Man, thx for showing me the other way to solve this problem. The way you use enumerate is genius!

I agree. Replace "/" with "//" and this solution is pretty close to perfect.

Nice

SHIT! So smart! I was wondering how the p+i didn't make it start from 0. Then I remembered the first one will make p default to 1 since p^i is n^0. I am beginning to understand enumerate() more and more

thanks

aw I was so over thinking this

type casting should have been done in the return statement, as it may not result any remainder, but the result of the division d always be a float. # Just a precaution

I agree with both, but for people who are still learning python as they go, a little naming readability on the solution goes a long way.

i and c are quite straightforward though?
i = index, which has always been a convention
c = character, which is quite obvious when you look at str(n)

It isn't as big of a deal as you make it out to be, this is about problem solving, not about following pep8 or any naming conventions.
If it was a software then I would 100% agree with you.

This is 1:1 with mine but this is not as readable. I think we should stop giving "Best Practices" to people that use one letter variable names like n, p, s, i... instead of number, power, sum ("result" would be better), index...

I flashed with this brief resolution! Tks!!

Enumerate also automatically iterates through each of the digits, so you don't need to cast the number to a string, and loop through each letter (like I did lol).

I did:

str_number = str(number)
digit_list = []
for digit in range(0, len(str_number)):
digit_list.append(int(str_number[digit]))

and then enumerated through that digit_list, but the "enumerate(str(n))" in

for i, c in enumerate(str(n)):

gives you access to each digit in the number as a string in a much more elegant way.

You can also start enumerating from any int you want
in this case it would be
enumerate(str(n),p)
I recommend always to check the documentation whenever you see something new

I'm new too starts about a week ago, for what I understand in enumerate the first i is index starting from 0. Each time it iterates the i will +1. So by using enumerate we have a default i that starts at 0 and +1 everytime. Since the i comes with the enumerate function itself thus we don't need to create another variable to increase p.

so s += pow(int(c),p+i), p+i = 0 at the first time which is p itself, then p+1, p+2 ....