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S1Q1 · Three-Digit Number with Digit-Sum Divisible by k

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Function: is_three_digit_and_digit_sum_divisible_by_k(n: int, k: int) -> bool

Core idea: check the range, sum the digits, check divisibility.

def is_three_digit_and_digit_sum_divisible_by_k(n: int, k: int) -> bool:
    if not (100 <= n <= 999):
        return False
    digit_sum = sum(int(d) for d in str(n))
    return digit_sum % k == 0

Key rules: - Three-digit: 100 <= n <= 999 - Digit sum: sum all digits of n - Divisible by k: digit_sum % k == 0 - Both conditions must hold

Problem Statement

Problem

Write a function is_three_digit_and_digit_sum_divisible_by_k(n, k) that returns True only if n is a three-digit number AND the sum of its digits is divisible by k.

Examples:

Input 
n=145, k=5
Output 
True
Input 
n=123, k=5
Output 
False
Input 
n=1040, k=5
Output 
False

Tracing all examples

n Three-digit? Digit sum % k Result
145, k=5 ✅ 100–999 1+4+5=10 10%5=0 ✅ True
123, k=5 ✅ 1+2+3=6 6%5=1 ❌ False
450, k=10 ✅ 4+5+0=9 9%10=9 ❌ False
999, k=9 ✅ 9+9+9=27 27%9=0 ✅ True
1040, k=5 ❌ (4 digits) - - False

Solution approaches

def is_three_digit_and_digit_sum_divisible_by_k(n: int, k: int) -> bool:
    if not (100 <= n <= 999):
        return False
    return sum(int(d) for d in str(n)) % k == 0

def is_three_digit_and_digit_sum_divisible_by_k(n: int, k: int) -> bool:
    if not (100 <= n <= 999):
        return False
    hundreds = n // 100
    tens     = (n // 10) % 10
    units    = n % 10
    digit_sum = hundreds + tens + units
    return digit_sum % k == 0

Extracts each digit using integer arithmetic - no string conversion needed.

def is_three_digit_and_digit_sum_divisible_by_k(n: int, k: int) -> bool:
    if not (100 <= n <= 999):
        return False
    total, temp = 0, n
    while temp:
        temp, digit = divmod(temp, 10)
        total += digit
    return total % k == 0

is_three_digit_and_digit_sum_divisible_by_k = lambda n, k: (
    100 <= n <= 999 and sum(int(d) for d in str(n)) % k == 0
)

Key takeaways

01

Check range before digit sum

100 <= n <= 999 is a cheap O(1) check. Return False early so the digit-sum computation only runs for valid three-digit numbers - short-circuit logic at its simplest.

02

sum(int(d) for d in str(n)) - clean digit sum

Converting to a string gives access to each digit character; int(d) converts back to an integer for summing. Works for any number of digits without manual division.

03

Arithmetic extraction for three digits: // and %

For exactly three digits: n//100 = hundreds, (n//10)%10 = tens, n%10 = units. Direct and fast - no string conversion needed when you know the exact digit count.