Smallest n digit number divisible by given three numbers

Given x, y, z and n, find smallest n digit number which is divisible by x, y and z.

Examples:

Input : x = 2, y = 3, z = 5
        n = 4
Output : 1020

Input : x = 3, y = 5, z = 7
        n = 2
Output : Not possible



1) Find smallest n digit number is pow(10, n-1).
2) Find LCM of given 3 numbers x, y and z.
3) Find remainder of the LCM when divided by pow(10, n-1).
4) Add the “LCM – remainder” to pow(10, n-1). If this addition is still a n digit number, we return the result. Else we return Not possible.

Illustration :
Suppose n = 4 and x, y, z are 2, 3, 5 respectively.
1) First find the least four digit number i.e. 1000,
2) LCM of 2, 3, 5 so the LCM is 30.
3) Find the reminder of 1000 % 30 = 10
4) Subtract the remainder from LCM, 30 – 10 = 20. Result is 1000 + 20 = 1020.

Below is the implementation of above approach:

C++

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// C++ program to find smallest n digit number
// which is divisible by x, y and z.
#include <bits/stdc++.h>
using namespace std;
  
// LCM for x, y, z
int LCM(int x, int y, int z)
{
    int ans = ((x * y) / (__gcd(x, y)));
    return ((z * ans) / (__gcd(ans, z)));
}
  
// returns smallest n digit number divisible
// by x, y and z
int findDivisible(int n, int x, int y, int z)
{
    // find the LCM
    int lcm = LCM(x, y, z);
  
    // find power of 10 for least number
    int ndigitnumber = pow(10, n-1);
      
    // reminder after
    int reminder = ndigitnumber % lcm;
  
    // If smallest number itself divides
    // lcm.
    if (reminder == 0)
         return ndigitnumber;
  
    // add lcm- reminder number for
    // next n digit number
    ndigitnumber += lcm - reminder;
  
    // this condition check the n digit
    // number is possible or not
    // if it is possible it return 
    // the number else return 0
    if (ndigitnumber < pow(10, n))
        return ndigitnumber;
    else
        return 0;
}
  
// driver code
int main()
{
    int n = 4, x = 2, y = 3, z = 5;
    int res = findDivisible(n, x, y, z);
  
    // if number is possible then 
    // it print the number
    if (res != 0)
        cout << res;
    else
        cout << "Not possible";
  
    return 0;
}

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Java














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// Java program to find smallest n digit number 


// which is divisible by x, y and z. 


import java.io.*; 


  


public class GFG { 


  


    static int __gcd(int a, int b) 


    { 


  


        if (b == 0) { 


            return a; 


        } 


        else { 


            return __gcd(b, a % b); 


        } 


    } 


  


    // LCM for x, y, z 


    static int LCM(int x, int y, int z) 


    { 


        int ans = ((x * y) / (__gcd(x, y))); 


        return ((z * ans) / (__gcd(ans, z))); 


    } 


  


    // returns smallest n digit number  


    // divisible by x, y and z 


    static int findDivisible(int n, int x,  


                                  int y, int z) 


    { 


          


        // find the LCM 


        int lcm = LCM(x, y, z); 


  


        // find power of 10 for least number 


        int ndigitnumber = (int)Math.pow(10, n - 1); 


  


        // reminder after 


        int reminder = ndigitnumber % lcm; 


  


        // If smallest number itself divides 


        // lcm. 


        if (reminder == 0) 


            return ndigitnumber; 


  


        // add lcm- reminder number for 


        // next n digit number 


        ndigitnumber += lcm - reminder; 


  


        // this condition check the n digit 


        // number is possible or not 


        // if it is possible it return 


        // the number else return 0 


        if (ndigitnumber < Math.pow(10, n)) 


            return ndigitnumber; 


        else


            return 0; 


    } 


  


    // driver code 


    static public void main(String[] args) 


    { 


  


        int n = 4, x = 2, y = 3, z = 5; 


        int res = findDivisible(n, x, y, z); 


  


        // if number is possible then 


        // it print the number 


        if (res != 0) 


            System.out.println(res); 


        else


            System.out.println("Not possible"); 


    } 


} 


  


// This code is contributed by vt_m. 



















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Python3














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# Python3 code to find smallest n digit  


# number which is divisible by x, y and z. 


from fractions import gcd 


import math 


  


# LCM for x, y, z 


def LCM( x , y , z ): 


    ans = int((x * y) / (gcd(x, y))) 


    return int((z * ans) / (gcd(ans, z))) 


      


# returns smallest n digit number  


# divisible by x, y and z 


def findDivisible (n, x, y, z): 


      


    # find the LCM 


    lcm = LCM(x, y, z) 


      


    # find power of 10 for least number 


    ndigitnumber = math.pow(10, n-1) 


      


    # reminder after 


    reminder = ndigitnumber % lcm 


      


    # If smallest number itself  


    # divides lcm. 


    if reminder == 0: 


        return ndigitnumber 


          


    # add lcm- reminder number for 


    # next n digit number 


    ndigitnumber += lcm - reminder 


      


    # this condition check the n digit 


    # number is possible or not 


    # if it is possible it return 


    # the number else return 0 


    if ndigitnumber < math.pow(10, n): 


        return int(ndigitnumber) 


    else: 


        return 0


  


# driver code 


n = 4


x = 2


y = 3


z = 5


res = findDivisible(n, x, y, z) 


  


# if number is possible then  


# it print the number 


if res != 0: 


    print( res) 


else: 


    print("Not possible") 


      


# This code is contributed by "Sharad_Bhardwaj".  



















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C#














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// C# program to find smallest n digit number 


// which is divisible by x, y and z. 


using System; 


  


public class GFG 


{ 


      


    static int __gcd(int a, int b) 


        { 


          


            if(b == 0)  


            { 


                return a; 


            } 


            else


            { 


                return __gcd(b, a % b); 


            } 


        } 


      


    // LCM for x, y, z 


    static int LCM(int x, int y, int z) 


    { 


        int ans = ((x * y) / (__gcd(x, y))); 


        return ((z * ans) / (__gcd(ans, z))); 


    } 


      


    // returns smallest n digit number divisible 


    // by x, y and z 


    static int findDivisible(int n, int x, int y, int z) 


    { 


        // find the LCM 


        int lcm = LCM(x, y, z); 


      


        // find power of 10 for least number 


        int ndigitnumber =(int)Math. Pow(10, n - 1); 


          


        // reminder after 


        int reminder = ndigitnumber % lcm; 


      


        // If smallest number itself divides 


        // lcm. 


        if (reminder == 0) 


            return ndigitnumber; 


      


        // add lcm- reminder number for 


        // next n digit number 


        ndigitnumber += lcm - reminder; 


      


        // this condition check the n digit 


        // number is possible or not 


        // if it is possible it return  


        // the number else return 0 


        if (ndigitnumber < Math.Pow(10, n)) 


            return ndigitnumber; 


        else


            return 0; 


    } 


      


    // Driver code 


  


    static public void Main () 


    { 


        int n = 4, x = 2, y = 3, z = 5; 


        int res = findDivisible(n, x, y, z); 


      


        // if number is possible then  


        // it print the number 


        if (res != 0) 


            Console.WriteLine(res); 


        else


            Console.WriteLine("Not possible"); 


              


    } 


} 


// This code is contributed by vt_m. 



















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PHP














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<?php 


// php program to find smallest n digit number 


// which is divisible by x, y and z. 


  


// gcd function 


function gcd($a, $b) { 


    return ($a % $b) ? gcd($b, $a % $b) : $b; 


} 


  


// LCM for x, y, z 


function LCM($x, $y, $z) 


{ 


    $ans = floor(($x * $y) / (gcd($x, $y))); 


    return floor(($z * $ans) / (gcd($ans, $z))); 


} 


  


// returns smallest n digit number divisible 


// by x, y and z 


function findDivisible($n, $x, $y, $z) 


{ 


      


    // find the LCM 


    $lcm = LCM($x, $y, $z); 


  


    // find power of 10 for least number 


    $ndigitnumber = pow(10, $n-1); 


      


    // reminder after 


    $reminder = $ndigitnumber % $lcm; 


  


    // If smallest number itself divides 


    // lcm. 


    if ($reminder == 0) 


        return $ndigitnumber; 


  


    // add lcm- reminder number for 


    // next n digit number 


    $ndigitnumber += $lcm - $reminder; 


  


    // this condition check the n digit 


    // number is possible or not 


    // if it is possible it return  


    // the number else return 0 


    if ($ndigitnumber < pow(10, $n)) 


        return $ndigitnumber; 


    else


        return 0; 


} 


  


// driver code 


    $n = 4; 


    $x = 2; 


    $y = 3; 


    $z = 5; 


    $res = findDivisible($n, $x, $y, $z); 


  


    // if number is possible then  


    // it print the number 


    if ($res != 0) 


        echo $res; 


    else


        echo "Not possible"; 


  


// This code is contributed by mits. 


?> 



















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Output:


1020





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