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

(→‎Solution 2 (bash))
(→‎Solution 3 (some guessing involved)/"Engineer's Induction")
 
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<math>
 
<math>
 
a_{1}\equiv 1 \pmod {1000} \\
 
a_{1}\equiv 1 \pmod {1000} \\
−
 
 
a_{2}\equiv 1 \pmod {1000} \\
 
a_{2}\equiv 1 \pmod {1000} \\
−
 
 
a_{3}\equiv 1 \pmod {1000} \\
 
a_{3}\equiv 1 \pmod {1000} \\
−
 
 
a_{4}\equiv 3 \pmod {1000} \\
 
a_{4}\equiv 3 \pmod {1000} \\
−
 
 
a_{5}\equiv 5 \pmod {1000} \\
 
a_{5}\equiv 5 \pmod {1000} \\
−
 
 
\cdots \\
 
\cdots \\
−
 
 
a_{25} \equiv 793 \pmod {1000} \\
 
a_{25} \equiv 793 \pmod {1000} \\
−
 
 
a_{26} \equiv 281 \pmod {1000} \\
 
a_{26} \equiv 281 \pmod {1000} \\
−
 
 
a_{27} \equiv 233 \pmod {1000} \\
 
a_{27} \equiv 233 \pmod {1000} \\
−
 
 
a_{28} \equiv 307 \pmod {1000}
 
a_{28} \equiv 307 \pmod {1000}
−
 
 
</math>
 
</math>
  
Line 45: Line 35:
  
 
=== Solution 3 (some guessing involved)/"Engineer's Induction" ===
 
=== Solution 3 (some guessing involved)/"Engineer's Induction" ===
−
All terms in the sequence are sums of previous terms, so the sum of all terms up to a certain point must be some linear combination of the first three terms. Also, we are given <math>a_{28}, a_{29}, </math> and <math>a_{30}</math>, so we can guess that there is some way to use them in a formula. Namely, we guess that there exists some <math>p, q, r</math> such that <math>\sum_{k=1}^{n}{a_k} = pa_n+qa_{n+1}+ra_{n+2}</math>. From here, we list out the first few terms of the sequence and the cumulative sums, and with a little bit of substitution and algebra we see that <math>(p, q, r) = (\frac{1}{2}, 0, \frac{1}{2})</math>, at least for the first few terms. From this, we have that <math>\sum_{k=1}^{28}{a_k} = \frac{a_{28}+a_{30}}{2} \equiv{\boxed{834}}(\mod 1000)</math>.
+
All terms in the sequence are sums of previous terms, so the sum of all terms up to a certain point must be some linear combination of the first three terms. Also, we are given <math>a_{28}, a_{29}, </math> and <math>a_{30}</math>, so we can guess that there is some way to use them in a formula. Namely, we guess that there exists some <math>p, q, r</math> such that <math>\sum_{k=1}^{n}{a_k} = pa_n+qa_{n+1}+ra_{n+2}</math>. From here, we list out the first few terms of the sequence and the cumulative sums, and with a little bit of substitution and algebra we see that <math>(p, q, r) = (\frac{1}{2}, 0, \frac{1}{2})</math>, at least for the first few terms. From this, we have that <math>\sum_{k=1}^{28}{a_k} = \frac{a_{28}+a_{30}}{2} \equiv{\boxed{834}}\pmod {1000}</math>.
  
 
Solution by zeroman; clarified by srisainandan6
 
Solution by zeroman; clarified by srisainandan6
 +
 +
 +
=== Solution 4 (if you did not know how to use the numbers given in the problem) ===
 +
By the Chinese remainder theorem, each number under 1000 is uniquely determined by its mod 8 and mod 125.
 +
 +
We list a few terms of the sequence mod 8:
 +
 +
<cmath>1,1,1,3,5,1,1,7,1,1,1,...</cmath>
 +
 +
Therefore, the cycle repeats every 8 numbers, and each cycle has a sum of 4 mod 8. Therefore, the sum mod 8 is <cmath>3 \cdot 4 + 1 + 1 + 1 + 3 = 2 \mod 8.</cmath>
 +
 +
Denote <cmath>s_n = \sum _{i=1} ^{n} a_i.</cmath> It is easy to prove that <math>s_{n+3} = s_{n+2} + s_{n+1} + s_{n}.</math>
 +
 +
We write the sum (<math>s_n</math>) of the first terms mod 125:
 +
 +
<cmath>1,2,3,6,11,20,37,68,0,-20,48,28,56,7,91,29,2,-3,28,27,52,-18,61,-30,13,44,27,84.</cmath>
 +
 +
Therefore the desired number is <math>84 \mod 125.</math>
 +
 +
From here, we can determine the number we are looking for is <math>750 + 84 = \boxed{834}.</math>
 +
-sd8
  
 
== See also ==
 
== See also ==

Latest revision as of 19:54, 31 July 2025

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.

Solutions

Solution 1

Define the sum as $s$. Since $a_n\ = a_{n + 3} - a_{n + 2} - a_{n + 1}$, the sum will be:

$s = a_{28} + \sum^{27}_{k=1} (a_{k+3}-a_{k+2}-a_{k+1}) \\ s = a_{28} + \left(\sum^{30}_{k=4} a_{k} - \sum^{29}_{k=3} a_{k}\right) - \left(\sum^{28}_{k=2} a_{k}\right)\\ s = a_{28} + (a_{30} - a_{3}) - \left(\sum^{28}_{k=2} a_{k}\right) = a_{28} + a_{30} - a_{3} - (s - a_{1})\\ s = -s + a_{28} + a_{30}$

Thus $s = \frac{a_{28} + a_{30}}{2}$, and $a_{28},\,a_{30}$ are both given; the last four digits of their sum is $3668$, and half of that is $1834$. Therefore, the answer is $\boxed{834}$.−

Solution 2 (bash)

Since the problem only asks for the first 28 terms and we only need to calculate mod 1000, we simply bash the first 28 terms:


$a_{1}\equiv 1 \pmod {1000} \\ a_{2}\equiv 1 \pmod {1000} \\ a_{3}\equiv 1 \pmod {1000} \\ a_{4}\equiv 3 \pmod {1000} \\ a_{5}\equiv 5 \pmod {1000} \\ \cdots \\ a_{25} \equiv 793 \pmod {1000} \\ a_{26} \equiv 281 \pmod {1000} \\ a_{27} \equiv 233 \pmod {1000} \\ a_{28} \equiv 307 \pmod {1000}$

Adding all the residues shows the sum is congruent to $\boxed{834}$ mod $1000$.

~ I-_-I

Solution 3 (some guessing involved)/"Engineer's Induction"

All terms in the sequence are sums of previous terms, so the sum of all terms up to a certain point must be some linear combination of the first three terms. Also, we are given $a_{28}, a_{29},$ and $a_{30}$, so we can guess that there is some way to use them in a formula. Namely, we guess that there exists some $p, q, r$ such that $\sum_{k=1}^{n}{a_k} = pa_n+qa_{n+1}+ra_{n+2}$. From here, we list out the first few terms of the sequence and the cumulative sums, and with a little bit of substitution and algebra we see that $(p, q, r) = (\frac{1}{2}, 0, \frac{1}{2})$, at least for the first few terms. From this, we have that $\sum_{k=1}^{28}{a_k} = \frac{a_{28}+a_{30}}{2} \equiv{\boxed{834}}\pmod {1000}$.

Solution by zeroman; clarified by srisainandan6


Solution 4 (if you did not know how to use the numbers given in the problem)

By the Chinese remainder theorem, each number under 1000 is uniquely determined by its mod 8 and mod 125.

We list a few terms of the sequence mod 8:

\[1,1,1,3,5,1,1,7,1,1,1,...\]

Therefore, the cycle repeats every 8 numbers, and each cycle has a sum of 4 mod 8. Therefore, the sum mod 8 is \[3 \cdot 4 + 1 + 1 + 1 + 3 = 2 \mod 8.\]

Denote \[s_n = \sum _{i=1} ^{n} a_i.\] It is easy to prove that $s_{n+3} = s_{n+2} + s_{n+1} + s_{n}.$

We write the sum ($s_n$) of the first terms mod 125:

\[1,2,3,6,11,20,37,68,0,-20,48,28,56,7,91,29,2,-3,28,27,52,-18,61,-30,13,44,27,84.\]

Therefore the desired number is $84 \mod 125.$

From here, we can determine the number we are looking for is $750 + 84 = \boxed{834}.$ -sd8

See also

2006 AIME II (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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