2006 AIME II Problems/Problem 11
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:
$a_{1}\equiv 1 \pmod {1000} \\
a_{2}\equiv 1 \pmod {1000} \\
a_{3}\equiv 1 \pmod {1000} \\
a_{4}\equiv 3 \pmod {1000} \\
a_{5}\equiv 5 \pmod {1000} \\
\cdots \\
a_{25} \equiv 793 \pmod {1000} \\
a_{26} \equiv 281 \pmod {1000} \\
a_{27} \equiv 233 \pmod {1000} \\
a_{28} \equiv 307 \pmod {1000}$ (Error compiling LaTeX. Unknown error_msg)
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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