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Largest N digit number divisible by given three numbers

Last Updated : 21 Mar, 2023
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Given four integers x, y, z, and n, the task is to find the largest n digit number which is divisible by x, y, and z

Examples:

Input: x = 2, y = 3, z = 5, n = 4 
Output: 9990 
9990 is the largest 4-digit number which is divisible by 2, 3 and 5.
Input: x = 3, y = 23, z = 6, n = 2 
Output: Not possible 

Approach:  

  • Find the largest n digit number i.e. pow(10, n) - 1 and store it in a variable largestN.
  • Find LCM of the given three numbers x, y and z say LCM.
  • Calculate the remainder when largestN is divided by LCM i.e. largestN % LCM and store it in a variable remainder.
  • Subtract remainder from largestN. If the result is still an n digit number then print the result.
  • Else print Not possible.


Below is the implementation of the above approach:
 

C++
// C++ program to find largest n digit number
// which is divisible by x, y and z.
#include <bits/stdc++.h>
using namespace std;

// Function to return the LCM of three numbers
int LCM(int x, int y, int z)
{
    int ans = ((x * y) / (__gcd(x, y)));
    return ((z * ans) / (__gcd(ans, z)));
}

// Function to return the largest n-digit
// number which is 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 largest n-digit number
    int largestNDigitNum = pow(10, n) - 1;

    int remainder = largestNDigitNum % lcm;

    // If largest number is the answer
    if (remainder == 0)
        return largestNDigitNum ;

    // find closest smaller number
    // divisible by LCM
    largestNDigitNum -= remainder;

    // if result is an n-digit number
    if (largestNDigitNum >= pow(10, n - 1))
        return largestNDigitNum;
    else
        return 0;
}

// Driver code
int main()
{
    int n = 2, x = 3, y = 4, z = 6;
    int res = findDivisible(n, x, y, z);

    // if the number is found
    if (res != 0)
        cout << res;
    else
        cout << "Not possible";

    return 0;
}
Java
// Java program to find largest n digit number
// which is divisible by x, y and z.
import java.math.*;
class GFG {
    
// Recursive function to return gcd of a and b 
    static int gcd(int a, int b) 
    { 
        // Everything divides 0  
        if (a == 0) 
          return b; 
        if (b == 0) 
          return a; 
       
        // base case 
        if (a == b) 
            return a; 
       
        // a is greater 
        if (a > b) 
            return gcd(a-b, b); 
        return gcd(a, b-a); 
    } 
    
// Function to return the LCM of three numbers
static int LCM(int x, int y, int z)
{
    int ans = ((x * y) / (gcd(x, y)));
    return ((z * ans) / (gcd(ans, z)));
}

// Function to return the largest n-digit
// number which is 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 largest n-digit number
    int largestNDigitNum = (int)Math.pow(10, n) - 1;

    int remainder = largestNDigitNum % lcm;

    // If largest number is the answer
    if (remainder == 0)
        return largestNDigitNum ;

    // find closest smaller number
    // divisible by LCM
    largestNDigitNum -= remainder;

    // if result is an n-digit number
    if (largestNDigitNum >= (int)Math.pow(10, n - 1))
        return largestNDigitNum;
    else
        return 0;
}

// Driver code
public static void main(String args[])
{
    int n = 2, x = 3, y = 4, z = 6;
    int res = findDivisible(n, x, y, z);

    // if the number is found
    if (res != 0)
        System.out.println(res);
    else
        System.out.println("Not possible");

}
}
Python3
# Python3 program to find largest n digit
# number which is divisible by x, y and z.

# Recursive function to return 
# gcd of a and b 
def gcd(a, b):

    # Everything divides 0 
    if (a == 0): 
        return b; 
    if (b == 0): 
        return a; 
    
    # base case 
    if (a == b): 
        return a; 
    
    # a is greater 
    if (a > b): 
        return gcd(a - b, b); 
    return gcd(a, b - a); 

# Function to return the LCM 
# of three numbers
def LCM(x, y, z):
    ans = ((x * y) / (gcd(x, y)));
    return ((z * ans) / (gcd(ans, z)));

# Function to return the largest n-digit
# number which is divisible by x, y and z
def findDivisible(n, x, y, z):
    
    # find the LCM
    lcm = LCM(x, y, z);
    
    # find largest n-digit number
    largestNDigitNum = int(pow(10, n)) - 1;
    
    remainder = largestNDigitNum % lcm;
    
    # If largest number is the answer
    if (remainder == 0):
        return largestNDigitNum ;
    
    # find closest smaller number
    # divisible by LCM
    largestNDigitNum -= remainder;
    
    # if result is an n-digit number
    if (largestNDigitNum >= int(pow(10, n - 1))):
        return largestNDigitNum;
    else:
        return 0;

# Driver code
n = 2; x = 3; 
y = 4; z = 6;
res = int(findDivisible(n, x, y, z));

# if the number is found
if (res != 0):
    print(res);
else:
    print("Not possible");

# This code is contributed 
# by mits
C#
// C# program to find largest n 
// digit number which is divisible 
// by x, y and z.
using System;

class GFG 
{
// Recursive function to return
// gcd of a and b 
static int gcd(int a, int b) 
{ 
    // Everything divides 0 
    if (a == 0) 
        return b; 
    if (b == 0) 
        return a; 
    
    // base case 
    if (a == b) 
        return a; 
    
    // a is greater 
    if (a > b) 
        return gcd(a - b, b); 
    return gcd(a, b - a); 
} 

// Function to return the 
// LCM of three numbers
static int LCM(int x, int y, int z)
{
    int ans = ((x * y) / (gcd(x, y)));
    return ((z * ans) / (gcd(ans, z)));
}

// Function to return the largest 
// n-digit number which is 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 largest n-digit number
    int largestNDigitNum = (int)Math.Pow(10, n) - 1;

    int remainder = largestNDigitNum % lcm;

    // If largest number is the answer
    if (remainder == 0)
        return largestNDigitNum ;

    // find closest smaller number
    // divisible by LCM
    largestNDigitNum -= remainder;

    // if result is an n-digit number
    if (largestNDigitNum >= (int)Math.Pow(10, n - 1))
        return largestNDigitNum;
    else
        return 0;
}

// Driver code
static void Main()
{
    int n = 2, x = 3, y = 4, z = 6;
    int res = findDivisible(n, x, y, z);

    // if the number is found
    if (res != 0)
        Console.WriteLine(res);
    else
        Console.WriteLine("Not possible");
}
}

// This code is contributed by ANKITRAI1
PHP
<?php
// PHP program to find largest n digit number
// which is divisible by x, y and z.

// Recursive function to return gcd of a and b 
function gcd($a, $b) 
{ 
    // Everything divides 0 
    if ($a == 0) 
        return $b; 
    if ($b == 0) 
        return $a; 
    
    // base case 
    if ($a == $b) 
        return $a; 
    
    // a is greater 
    if ($a > $b) 
        return gcd($a - $b, $b); 
    return gcd($a, $b - $a); 
} 

// Function to return the LCM 
// of three numbers
function LCM($x, $y, $z)
{
$ans = (($x * $y) / (gcd($x, $y)));
return (($z * $ans) / (gcd($ans, $z)));
}

// Function to return the largest n-digit
// number which is divisible by x, y and z
function findDivisible($n, $x, $y, $z)
{
    
    // find the LCM
    $lcm = LCM($x, $y, $z);
    
    // find largest n-digit number
    $largestNDigitNum = (int)pow(10, $n) - 1;
    
    $remainder = $largestNDigitNum % $lcm;
    
    // If largest number is the answer
    if ($remainder == 0)
        return $largestNDigitNum ;
    
    // find closest smaller number
    // divisible by LCM
    $largestNDigitNum -= $remainder;
    
    // if result is an n-digit number
    if ($largestNDigitNum >= (int)pow(10, $n - 1))
        return $largestNDigitNum;
    else
        return 0;
}

// Driver code
$n = 2; $x = 3; $y = 4; $z = 6;
$res = findDivisible($n, $x, $y, $z);

// if the number is found
if ($res != 0)
    echo $res;
else
    echo "Not possible";

// This code is contributed 
// by Akanksha Rai
JavaScript
<script>

// Javascript program to find largest n 
// digit number which is divisible 
// by x, y and z.

// Recursive function to return
// gcd of a and b 
function gcd(a, b) 
{ 
    // Everything divides 0 
    if (a == 0) 
        return b; 
    if (b == 0) 
        return a; 
    
    // base case 
    if (a == b) 
        return a; 
    
    // a is greater 
    if (a > b) 
        return gcd(a - b, b); 
    return gcd(a, b - a); 
} 

// Function to return the 
// LCM of three numbers
function LCM(x, y, z)
{
    var ans = parseInt((x * y) / (gcd(x, y)));
    return parseInt((z * ans) / (gcd(ans, z)));
}

// Function to return the largest 
// n-digit number which is divisible
// by x, y and z
function findDivisible(n, x, y, z)
{

    // find the LCM
    var lcm = LCM(x, y, z);

    // find largest n-digit number
    var largestNDigitNum = Math.pow(10, n) - 1;

    var remainder = largestNDigitNum % lcm;

    // If largest number is the answer
    if (remainder == 0)
        return largestNDigitNum ;

    // find closest smaller number
    // divisible by LCM
    largestNDigitNum -= remainder;

    // if result is an n-digit number
    if (largestNDigitNum >= Math.pow(10, n - 1))
        return largestNDigitNum;
    else
        return 0;
}

// Driver code
var n = 2, x = 3, y = 4, z = 6;
var res = findDivisible(n, x, y, z);
// if the number is found
if (res != 0)
    document.write(res);
else
    document.write("Not possible");

</script>

Output: 
96

 

Time Complexity: O(log(min(x, y,z ))) + O(log(n)) as we are doing lcm of x,y,z we need log(min(x,y,z)) time complexity for that + log(n) for doing pow(10,n-1) so overall time complexity will be O(log(min(x, y,z ))) + O(log(n))

Auxiliary Space: O(log(min(x, y, z))) + O(log(n)) as we are doing lcm of x,y,z this lcm will be done in recursively manner so recursion need extra O(log(min(x, y, z))) auxiliary stack space, and addition for doing pow(10,n-1) which is also in recursive manner which also need log(n) extra auxiliary stack space


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