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

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

Revision as of 18:10, 10 May 2015

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

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 $280+280-80+16-2=494$.

See Also

2015 AIME I (ProblemsAnswer KeyResources)
Preceded by
Problem 8
Followed by
Problem 10
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15
All AIME Problems and Solutions

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