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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.

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: $S(1) + S(2) + S(3) + S(5) + S(8) + ... + S(f_{90})$.

hk wrote: ↑Mon Apr 13, 2020 1:13 pm
Please have a look at the fastest solvers table....

True, there are some quick solutions but they all revolve around figuring out the mathematics. Once done the coding is simple. How do you balance mathematical know-how against coding ability? Problem 243 was similar - it was easy if you got your head around the maths. (I was much more familiar with the mathematics required for that problem!)

I still wouldn't put this problem in the same category with problems 53 and 55 for example!

I have what I think is a correct algorithm for this problem but my answer is not being accepted.

RobertStanforth wrote: ↑Sat Jan 04, 2020 9:04 am
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: $S(1) + S(2) + S(3) + S(5) + S(8) + ... + S(f_{90})$.

Is this right? From the problem description the answer should be given by
$S(f_2) + S(f_3) + ... + S(f_{90})$

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: $S(1) + S(2) + S(3) + S(5) + S(8) + ... + S(f_{90})$.

codingclubwaseesucks wrote: ↑Sat Jul 04, 2020 6:51 pm
Could you clarify what you mean about "Instead, s(n) is the inverse digit sum, so digitsum(s(n))=n"?
What is an Inverse Digit Sum?

Let digitsum(x) be the sum of the digits of the number x. For example, digitsum(19) = 10. An inverse function would reverse that. If the sum of the digits is 10, what is the number. Without the restriction, "the smallest number," there would be lots answers With the restriction, we can say 19 is the smallest number with a digit sum of 10. That is what is meant by an Inverse Digit Sum.

In the problem, n is the digit sum, and s(n) is the smallest number whose digits sum to n. digitsum(19) = 10 and s(10) = 19.