Ok, another problem I have - I want to believe I got the correct number and the answer but it is not accepted.
Anyway, could please someone tell me if the number I have is correct, starting with 11742003035506558... and ending with ...32895630172796932853738956168354898000000
the six zeros at the end tell me that I might have a glitch in my computation...
Big Fib
Re: Big Fib
Can't be a prime and end in zerosrmplpmpl wrote:Ok, another problem I have - I want to believe I got the correct number and the answer but it is not accepted.
Anyway, could please someone tell me if the number I have is correct, starting with 11742003035506558... and ending with ...32895630172796932853738956168354898000000
the six zeros at the end tell me that I might have a glitch in my computation...
It would trivially be divisible by 2 or 5 -
the_impaler
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Re: Big Fib
[quote="MerickOWA"]
Can't be a prime and end in zeros ;) It would trivially be divisible by 2 or 5 ;)[/quote]
How do you know that it "can't be a prime" ? 13 is prime and so is 233 and there are more.
Of course you know because you have found it but off the bat it's hard to say.
That would probably give adum another crazy number crunching idea like "find the prime fib that has last 100 digits matching pi in reverse"
Can't be a prime and end in zeros ;) It would trivially be divisible by 2 or 5 ;)[/quote]
How do you know that it "can't be a prime" ? 13 is prime and so is 233 and there are more.
Of course you know because you have found it but off the bat it's hard to say.
That would probably give adum another crazy number crunching idea like "find the prime fib that has last 100 digits matching pi in reverse"
I have no idea what Merick is talking about... anyway: you're supposed to give every 20,000th digit of the 1.5 millionth Fibonacci number -- and your answer is having a string of zeroes at the end. I would like to say that it's extremely unlikely that you answer is correct.
I'm not a number theorist but I'm guessing that the answer to this challenge pretty much amounts to a random number -- meaning that every digit is roughly equally likely.
There's not that much that can go wrong with this challenge (or the coding involved). Just re-check your code.
I'm not a number theorist but I'm guessing that the answer to this challenge pretty much amounts to a random number -- meaning that every digit is roughly equally likely.
There's not that much that can go wrong with this challenge (or the coding involved). Just re-check your code.
Yupp, I know that 20,000th digit part and I could extract THAT part easily, but since I obviously got the wrong number, the resulting digits were wrong, too. Bah, back to computation.sigi wrote:I have no idea what Merick is talking about... anyway: you're supposed to give every 20,000th digit of the 1.5 millionth Fibonacci number -- and your answer is having a string of zeroes at the end. I would like to say that it's extremely unlikely that you answer is correct.
I'm not a number theorist but I'm guessing that the answer to this challenge pretty much amounts to a random number -- meaning that every digit is roughly equally likely.
There's not that much that can go wrong with this challenge (or the coding involved). Just re-check your code.
Thumbs up to these challenges who make me learning stuff!
Re: Big Fib
It can't be prime if the last digit is a zero. Any number that ends in zero is divisible by 10, and thus NOT prime.the_impaler wrote:How do you know that it "can't be a prime" ? 13 is prime and so is 233 and there are more.
Of course you know because you have found it but off the bat it's hard to say.
That would probably give adum another crazy number crunching idea like "find the prime fib that has last 100 digits matching pi in reverse"
Same way I can trivially prove that every prime (except 2 and 5) must end with either 1, 3, 7 or 9. If it ended in 0,2,4,6,8 it would be even and thus divisible by 2. If it ends in 0 or 5 its divisible by 5.
I have the same value... either its correct, or we're using the same codebase:
Used this one: http://www.bigzaphod.org/fibonacci/
My own perl script was too damn slow and I was to lazy to write C code
Used this one: http://www.bigzaphod.org/fibonacci/
My own perl script was too damn slow and I was to lazy to write C code
hm, I used another tool (Formula One IDE) for computation of the number and a generic tool to extract the digits asked for. Length of "my" number is 67170, so the solution had four digits.eike42 wrote:I have the same value... either its correct, or we're using the same codebase:
Used this one: http://www.bigzaphod.org/fibonacci/
My own perl script was too damn slow and I was to lazy to write C code
I'm kinda stuck here. First I computed what I thought was the right number using bc. Then I noticed it asked for the 1,500,000th member of the Fibonacci sequence instead of fib(1.5m), so I recalculated. Still my solutions don't seem to be correct.
My number starts with "19768781" and ends with "345698". It has 313481 digits.
Did I get it right that you want the first digit, then take a 20,000-step, that digit (20001th) and so on? Or do you want the first and the k*20,0000th digits, with k = 1,2,...
Bye,
hoki
My number starts with "19768781" and ends with "345698". It has 313481 digits.
Did I get it right that you want the first digit, then take a 20,000-step, that digit (20001th) and so on? Or do you want the first and the k*20,0000th digits, with k = 1,2,...
Bye,
hoki