fibonacci problems? "If 8th number is 2017, what is largest possible value of 1st?"

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nanase

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we have eight positive integers in a row. Starting from the third, each is the sum of the two numbers before it. If the eighth number is 2017, what is the largest possible value of the first one?

I am approaching the question as follow :
1st a
2nd b
3rd a+b
4th a + 2b
5th 2a + 3b
6th 3a + 5b
7th 5a + 8b
8th 8a + 13b =2017

How do we solve from one equation only?
appreciate the help
 
nanaseailie said:
we have eight positive integers in a row. Starting from the third, each is the sum of the two numbers before it. If the eighth number is 2017, what is the largest possible value of the first one?

I am approaching the question as follow :
1st a
2nd b
3rd a+b
4th a + 2b
5th 2a + 3b
6th 3a + 5b
7th 5a + 8b
8th 8a + 13b =2017

How do we solve from one equation only?
appreciate the help
Click to expand...
[imath]8a + 13b = 2017[/imath]

[imath]b = \dfrac{2017 - 8a}{13}[/imath]

b is a positive integer, so the numerator must be divisible by 13. What is the largest such that this will happen?

-Dan
 
nanaseailie said:
[imath]8a+13b=2017 \implies 2017-13b[/imath]we have eight positive integers in a row. Starting from the third, each is the sum of the two numbers before it. If the eighth number is 2017, what is the largest possible value of the first one?

I am approaching the question as follow :
1st a
2nd b
3rd a+b
4th a + 2b
5th 2a + 3b
6th 3a + 5b
7th 5a + 8b
8th 8a + 13b =2017

How do we solve from one equation only?
appreciate the help
Click to expand...
To maximize a, you need to minimize b.

[imath]8a+13b = 2017 \implies 2017 -13b[/imath] must be a multiple of [imath]8[/imath] or [imath]2017-13b \equiv 0 \mod 8[/imath].

Solve for [imath]b.[/imath]
 
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@BigBeachBanana & @topsquark I am able to get a = 10 and b = 149 through selective or smart trials but still consuming too much time.
how do I use that mod 8 ? can you explain/hint more?
 
nanaseailie said:
@BigBeachBanana & @topsquark I am able to get a = 10 and b = 149 through selective or smart trials but still consuming too much time.
how do I use that mod 8 ? can you explain/hint more?
Click to expand...
This does not look like a solution with the largest value of 'a'.

I would use 'mod 13' here. I.e. [math]8a \equiv 2017 \mod 13[/math]. Can you find a nice expression for all such 'a's ?
 
Using Linear Diophantine for 8a + 13b = 2017. We get the following.
Since gcd(8, 13) = 1 so we can find integers a and b such that 8a + 13b = 1. 8(5) + 13(-3) = 1. Now multiply both sides by 2017. This gives 8(10,085)+13(-6051) =2017.
So one solution will be (a,b) = (10,085, -6051). ALL (integer) solutions will be in the form (a,b)= (10085 - 13t, -6051+ 8t) where t is an integer.
Continue from here.
 
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blamocur said:
Which question?
Click to expand...
we have eight positive integers in a row. Starting from the third, each is the sum of the two numbers before it. If the eighth number is 2017, what is the largest possible value of the first one?
 
nanaseailie said:
we have eight positive integers in a row. Starting from the third, each is the sum of the two numbers before it. If the eighth number is 2017, what is the largest possible value of the first one?
Click to expand...
I don't see "all positive integers" mentioned in there.
 
Steven G said:
ALL (integer) solutions will be in the form (a,b)= (10085 - 13t, -6051+ 8t) where t is an integer.
Continue from here.
Click to expand...
nanaseailie said:
The question says all positive integers though? right?
Click to expand...
You want to minimize a and yes, a>0.
If a = 10085-13t, then you want to find the smallest t (since smaller t's make a smaller) so a is still positive.
How do you plan on doing that?
Hint:
10085-13*1=10072>0
10085-13*2=10059>0
10085-13*3=10046>0
....
 
Steven G said:
You want to minimize a and yes, a>0.
If a = 10085-13t, then you want to find the smallest t (since smaller t's make a smaller) so a is still positive.
How do you plan on doing that?
Hint:
10085-13*1=10072>0
10085-13*2=10059>0
10085-13*3=10046>0
....
Click to expand...
I meant to say that you want to minimize b.
 
blamocur said:
I see. Are you having any progress with this problem?
Click to expand...
I can get the answers as in post #4 through limited trials & error, but I am still intrigued and working out the method on mod (I am totally noob on this method) and also from your tips as well as Steven G's hint.
 
nanaseailie said:
I can get the answers as in post #4 through limited trials & error, but I am still intrigued and working out the method on mod (I am totally noob on this method) and also from your tips as well as Steven G's hint.
Click to expand...
Any more tips would simply be showing a solution that you wouldn't understand. Your reply suggests that you don't have a foundation to solve this problem.

Read about the Diophantine Equation, Bezout's Theorem, and Extended Euclidean Algorithm. These are often taught in Elementary Number Theory courses at the university level.

PS: Your answer isn't correct as mentioned above.
 
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BigBeachBanana said:
Any more tips would simply be showing a solution that you wouldn't understand. Your reply suggests that you don't have a foundation to solve this problem.

Read about the Diophantine Equation, Bezout's Theorem, and Extended Euclidean Algorithm. These are often taught in Elementary Number Theory courses at the university level.

PS: Your answer isn't correct as mentioned above.
Click to expand...
seriously my answer is still wrong?
I thought with a = 10 and b = 149
then 8a +13b = 2017
and all my eight values are all positive
 
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