Problem Description

You need to find the smallest number that meets two conditions:

1. It must be greater than or equal to a given number n
2. The product of its digits must be divisible by a given number t

For example, if n = 10 and t = 2, you would check numbers starting from 10:

• For 10: digit product = 1 × 0 = 0, and 0 is divisible by 2, so the answer is 10
• If n = 15 and t = 3, you would check:
  • 15: digit product = 1 × 5 = 5, not divisible by 3
  • 16: digit product = 1 × 6 = 6, divisible by 3, so the answer is 16

The solution works by checking each number starting from n and calculating the product of its digits. For each number, it extracts digits one by one using modulo 10 operation (x % 10) and integer division (x // 10), multiplying them together. Once it finds a number whose digit product is divisible by t (checked using p % t == 0), it returns that number.

The key insight is that you don't need to check too many numbers because within any consecutive 10 numbers, at least one will contain the digit 0, making its digit product 0, which is divisible by any non-zero t.

Intuition

The key observation is that we're looking for a number whose digit product is divisible by t. Since we want the smallest such number that's at least n, the natural approach is to start from n and check each consecutive number.

Why can we use brute force enumeration? The critical insight is that we won't need to check many numbers. Consider any sequence of 10 consecutive integers - at least one of them will contain the digit 0. For example, in the range [20, 29], the number 20 has a 0. When any number contains a 0 digit, the product of all its digits becomes 0, and 0 is divisible by any non-zero integer t.

This means in the worst case, we only need to check at most 10 numbers before finding one that works. Even if we start at a number like 11 where the digit product is 1 × 1 = 1 (not divisible by most values of t), by the time we reach 20, we'll have a digit product of 2 × 0 = 0 which satisfies our condition.

Therefore, instead of trying to construct the answer digit by digit or using complex mathematical formulas, we can simply iterate through numbers starting from n, calculate each number's digit product, and return the first one that's divisible by t. The efficiency is guaranteed by the fact that zeros appear frequently in decimal representations.

Learn more about Math patterns.

Solution Approach

The solution uses a straightforward enumeration approach. We iterate through integers starting from n and check each one until we find a number whose digit product is divisible by t.

The implementation follows these steps:

1. Start enumeration from n: Use count(n) to generate consecutive integers starting from n. This creates an infinite iterator that yields n, n+1, n+2, ...

2. Calculate digit product for each number: For each number i in the sequence:

   • Initialize a product variable p = 1
   • Create a copy of the current number x = i
   • Extract digits using a while loop:
     • Get the last digit using x % 10
     • Multiply it with the running product: p *= x % 10
     • Remove the last digit using integer division: x //= 10
     • Continue until all digits are processed (when x becomes 0)

3. Check divisibility: After calculating the digit product p, check if p % t == 0. If true, we've found our answer and return i.

4. Return the first valid number: The loop continues until it finds the first number whose digit product is divisible by t, which is guaranteed to happen within at most 10 iterations due to the presence of zeros in decimal representations.

For example, with n = 23 and t = 6:

• Check 23: digits are 2 and 3, product = 2 × 3 = 6, 6 % 6 = 0 ✓ Return 23
• If it wasn't divisible, we'd check 24, 25, ..., until finding a valid number

The time complexity is O(d) per number checked, where d is the number of digits, and we check at most 10 numbers, making it very efficient.

Example Walkthrough

Let's walk through finding the smallest number ≥ n = 234 with digit product divisible by t = 7.

Step 1: Check n = 234

• Extract digits: 234 → 4 (234 % 10), then 3 (23 % 10), then 2 (2 % 10)
• Calculate product: 2 × 3 × 4 = 24
• Check divisibility: 24 % 7 = 3 (not divisible by 7)
• Continue to next number

Step 2: Check 235

• Extract digits: 235 → 5, 3, 2
• Calculate product: 2 × 3 × 5 = 30
• Check divisibility: 30 % 7 = 2 (not divisible by 7)
• Continue to next number

Step 3: Check 236

• Extract digits: 236 → 6, 3, 2
• Calculate product: 2 × 3 × 6 = 36
• Check divisibility: 36 % 7 = 1 (not divisible by 7)
• Continue to next number

Step 4: Check 237

• Extract digits: 237 → 7, 3, 2
• Calculate product: 2 × 3 × 7 = 42
• Check divisibility: 42 % 7 = 0 ✓ (divisible by 7!)
• Return 237

The algorithm found the answer after checking only 4 numbers. Note that if we hadn't found a match yet, by the time we reached 240, its digit product would be 2 × 4 × 0 = 0, which is divisible by any non-zero t, guaranteeing we find an answer within 10 iterations.

Solution Implementation

Time and Space Complexity

The time complexity is O(t * log n) where n is the starting number and t is the target divisor. The algorithm iterates through numbers starting from n until it finds one whose digit product is divisible by t. In the worst case, we need to check up to t numbers (since at most every t-th number could have a digit product divisible by t). For each number checked, we calculate the product of its digits, which takes O(log n) time (the number of digits in the number). Therefore, the overall time complexity is O(t * log n).

The space complexity is O(1) as we only use a constant amount of extra space for variables i, p, and x, regardless of the input size.

Note: The reference answer states O(log n) for time complexity, which appears to consider only the digit product calculation for a single number, not accounting for the iteration through potentially multiple numbers until finding one that satisfies the condition.

Learn more about how to find time and space complexity quickly.

Common Pitfalls

1. Not Handling Zero Digits Properly

The most critical pitfall is forgetting that when a number contains the digit 0, the entire product becomes 0. This is actually beneficial since 0 is divisible by any non-zero number, but failing to recognize this can lead to unnecessary complexity or incorrect edge case handling.

Problem Example:

# Incorrect approach that might skip numbers with 0
def incorrect_solution(n, t):
    for candidate in count(n):
        if '0' in str(candidate):  # Wrong: skipping zeros
            continue
        # ... calculate product

Solution: Embrace zeros as valid solutions. When the digit product is 0, it's divisible by any non-zero t.

2. Integer Overflow in Large Digit Products

For numbers with many large digits, the product can grow extremely large, potentially causing overflow issues in languages with fixed integer sizes (though Python handles arbitrary precision).

Problem Example:

# For n = 9999999 and t = 2
# Digit product = 9^7 = 4,782,969 (very large)

Solution: Consider early termination when a 0 is encountered, or use modular arithmetic:

def optimized_solution(n, t):
    for candidate in count(n):
        digit_product = 1
        temp_num = candidate
      
        while temp_num > 0:
            digit = temp_num % 10
            if digit == 0:  # Early termination
                digit_product = 0
                break
            digit_product = (digit_product * digit) % t  # Keep product modulo t
            temp_num //= 10
      
        if digit_product % t == 0:
            return candidate

3. Inefficient String Conversion Approach

Converting numbers to strings to extract digits is less efficient than using arithmetic operations.

Problem Example:

# Inefficient approach
def string_based_solution(n, t):
    for candidate in count(n):
        digits = [int(d) for d in str(candidate)]  # String conversion overhead
        product = 1
        for digit in digits:
            product *= digit

Solution: Stick with the arithmetic approach using modulo and integer division as shown in the original solution.

4. Not Considering t = 1 Edge Case

When t = 1, every number's digit product is divisible by 1, so the answer is always n itself. While the general solution handles this correctly, recognizing this case can lead to immediate optimization.

Solution: Add a special case check:

def optimized_solution(n, t):
    if t == 1:
        return n  # Every digit product is divisible by 1
  
    # ... rest of the solution

5. Misunderstanding the Problem Requirements

Some might mistakenly think they need to find a number whose digits sum (not product) is divisible by t, or that the number itself (not its digit product) should be divisible by t.

Solution: Carefully read the problem - we need the PRODUCT of digits to be divisible by t, not the sum or the number itself.