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Difference between revisions of "2004 AIME II Problems/Problem 9"

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A [[sequence]] of positive integers with <math> a_1=1 </math> and <math> a_9+a_{10}=646 </math> is formed so that the first three terms are in [[geometric sequence|geometric progression]], the second, third, and fourth terms are in [[arithmetic sequence|arithmetic progression]], and, in general, for all <math> n\ge1, </math> the terms <math> a_{2n-1}, a_{2n}, a_{2n+1} </math> are in geometric progression, and the terms <math> a_{2n}, a_{2n+1}, </math> and <math> a_{2n+2} </math> are in arithmetic progression. Let <math> a_n </math> be the greatest term in this sequence that is less than <math>1000</math>. Find <math> n+a_n. </math>
 
A [[sequence]] of positive integers with <math> a_1=1 </math> and <math> a_9+a_{10}=646 </math> is formed so that the first three terms are in [[geometric sequence|geometric progression]], the second, third, and fourth terms are in [[arithmetic sequence|arithmetic progression]], and, in general, for all <math> n\ge1, </math> the terms <math> a_{2n-1}, a_{2n}, a_{2n+1} </math> are in geometric progression, and the terms <math> a_{2n}, a_{2n+1}, </math> and <math> a_{2n+2} </math> are in arithmetic progression. Let <math> a_n </math> be the greatest term in this sequence that is less than <math>1000</math>. Find <math> n+a_n. </math>
  
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== Solution ==
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== Solution 1==
 
Let <math>x = a_2</math>; then solving for the next several terms, we find that <math>a_3 = x^2,\ a_4 = x(2x-1),\ a_5</math> <math>= (2x-1)^2,\ a_6</math> <math>= (2x-1)(3x-2)</math>, and in general, <math>a_{2n} = f(n-1)f(n)</math>, <math>a_{2n+1} = f(n)^2</math>, where <math>f(n) = nx - (n-1)</math>.{{ref|1}}  
 
Let <math>x = a_2</math>; then solving for the next several terms, we find that <math>a_3 = x^2,\ a_4 = x(2x-1),\ a_5</math> <math>= (2x-1)^2,\ a_6</math> <math>= (2x-1)(3x-2)</math>, and in general, <math>a_{2n} = f(n-1)f(n)</math>, <math>a_{2n+1} = f(n)^2</math>, where <math>f(n) = nx - (n-1)</math>.{{ref|1}}  
  
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<cmath>\begin{align*}a_{2n+1} &= \frac{a_{2n}^2}{a_{2n-1}} = \frac{f(n-1)^2f(n)^2}{f(n-1)^2} = f(n)^2 \\
 
<cmath>\begin{align*}a_{2n+1} &= \frac{a_{2n}^2}{a_{2n-1}} = \frac{f(n-1)^2f(n)^2}{f(n-1)^2} = f(n)^2 \\
 
\end{align*}</cmath>
 
\end{align*}</cmath>
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==Solution 2 (Cheese)==
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Let <math>x = a_2</math>. It is apparent that the sequence grows relatively fast, so we start trying positive integers to see what <math>x</math> can be. Finding that <math>x = 5</math> works, after bashing out the rest of the terms we find that <math>a_{16} = 957</math> and <math>a_{17} = 1089</math>, hence our answer is <math>957 + 16 = \boxed{973}</math>.
  
 
== See also ==
 
== See also ==

Revision as of 18:43, 6 September 2022

Problem

A sequence of positive integers with $a_1=1$ and $a_9+a_{10}=646$ is formed so that the first three terms are in geometric progression, the second, third, and fourth terms are in arithmetic progression, and, in general, for all $n\ge1,$ the terms $a_{2n-1}, a_{2n}, a_{2n+1}$ are in geometric progression, and the terms $a_{2n}, a_{2n+1},$ and $a_{2n+2}$ are in arithmetic progression. Let $a_n$ be the greatest term in this sequence that is less than $1000$. Find $n+a_n.$

Solution 1

Let $x = a_2$; then solving for the next several terms, we find that $a_3 = x^2,\ a_4 = x(2x-1),\ a_5$ $= (2x-1)^2,\ a_6$ $= (2x-1)(3x-2)$, and in general, $a_{2n} = f(n-1)f(n)$, $a_{2n+1} = f(n)^2$, where $f(n) = nx - (n-1)$.[1]

From \[a_9 + a_{10} = f(4)^2 + f(4)f(5) = (4x-3)(9x-7) = 646 = 2\cdot 17 \cdot 19\], we find that by either the quadratic formula or trial-and-error/modular arithmetic that $x=5$. Thus $f(n) = 4n+1$, and we need to find the largest $n$ such that either $f(n)^2\, \mathrm{or}\, f(n)f(n-1) < 1000$. This happens with $f(7)f(8) = 29 \cdot 33 = 957$, and this is the $2(8) = 16$th term of the sequence.

The answer is $957 + 16 = \boxed{973}$.


^ We can show this by simultaneous induction: since \begin{align*}a_{2n} &= 2a_{2n-1} - a_{2n-2} = 2a_{2(n-1)+1} - a_{2(n-1)} \\ &= 2f(n-1)^2 - f(n-2)f(n-1) \\ &= f(n-1)[2f(n-1) - f(n-2)] \\ &= f(n-1)[(2n-2-n+2)x-(2n-4-n+3)] \\ &= f(n-1)f(n) \end{align*} and \begin{align*}a_{2n+1} &= \frac{a_{2n}^2}{a_{2n-1}} = \frac{f(n-1)^2f(n)^2}{f(n-1)^2} = f(n)^2 \\ \end{align*}

Solution 2 (Cheese)

Let $x = a_2$. It is apparent that the sequence grows relatively fast, so we start trying positive integers to see what $x$ can be. Finding that $x = 5$ works, after bashing out the rest of the terms we find that $a_{16} = 957$ and $a_{17} = 1089$, hence our answer is $957 + 16 = \boxed{973}$.

See also

2004 AIME II (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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