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

Latest revision as of 17:34, 2 October 2024

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

@@ 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:
-:<math>s = (a_4 - a_3 - a_2) + (a_5 - a_4 - a_3) + \ldots (a_{30} - a_{29} - a_{28}) + a_{28}</math>
+<center><math>s = a_{28} + \sum^{27}_{k=1} (a_{k+3}-a_{k+2}-a_{k+1}) \\
-:<math>s = (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>
+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}
+</math></center>
-The first two groups [[telescope]]. The third resembles <math>s</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:
-:<math>s\ = a_1 - a_3 + a_{28} + a_{30} - s</math>
-:<math>2s\ = a_{28} + a_{30}</math>
-:<math>s\ = \frac{a_{28} + a_{30}}{2}</math>
-<math>a_{28}</math> and <math>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>834</math>.
+<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
 == See also ==
@@ Line 19: / Line 43: @@
 [[Category:Intermediate Algebra Problems]]
+{{MAA Notice}}

2006 AIME II (Problems • Answer Key • Resources)
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