Smallest Divisible Digit Product I

Smallest Divisible Digit Product I - Problem

Math Enumeration Easy

You are given two integers n and t.

Return the smallest number greater than or equal to n such that the product of its digits is divisible by t.

For example, if n = 15 and t = 3, we need to find the smallest number ≥ 15 where the product of digits is divisible by 3. The number 16 has digit product 1×6 = 6, which is divisible by 3, so the answer is 16.

Input & Output

Example 1 — Basic Case

$ Input: n = 15, t = 3

› Output: 16

💡 Note: Starting from 15: digit product = 1×5 = 5, not divisible by 3. Next number 16: digit product = 1×6 = 6, which is divisible by 3, so return 16.

Example 2 — Zero Digit

$ Input: n = 19, t = 7

› Output: 20

💡 Note: Starting from 19: digit product = 1×9 = 9, not divisible by 7. Next number 20: digit product = 2×0 = 0, which is divisible by 7, so return 20.

Example 3 — Answer is n itself

$ Input: n = 12, t = 2

› Output: 12

💡 Note: Starting from 12: digit product = 1×2 = 2, which is divisible by 2, so return 12 itself.

Constraints

1 ≤ n ≤ 10⁶
1 ≤ t ≤ 10⁶

Visualization

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Asked in

G Google 25 M Microsoft 15

The key insight is to iterate from n upwards and check each number's digit product for divisibility by t. The brute force approach works well for this problem since the search space is typically small. Best approach is brute force enumeration. Time: O(k×d), Space: O(1)

Common Approaches

✓ Brute Force Enumeration

⏱️ Time: O(k × d) Space: O(1)

Start from n and check each subsequent number. For each number, calculate the product of its digits and check if it's divisible by t. Return the first number that satisfies the condition.

Early Termination Optimization

⏱️ Time: O(k × d) Space: O(1)

Similar to brute force but we can optimize by recognizing that if any digit is 0, the product is 0, and 0 is divisible by any non-zero t. This gives us early termination in some cases.

Brute Force Enumeration — Algorithm Steps

Step 1: Start with current number = n
Step 2: Calculate product of digits for current number
Step 3: Check if product is divisible by t, if yes return current number
Step 4: Otherwise increment current number and repeat

Visualization

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Step-by-Step Walkthrough

Start at n=15

Begin checking from the given number

Check digit products

15: 1×5=5, not divisible by 3

Find answer

16: 1×6=6, divisible by 3 ✓

Code -

solution.c — C

#include <stdio.h>

int solution(int n, int t) {
    int current = n;
    while (1) {
        // Calculate product of digits
        int product = 1;
        int temp = current;
        while (temp > 0) {
            int digit = temp % 10;
            product *= digit;
            temp /= 10;
        }
        
        // Check if divisible by t
        if (product % t == 0) {
            return current;
        }
        
        current++;
    }
}

int main() {
    int n, t;
    scanf("%d", &n);
    scanf("%d", &t);
    int result = solution(n, t);
    printf("%d\n", result);
    return 0;
}

Time & Space Complexity

Time Complexity

⏱️

O(k × d)

k is the gap between n and answer, d is average number of digits

✓ Linear Growth

Space Complexity

O(1)

Only using constant extra variables

✓ Linear Space

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Input & Output

Constraints

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Related Problems

Common Approaches

Brute Force Enumeration — Algorithm Steps

Visualization

Code -

Time & Space Complexity

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