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Difference between revisions of "Mock AIME 2 2006-2007 Problems/Problem 8"

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== Problem ==
 
== Problem ==
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The [[positive integer]]s <math>\displaystyle x_1, x_2, ... , x_7</math> satisfy <math>\displaystyle x_6 = 144</math> and <math>\displaystyle x_{n+3} = x_{n+2}(x_{n+1}+x_n)</math> for <math>\displaystyle n = 1, 2, 3, 4</math>. Find the last three [[digit]]s of <math>\displaystyle x_7</math>.
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The [[positive integer]]s <math>x_1, x_2, ... , x_7</math> satisfy <math>x_6 = 144</math> and <math>x_{n+3} = x_{n+2}(x_{n+1}+x_n)</math> for <math>n = 1, 2, 3, 4</math>. Find the last three [[digit]]s of <math>x_7</math>.
 
==Solution==
 
==Solution==
  
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Thus we see there are two possible sequences, but in both cases the answer is 456.
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Thus we see there are two possible sequences, but in both cases the answer is <math>\boxed{456}</math>.
  
 
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Revision as of 15:49, 16 March 2009

Problem

The positive integers $x_1, x_2, ... , x_7$ satisfy $x_6 = 144$ and $x_{n+3} = x_{n+2}(x_{n+1}+x_n)$ for $n = 1, 2, 3, 4$. Find the last three digits of $x_7$.

Solution

This solution is rather long and unpleasant, so a nicer solution may exist:

From the givens, $x_4 = x_3(x_2 + x_1)$ and so $x_5 = x_4(x_3 + x_2) = x_3(x_2 + x_1)(x_3 + x_2)$ and $x_6 = x_5(x_4 + x_3) = x_3(x_2 + x_1)(x_3 + x_2)(x_3(x_2 + x_1) + x_3) = x_3^2(x_3 + x_2)(x_2 + x_1)(x_2 + x_1 + 1) = 144 = 2^4\cdot 3^2$.

Note that this factorization of 144 contains a pair of consecutive integers, $x_2 + x_1$ and $x _2 + x_1 + 1$. The factors of 144 are 1, 2, 3, 4, 6, 8, 9, 12, 16, 18, 24, 36, 48, 72 and 144 itself. As both $x_1$ and $x_2$ are positive integers, $x_1 + x_2 \geq 2$, so we must have $x_1 + x_2$ equal to one of 2, 3 and 8.

If $x_1 + x_2 = 2$ then $x_1 = x_2 = 1$ and so $x_3^2(x_3 + 1)\cdot 2 \cdot 3 = 144$ from which $x_3^2(x_3 + 1) = 24$. It is clear that this equation has no solutions if $x_3 \geq 3$, and neither $x_3 = 1$ nor $x_3 = 2$ is a solution, so in this case we have no solutions.

If $x_1 + x_2 = 8$ then $x_3^2(x_3 + x_2)\cdot 8 \cdot 9 = 144$ so $x_3^2(x_3 + x_2) = 2$. It is clear that $x_3 = x_2 = 1$ is the unique solution to this equation in positive integers. Then $x_1 = 8 - x_2 = 7$ and our sequence is $7, 1, 1, 8, 16, 144, 144(16 + 8) = 3456$.

If $x_1 + x_2 = 3$ then either:

a) $x_1 = 1, x_2 = 2$ and so $x_3^2(x_3 + 2)\cdot 3\cdot 4 = 144$ so $x_3^2(x_3 + 2) = 12$, which has no solutions in positive integers

or

b) $x_1 = 2, x_2 = 1$ and so $x_3^2(x_3 + 1)\cdot 3\cdot 4 = 144$ so $x_3^2(x_3 + 1) = 12$ which has solution $x_3 = 2$. Then our sequence becomes $2, 1, 2, 6, 18, 144, 144(18 + 6) = 3456$.


Thus we see there are two possible sequences, but in both cases the answer is $\boxed{456}$.

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