Numbers and Math Basics

A huge share of number problems reduce to one loop that peels digits off a number, right to left. Two operations do all the work, both from earlier modules:

• n % 10 gives the last digit.
• n / 10 (integer division) drops the last digit.

Repeat until n becomes 0 and you have visited every digit:

def solve(n):
    count = 0
    while n > 0:
        digit = n % 10     # grab last digit
        count += 1         # ... do something with it
        n = n // 10        # drop last digit
    return count           # here: how many digits n has

long long solve(long long n) {
    long long count = 0;
    while (n > 0) { count++; n /= 10; }
    return count;
}

This digit-peeling loop is the running-total pattern applied to digits: keep an accumulator, update it with each digit, shrink n each pass. Counting digits, summing digits, finding the largest digit — all are the same loop with a different update. For inputs that might be 0 or negative, handle those as small special cases (0 has one digit; take the absolute value first if negatives are allowed).