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

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== Problem ==
== Problem ==
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>.
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.
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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