Peel the last digit off
Think of a number as a stack of digit cards. Divide 4729 by 10 and the remainder is 9, which is exactly the last digit. Whole-number division by 10 gives 472, which is the same number with that last digit torn off.
Two small moves give you everything: read the last digit, then drop it.
Remainder after dividing by 10: is the last digit.
Whole-number division by 10: drops the last digit.
4729 % 10 # 9 4729 // 10 # 472
Digits of 4729, right to left
4
07
12
29
3digit
4729 % 10 = 9. total = 9, n becomes 472.
Move 1 of 4
Repeat until nothing is left
Do it again and again. Read the last digit, add it to a running total, then drop it. When the number reaches 0 there are no digits left, so you have met every one of them, from right to left.
What if the number is 0 to begin with? The loop never runs and the total stays 0, which is the right answer for a sum.
In Java and C++: dividing two whole numbers already rounds down for positive values.
total = 0
while n > 0:
total += n % 10
n //= 10
return total