Each new term in the Fibonacci sequence is generated by adding the previous two terms. By starting with 1 and 2, the first 10 terms will be:
1, 2, 3, 5, 8, 13, 21, 34, 55, 89, ...
By considering the terms in the Fibonacci sequence whose values do not exceed the input value, find the sum of the even-valued terms.
Solution Stats
Problem Comments
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4 Comments
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1 older comment
Aurelien Queffurust
on 20 Feb 2012
I wanted to use : z = feval(symengine,'numlib::fibonacci',10)
to get the 10th element for example but CODY returns:
Could not check out Symbolic Math Toolbox license.
pas de chance!
yurenchu
on 13 Mar 2017
You should add test cases where x or x+1 equals an even fibonacci number (either small or large).
Majid Farzaneh
on 15 Oct 2018
Nice problem. I always have problem with big numbers.
Mehdi BENSEDDIK
on 10 Apr 2019
I solved the problem until x = 4000000 but x = 97455000 it gives me an error of exceeding the memory(overflow) because the number is too large.
Solution Comments
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1 Comment
Ankith Rathod
on 19 Jul 2018
y is it not working when it even works on octave and get results in a sec
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1 Comment
Ankith Rathod
on 19 Jul 2018
While evaluating the solution, the server encountered an error caused by temporary unavailability of MATLAB Service. Wait a few minutes for the MATLAB Service to return, and then rescore.
those test numbers are too big
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1 Comment
Jitesh Kumar
on 15 Oct 2017
it is correct but by server error its not showing
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1 Comment
Harish Maradana
on 13 Mar 2013
test for cheating
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1 Comment
Peter Wittenberg
on 14 Apr 2012
OK, I gamed this one so it only works up to about 1 billion. I wanted to see everyone else's interesting solutions.
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2 Comments
Alfonso Nieto-Castanon
on 3 Feb 2012
this trick (c.f. Solution 26456) works only for these particular test cases, it would fail for example with euler002(6)...
Ankith Rathod
on 19 Jul 2018
While evaluating the solution, the server encountered an error caused by temporary unavailability of MATLAB Service. Wait a few minutes for the MATLAB Service to return, and then rescore.
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