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 numbersint 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 zint 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 codeint 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 numbersstatic 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 zstatic 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 codepublic 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 numbersdef 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 zdef 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 coden = 2; x = 3; y = 4; z = 6;res = int(findDivisible(n, x, y, z));# if the number is foundif (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 numbersstatic 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 zstatic 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 codestatic 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 - $a); } // Function to return the LCM // of three numbersfunction LCM($x, $y, $z){$ans = (($x * $y) / (gcd($x, $y)));return (($z * $ans) / (gcd($ans, $z)));}// Function to return the largest n-digit// number which is divisible by x, y and zfunction findDivisible($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, $z);// if the number is foundif ($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 numbersfunction 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 zfunction 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 codevar n = 2, x = 3, y = 4, z = 6;var res = findDivisible(n, x, y, z);// if the number is foundif (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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