-1 , -1 \rightarrow 8\] \[+1 , -1 \rightarrow 9\] \[-1 , +1 \rightarrow 9\] Therefore this gives $224 + 8 + 8 + 9 + 9 = 258$. However, note that we have to add the case where you have $3$ consecutive numbers in arrangement such that only $2$ consecutive numbers have difference $1$. For example, $1,3,2$ is one such triple. There are $8$ triples of consecutive numbers and $4$ ways to arrange each one(e.x: $(1,3,2);(3,1,2);(2,1,3);(2,3,1)$). This adds on 32 working cases, so this case gives $258 + 32 = 290$.

Note that there is an ultraspecial(Yes, I know that's not a word) case where we generate a pair of $a_i$ that have difference one. This can only happen if $a_3 = 1$ and $a_1$ and $a_2$ have difference $2$. This contributes $14$ cases($2 * 8$ and then subtract $2$ because of the cases $3,1,1$ and $4,2,1$).

Therefore, our answer is $190 + 290 + 14 = \boxed{494}$. Solution by hyxue


Video Solution

https://youtu.be/nxieVbkk0jY

~MathProblemSolvingSkills.com


See Also

2015 AIME I (Problems • Answer Key • Resources)
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Problem 10
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