Difference between revisions of "2006 AIME II Problems/Problem 11"
Rdfzchenyy (talk | contribs) m (→Solution 2 (bash)) |
Mathfun2012 (talk | contribs) m (→Solution 3 (some guessing involved)/"Engineer's Induction") |
||
| (4 intermediate revisions by 3 users not shown) | |||
| 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>.− | ||
=== Solution 2 (bash) === | === 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: | 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_{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_{2}\equiv 1 \pmod {1000} | + | a_{26} \equiv 281 \pmod {1000} \\ |
| − | + | a_{27} \equiv 233 \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} | 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>. | 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
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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
