Difference between revisions of "2006 AIME II Problems/Problem 11"
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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: | 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 | + | <math>a_{1}\equiv 1 \pmod {1000} |
− | + | \newline | |
− | + | a_{2}\equiv 1 \pmod {1000} | |
− | + | \newline | |
− | + | a_{3}\equiv 1 \pmod {1000} | |
− | + | \newline | |
− | + | a_{4}\equiv 3 \pmod {1000} | |
− | + | \newline | |
− | + | a_{5}\equiv 5 \pmod {1000} | |
− | + | \newline | |
+ | \cdots | ||
+ | \newline | ||
+ | a_{25} \equiv 793 \pmod {1000} | ||
+ | \newline | ||
+ | a_{26} \equiv 281 \pmod {1000} | ||
+ | \newline | ||
+ | a_{27} \equiv 233 \pmod {1000} | ||
+ | \newline | ||
+ | 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 1000. |
Revision as of 13:44, 19 October 2019
Contents
Problem
A sequence is defined as follows and, for all positive integers Given that and find the remainder when is divided by 1000.
Solution
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
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 1000.
~ I-_-I
Solution 3 (some guessing involved)
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
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.