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: | ||
Line 11: | Line 11: | ||
</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: | ||
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− | <math>a_{1}\equiv 1 \pmod {1000} | + | <math> |
− | \ | + | 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_{5}\equiv 5 \pmod {1000} \\ |
− | a_{4}\equiv 3 \pmod {1000} | + | \cdots \\ |
− | \ | + | a_{25} \equiv 793 \pmod {1000} \\ |
− | a_{5}\equiv 5 \pmod {1000} | + | a_{26} \equiv 281 \pmod {1000} \\ |
− | \ | + | a_{27} \equiv 233 \pmod {1000} \\ |
− | \cdots | + | a_{28} \equiv 307 \pmod {1000} |
− | \ | + | </math> |
− | 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 1000. | + | Adding all the residues shows the sum is congruent to <math>\boxed{834}</math> mod <math>1000</math>. |
~ I-_-I | ~ I-_-I | ||
=== 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}} | + | 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 |
Latest revision as of 17:34, 2 October 2024
Contents
[hide]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:
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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