If only you smarter fellas documented your code. THe world would be perfect. :/

When I tried a solution similar to this, I got a runtime error. Is Character.getNumericvalue(intString.charAt() substantially faster than Integer.parseInt(String.substring())? Or was there perhaps something else that I was missing?

Easier, yes.
Simpler, no: recursion is harder to read, understand and debug.
Faster, barely: log10(n) is only slightly faster than converting to String.
Space efficient, hard no: recursion will alocate more stack space than a string.

Personaly I prefer the solution above more, as it is easier work with(all be at slower).

I cannot see the solution since I have not solved this kata in Java, but the integer part of log10 of a positive integer + 1 is equal to the length of the decimal representation of this number:

Math.floor(Math.log10(n)+1)==n.toString().length

(more or less, sorry if this is not correct, but my Java is not fluent :) ).

can somebody explain why he used that "log10(n)" over there ?
I haven't done any math in like 2 years so I kinda forgot about logs and how they work.
Thanks a lot !

I wrote literally same code without looking at your code xd. Good job!

This can be done much easier without having to convert your int to a string.

good jop but you better get the chars form the string by charArray as .charAt method has a bad complexity :)

Im a little confused as too what your asking. If n is greater than the sum of its digits raised to the (p + i) power, then the final step of (sum % n) will never return 0, causing the method to return -1.

how you check whether n is bigger of sum