Difference between revisions of "2006 AIME II Problems/Problem 11"
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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. | 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. | ||
− | == | + | == Solutions == |
− | ==Solution 1 | + | === 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: | 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>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}) \\ | ||
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</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: | ||
− | |||
− | + | ||
+ | <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}}(\mod 1000)</math>. | ||
+ | |||
+ | Solution by zeroman; clarified by srisainandan6 | ||
== See also == | == See also == |
Latest revision as of 10:07, 12 October 2023
Contents
Problem
A sequence is defined as follows and, for all positive integers
Given that
and
find the remainder when
is divided by 1000.
Solutions
Solution 1
Define the sum as . Since
, the sum will be:
![$s = a_{28} + \sum^{27}_{k=1} (a_{k+3}-a_{k+2}-a_{k+1}) \\ s = a_{28} + \left(\sum^{30}_{k=4} a_{k} - \sum^{29}_{k=3} a_{k}\right) - \left(\sum^{28}_{k=2} a_{k}\right)\\ s = a_{28} + (a_{30} - a_{3}) - \left(\sum^{28}_{k=2} a_{k}\right) = a_{28} + a_{30} - a_{3} - (s - a_{1})\\ s = -s + a_{28} + a_{30}$](http://latex.artofproblemsolving.com/2/a/d/2ad278b512318ab58fc7ef7280e03bd129453fe1.png)
Thus , and
are both given; the last four digits of their sum is
, and half of that is
. Therefore, the answer is
.−
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:
Adding all the residues shows the sum is congruent to mod
.
~ 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 and
, so we can guess that there is some way to use them in a formula. Namely, we guess that there exists some
such that
. 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
, at least for the first few terms. From this, we have that
.
Solution by zeroman; clarified by srisainandan6
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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