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

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

@@ Line 2: / Line 2: @@
 A [[sequence]] is defined as follows <math> a_1=a_2=a_3=1, </math> and, for all positive integers <math> n, a_{n+3}=a_{n+2}+a_{n+1}+a_n. </math> Given that <math> a_{28}=6090307, a_{29}=11201821, </math> and <math> a_{30}=20603361, </math> find the [[remainder]] when <math>\sum^{28}_{k=1} a_k </math> is divided by 1000.
-== Solution ==
+== Solutions ==
+=== Solution 1 ===
 Define the sum as <math>s</math>. Since <math>a_n\ = a_{n + 3} - a_{n + 2} - a_{n + 1} </math>, the sum will be:
-<center><math>\begin{align*}s &= a_{28} + \sum^{27}_{k=1} (a_{k+3}-a_{k+2}-a_{k+1}) \\
+<center><math>s = a_{28} + \sum^{27}_{k=1} (a_{k+3}-a_{k+2}-a_{k+1}) \\
-&= 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} + \left(\sum^{30}_{k=4} a_{k} - \sum^{29}_{k=3} a_{k}\right) - \left(\sum^{28}_{k=2} a_{k}\right)\\
-&= a_{28} + (a_{30} - a_{3}) - \left(\sum^{28}_{k=2} a_{k}\right) = a_{28} + a_{30} - a_{3} - (s - a_{1})\\
+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 + a_{28} + a_{30}
+s = -s + a_{28} + a_{30}
-\end{align*}</math></center>
+</math></center>
-Thus <math>s = \frac{a_{28} + a_{30}}{2}</math>, and <math>a_{28},\,a_{30}</math> are both given; the last four digits of their sum is <math>3668</math>, and half of that is <math>1834</math>. Therefore, the answer is <math>\boxed{834}</math>.
+Thus <math>s = \frac{a_{28} + a_{30}}{2}</math>, and <math>a_{28},\,a_{30}</math> are both given; the last four digits of their sum is <math>3668</math>, and half of that is <math>1834</math>. Therefore, the answer is <math>\boxed{834}</math>.−
+=== 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:
-==Solution 2==
-Brute Force. Since the problem asks for the answer of the end value when divided by 1000, it wouldn't be that difficult because you only need to keep track of the last 3 digits.
+<math>
+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}
+</math>
+Adding all the residues shows the sum is congruent to <math>\boxed{834}</math> mod <math>1000</math>.
+~ 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 <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 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 ==

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
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