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

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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>>=2z, <math>a_5</math>>=4z, <math>a_6</math>>=8z, and <math>a_7</math>>=16z, by which point we see that this function diverges.
 
For the first one, <math>a_4</math>>=2z, <math>a_5</math>>=4z, <math>a_6</math>>=8z, and <math>a_7</math>>=16z, by which point we see that this function diverges.
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For the second one, <math>a_4</math>>=3, <math>a_5</math>>=6, <math>a_6</math>>=18, and <math>a_7</math>>=54, by which point we see that this function diverges.
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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:
 
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>|y-x|</math><2 (280 options)

Revision as of 17:49, 24 March 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, $a_4$>=2z, $a_5$>=4z, $a_6$>=8z, and $a_7$>=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. (14 options, none of which coincide with either option 1 or option 2) Adding the total number of such ordered triples yields $280+280-80+14=494$.

See Also

2015 AIME I (Problems • Answer Key • Resources)
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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