Brainsteem Mathematics Challenges: Fibonacci 2018
A general Fibonacci sequence is any sequence of numbers whereby each term is the sum of the previous two terms.
So, let G(n) = G(n-1) + G(n-2), such that G(0)=a and G(1)=b, where a, b are positive integers and a < b.
Now, if the tenth term of the sequence is equal to 2018, find the smallest possible value of (a+b).
As always, these questions are designed to be done by hand, without computational assistance. In this case, you should find some structure to the puzzle so that you can then generate many other similar problems.
The first correct answer and further interesting comments will be rewarded with an upvote.
Enjoy!
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All right. If we start with G(0) as first therm it would be G(9) which would be after the formula I pointed out in another comment
2018=21a+34b
again 2018 mod 55 is 38 2018-38=1980 1980/55=36
so the a has to be 36 or less.
2018-21a=34b
1009-21a/2=17b
a/2=c-> cmax=18
(1009-21c)/17=b
for a+b minimal a has to be as small as posible since b has a bigger impact (2ax=34 x has to be bigger than 1.5)
So you can either make a table or test for each number for c beginning at 1 and stopping as soon as b is an integer.
I dicided to go with the last option. I won't post the boring calculations.
It turns out it works for c=10 and so b is 47 and a is 20.
I checked even further and it turns out there is no other possible pair smaler than c=18 and if c larger 18 a smaller b would not apply.
Thanks, upvoted two of you!
Thanks for the problem. I wish they'd give us similarily hard problems in school.
Hope to see another one from you somtimes.
Most countries have national mathematics competitions at various levels, but it needs the school to actually supervise the tests. So you need to ask your head of maths to apply and get those interested students to practise such questions. :-)
Well I guess that could be an option. But I don't think they have time for that. Probably when we're finished with the Abitur here in Germany.
Just a note, the way the question is worded is not unusual for a maths competition question. But the solver just needs to be a bit careful in processing the algebra.
For example, the "tenth term" is not G(10) ;-)
Are you sure? https://www.varsitytutors.com/algebra_1-help/how-to-find-the-nth-term-of-an-arithmetic-sequence
This goes into the topic of "offsets" and how sometimes the same sequence of numbers is defined differently depending on how it is conceptually used, eg, traditionally, the first Fibonacci term is Fib(0)=0, but in combinatorial problems that makes no sense so Fib(0)=1.
School arithmetic progressions make life easy by letting the first term be a(1). But if, say, the sequence of odd numbers is defined by f(n)=2n+1, then it seems clear that the first term is f(0)=1.
Ultimately, it is a matter of context. It is also a matter of unresolved confusion that linguistic, mathematical and computing ordinals require that context to be unambiguously translated.
In this case, I hope it was clear that G(0)=a is the first term.
The 10th term in the sequence is 21a+34b = 2018 or 21(a+b) + 13b = 2018. Since 21 x 96 = 2016, which is 2018-2, our objective is to find b that will make 13b-2 divisible by 21. The smallest positive integer b satisfying this condition is 5. Therefore, 21(a+b) + 13b = 21(a+5) + 13x5 = 21a + 170 = 2018, i.e. 21a = 1848 or a = 88.
So the smallest possible value of (a+b) should be 88 + 5 = 93
We seem to be having fun with this question!
Nearly... but a < b.
As the smallest b satisfying the condition of 13b-2 diviaible by 21 is 5. 5+21n where n=0,1,... will also satisfy the condition of b. Followong this logic, we can fond the first pair of a and b such that a < b , which is when a = 20 and b = 47. In such case, a + b is 67. Also, the next possible value of b is 47 + 21 = 68 which is greater than the above a + b. Therefore, the smallest possible value of a + b is 67.
Cool, two of you solved it at about the same time!
In fact, with the additional condition of a < b, 20 and 47 is the only possible pair of a and b
Yes :-)
I made this question myself - the sequence is just an excuse to create the Diophantine equation, which I knew was a pain to solve, given the fairly small numbers.
Thanks a lot for taking part.
G(10)=2018=34G(0)+55G(1)
=>G(0)=(2018-55*G(1))/34;
=>G(0)=27;G(1)=20;
But 27>20 => there are no solution
Thanks, but please see my other comments.
This is a cute problem, if there are no answers soon I'll take a stab at it.
Sure, as you can see, everybody who has tried knows how it should be solved... but also needs to read the conditions very carefully and not rush into undeclared assumptions.
I'm surprised, but looks like a good lesson for anybody taking such tests in real life: read the question carefully and rewrite all assumptions and the expected final answer.
Oh. They are all really close now :P
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