Smallest N digit number divisible by N

Given a positive integers N, the task is to find the smallest N digit number divisible by N.

Examples:

Input: N = 4
Output: 1000
Explanation:
1000 is the smallest 4-digit number divisible by 4.

Input: N = 5
Output: 10000
Explanation:
10000 is the smallest 5-digit number divisible by 5.

Approach:

• A series will be formed like 1, 10, 102, 1000, 10000, 100002, 1000006, 10000000……… for n = 1, 2, 3…..
• So the Nth term will be = n*\lceil 10^\frac{n-1}{n} \rceil

Below is the implementation of the above approach

// C++ program to find the smallest 
// K digit number divisible by K.
#include <iostream>
#include <cmath>
using namespace std;

// Function to find the smallest 
// K digit number divisible by K.
void smallestDivisibleNumber(int K)
{  
    cout << K * ceil(pow(10, (K - 1)) / K);
}

// Driver code
int main()
{
    int K = 4;  
    smallestDivisibleNumber(K);
    
    return 0;
}

Output:

1000