Problem 684

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Kenya_A
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Problem 684

Post by Kenya_A »

Problem 684 (View Problem)
Hi,
I haven't solved this yet, but I noticed that the example number has the same quotient and remainder when divisible by 9, so it's impossible to tell using it alone if you have made a mistake in entering one of these into the solution.

I can't solve it because, try I might, I haven't found enough tricks to calculating the numbers in a reasonable time and without overflow. I have done modulo operations after each addition and multiplication, tried (and abandoned in favour of built-ins) a method of modular exponentiation which cancels out much of the multiplying by setting the relevant exponents mod x to 1, and found closed form expressions for the sums. The limit is the exponentiations. What, besides maybe the F's thm, can speed this up? It may be finished yet by this time next year at this rate.
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8-) 835410_BD2wXg2iurN8FEuwdp659qnINUNqNKbA :lol:

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hk
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Re: Problem 684

Post by hk »

In the big pink box at the top of the page it reads:

'Don't ask for hints how to solve a problem'
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nicolas.patrois
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Re: Problem 684

Post by nicolas.patrois »

Hi,
Is sum from i=2 to i=30 S(fi) (mod 1_000_000_007) equal to xxxxxxxxx? <value snipped by moderator>
Thank you.
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RudiGj
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Re: Problem 684

Post by RudiGj »

Just need a little bit of clarification on this problem.

Do you have to find the smallest digit sum for each term of the fibonacci sequence from the 2nd to the 90th?

Therefore the answer would be:

digitsum(1) + digitsum(2) + digitsum(3) + digitsum(5) + digitsum(8) + .... + digitsum(n) until the 90th term?

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RobertStanforth
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Re: Problem 684

Post by RobertStanforth »

RudiGj wrote:
Sat Jan 04, 2020 12:04 am
Do you have to find the smallest digit sum for each term of the fibonacci sequence from the 2nd to the 90th?
No, that is not correct. $S(n)$ is not digitsum.
Instead, $s(n)$ is the inverse digit sum, so $\mathrm{digitsum}(s(n)) = n$.
Note also that $S$ is the sum of $s$.

So, (given a Fibonacci number $f_i$), you have to find the smallest number whose digit sum is $n$, for each number $n$ up to $f_i$. Their sum gives you $S(f_i)$.
Then the answer would be: $S(1) + S(2) + S(3) + S(5) + S(8) + ... + S(f_{90})$.

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