Count of m digit integers that are divisible by an integer n

Given two numbers m and n, count number of m digit numbers that are divisible by n.

Examples:

Input : m = 2 
        n = 6
Output : 15
Two digit numbers that are divisible by 6
are 12, 18, 24, 30, 36, ....., 96.

Input : m = 3 
        n = 5
Output :180



A simple solution is two try all m digit numbers. For every number, check if it is divisible by n. If yes, we increment count.

An efficient solution involves following steps.
The idea is based on the fact that starting from first divisible number, every n-th number is divisible by n.

  1. Find largest m digit number.
  2. Find largest m-1 digit number.
  3. Divide both number by n and subtract later from prior.

Below is the implementation of above steps.

C++

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// C++ program to count m digit numbers having
// n as divisor.
#include<bits/stdc++.h>
using namespace std;
  
// Returns count of m digit numbers having n
// as divisor
int findCount(int m, int n)
{    
    // generating largest number of m digit
    int num1 = 0;
    for (int i = 0; i < m; i++)
        num1 = (num1 * 10) + 9;
  
    // generating largest number of m-1 digit
    int num2 = 0;
    for (int i = 0; i < (m - 1); i++)
        num2 = (num2 * 10) + 9;
  
    // returning number of dividend
    return ((num1 / n) - (num2 / n));
}
  
// Driver code
int main()
{
    int m = 2, n = 6;
    printf("%d\n", findCount(m, n));
    return 0;
}

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Java

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// Java program to count m digit numbers having
// n as divisor.
  
class Main
{
    // Returns count of m digit numbers having n
    // as divisor
    static int findCount(int m, int n)
    {    
        // generating largest number of m digit
        int num1 = 0;
        for (int i = 0; i < m; i++)
            num1 = (num1 * 10) + 9;
       
        // generating largest number of m-1 digit
        int num2 = 0;
        for (int i = 0; i < (m - 1); i++)
            num2 = (num2 * 10) + 9;
       
        // returning number of dividend
        return ((num1 / n) - (num2 / n));
    }
      
    // main function
    public static void main (String[] args) 
    {
        int m = 2, n = 6;
        System.out.println(findCount(m, n));
    }
}
  
/* This code is contributed by Harsh Agarwal */

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Python3

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# Python3 program to count m digit
# numbers having n as divisor.
  
# Returns count of m digit 
# numbers having n as divisor
def findCount(m, n):
  
    # Generating largest number of m digit
    num1 = 0
      
    for i in range(0, m):
        num1 = (num1 * 10) + 9
  
    # Generating largest number of m-1 digit
    num2 = 0
      
    for i in range(0, (m - 1)):
        num2 = (num2 * 10) + 9
  
    # returning number of dividend
    return int((num1 / n) - (num2 / n)) 
  
  
# Driver code
m = 2; n = 6
print(findCount(m, n)) 
  
# This code is contributed by Smitha Dinesh Semwal

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C#

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// C# program to count m digit numbers
// having n as divisor.
using System; 
  
class GfG {
      
    // Returns count of m digit numbers
    // having n as divisor
    static int findCount(int m, int n)
    
          
        // generating largest number 
        // of m digit
        int num1 = 0;
        for (int i = 0; i < m; i++)
            num1 = (num1 * 10) + 9;
      
        // generating largest number 
        // of m-1 digit
        int num2 = 0;
        for (int i = 0; i < (m - 1); i++)
            num2 = (num2 * 10) + 9;
      
        // returning number of dividend
        return ((num1 / n) - (num2 / n));
    }
      
    // main function
    public static void Main () 
    {
        int m = 2, n = 6;
          
        Console.Write(findCount(m, n));
    }
}
  
// This code is contributed by parashar.

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PHP

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<?php
// PHP program to count m digit
// numbers having n as divisor.
  
// Returns count of m digit numbers
// having n as divisor
function findCount($m, $n)
    // generating largest number 
    // of m digit
    $num1 = 0;
    for ($i = 0; $i < $m; $i++)
        $num1 = ($num1 * 10) + 9;
  
    // generating largest number 
    // of m-1 digit
    $num2 = 0;
    for ($i = 0; $i < ($m - 1); $i++)
        $num2 = ($num2 * 10) + 9;
  
    // returning number of dividend
    return (($num1 / $n) - ($num2 / $n));
}
  
// Driver code
$m = 2; $n = 6;
echo findCount($m, $n), "\n";
  
// This code is contributed by ajit
?>

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Output :

 
15

Time complexity : O(m)

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Improved By : parashar, jit_t



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