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

m (Solution 1)
(Solution 1)
 
Line 13: Line 13:
 
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>.
  
by an anonymous user
+
Solution by an anonymous user.
  
 
=== Solution 2 (bash) ===
 
=== Solution 2 (bash) ===

Latest revision as of 23:11, 25 October 2020

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.

Solution

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 by an anonymous user.

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:

$a_{1}\equiv 1 \pmod {1000} \newline a_{2}\equiv 1 \pmod {1000} \newline a_{3}\equiv 1 \pmod {1000} \newline a_{4}\equiv 3 \pmod {1000} \newline a_{5}\equiv 5 \pmod {1000} \newline \cdots  \newline a_{25} \equiv 793 \pmod {1000} \newline a_{26} \equiv 281 \pmod {1000} \newline a_{27} \equiv 233 \pmod {1000} \newline a_{28} \equiv 307 \pmod {1000}$

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 $\sum_{k=1}^{28}{a_k} = \frac{a_{28}+a_{30}}{2} \equiv{\boxed{834}}(\mod 1000)$.

Solution by zeroman; clarified by srisainandan6

See also

2006 AIME II (ProblemsAnswer KeyResources)
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. AMC logo.png

Invalid username
Login to AoPS