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Difference between revisions of "2006 AIME I Problems/Problem 11"

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== Solution ==
 
== Solution ==
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We proceed [[recursion | recursively]].  Suppose we can build <math>T_m</math> towers using blocks of size <math>1, 2, \ldots, m</math>.  How many towers can we build using blocks of size <math>1, 2, \ldots, m, m + 1</math>?  If we remove the block of size <math>m + 1</math> from such a tower (keeping all other blocks in order), we get a valid tower using blocks <math>1, 2, \ldots, m</math>.  Given a tower using blocks <math>1, 2, \ldots, m</math> (with <math>m \geq 2</math>), we can insert the block of size <math>m + 1</math> in exactly 3 places: at the beginning, immediately following the block of size <math>m - 1</math> or immediately following the block of size <math>m</math>.  Thus, there are 3 times as many towers using blocks of size <math>1, 2, \ldots, m, m + 1</math> as there are towers using only <math>1, 2, \ldots, m</math>.  There are 2 towers which use blocks <math>1, 2</math>, so there are <math>2\cdot 3^6 = 1458</math> towers using blocks <math>1, 2, \ldots, 8</math>, so the answer is <math>458</math>.
+
Define the sum as <math>x</math>. Notice that <math>a_n\ = a_{n + 3} - a_{n + 2} - a_{n + 1} </math>, so the sum will be:
 +
:<math>x = (a_4 - a_3 - a_2) + (a_5 - a_4 - a_3) + \ldots (a_{30} - a_{29} - a_{28}) + a_{28}</math>
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:<math>x = (a_4+ a_5 \ldots a_{30}) - (a_3 + a_4 + \ldots a_{29}) - (a_2 + a_3 + \ldots a_{28}) + a_{28} + (a_1 - a_1)</math>
  
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(Note that we cannot say, "there is one tower using the block <math>1</math>, so there are <math>3^7</math> towers using the blocks <math>1, 2, \ldots, 8</math>."  The reason this fails is that our recursion only worked when <math>m \geq 2</math>: when <math>m = 1</math>, there are only 2 places to insert a block of size <math>m + 1 = 2</math>, at the beginning or at the end, rather than the 3 places we have at later stages.  Also, note that this method generalizes directly to seeking the number of towers where we change the second rule to read, "The cube immediately on top of a cube with edge-length <math>k</math> must have edge-length at most <math>k + n</math>," where <math>n</math> can be any fixed integer.)
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The first two groupings almost completely cancel. The third resembles <math>x</math>.
 +
 
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:<math>x\ = a_1 - a_3 + a_{28} + a_{30} - x</math>
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:<math>2x\ = a_{28} + a_{30}</math>
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:<math>x\ = \frac{a_{28} + a_{30}}{2}</math>  
 +
 
 +
<math>a_{28}</math> and <math>a_{30}</math> are both given; the last four digits of the sum is <math>3668</math>, and half of that is <math>1834</math>. Therefore, the answer is <math>834</math>.
  
 
== See also ==
 
== See also ==

Revision as of 15:43, 25 September 2007

Problem

A sequence is defined as follows $a_1=a_2=a_3=1,$ and, for all positive integers $n, a_{n+3}=a_{n+2}+a_{n+1}+a_n.$ Given that $a_{28}=6090307, a_{29}=11201821,$ and $a_{30}=20603361,$ find the remainder when $\sum^{28}_{k=1} a_k$ is divided by 1000.

Solution

Define the sum as $x$. Notice that $a_n\ = a_{n + 3} - a_{n + 2} - a_{n + 1}$, so the sum will be:

$x = (a_4 - a_3 - a_2) + (a_5 - a_4 - a_3) + \ldots (a_{30} - a_{29} - a_{28}) + a_{28}$
$x = (a_4+ a_5 \ldots a_{30}) - (a_3 + a_4 + \ldots a_{29}) - (a_2 + a_3 + \ldots a_{28}) + a_{28} + (a_1 - a_1)$

The first two groupings almost completely cancel. The third resembles $x$.

$x\ = a_1 - a_3 + a_{28} + a_{30} - x$
$2x\ = a_{28} + a_{30}$
$x\ = \frac{a_{28} + a_{30}}{2}$

$a_{28}$ and $a_{30}$ are both given; the last four digits of the sum is $3668$, and half of that is $1834$. Therefore, the answer is $834$.

See also

2006 AIME I (Problems • Answer Key • Resources)
Preceded by
Problem 10 		Followed by
Problem 12
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15
All AIME Problems and Solutions
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