Given an integer N, the task is to find the smallest N digit number divisible by all possible prime digits, i.e, 2, 3, 5 and 7. Print -1 if no such number is possible.
Examples:
Input: N = 5
Output: 10080
Explanation: 10080 is the smallest five-digit number that is divisible by 2, 3, 5 and 7.
Input: N = 3
Output: 210
Approach:
Follow the steps given below to solve the problem:
- Since all the four numbers 2, 3, 5, 7 are prime it means N will also be divisible by their product 2 × 3 × 5 × 7 = 210
- For N < 3, no such number exists. So, print -1.
- For N = 3, the answer will be 210.
- For N > 3, the following computation needs to be done:
- Find Remainder R = 10N-1 % N.
- Add 210 – R to 10N-1.
Below is the implementation of the above approach:
C++
// C++ implementation of the above approach #include <bits/stdc++.h>using namespace std;// Function to find the minimum number of // n digits divisible by all prime digits void minNum(int n){ if (n < 3) cout << -1; else cout << (210 * ((int)(pow(10, n - 1) / 210) + 1));}// Driver Codeint main(){ int n = 5; minNum(n); return 0;}// This code is contributed by amal kumar choubey |
Java
// Java implementation of the above approachclass GFG{ // Function to find the minimum number of // n digits divisible by all prime digitsstatic void minNum(int n){ if(n < 3) System.out.println(-1); else System.out.println(210 * ( (int)(Math.pow(10, n - 1) / 210) + 1));}// Driver codepublic static void main(String[] args) { int n = 5; minNum(n);}}// This code is contributed by Stuti Pathak |
Python3
# Python3 implementation of the above approach from math import * # Function to find the minimum number of # n digits divisible by all prime digits.def minNum(n): if n < 3: print(-1) else: print(210 * (10**(n-1) // 210 + 1)) # Driver Code n = 5minNum(n) |
C#
// C# implementation of the above approachusing System;class GFG{// Function to find the minimum number of// n digits divisible by all prime digitsstatic void minNum(int n){ if (n < 3) Console.WriteLine(-1); else Console.WriteLine(210 * ((int)(Math.Pow(10, n - 1) / 210) + 1));}// Driver codepublic static void Main(String[] args){ int n = 5; minNum(n);}}// This code is contributed by amal kumar choubey |
Javascript
<script>// Javascript implementation of the above approach // Function to find the minimum number of // n digits divisible by all prime digits function minNum(n){ if (n < 3) document.write( -1); else document.write((210 * (parseInt(Math.pow(10, n - 1) / 210) + 1)));}// Driver Codevar n = 5;minNum(n);// This code is contributed by rrrtnx.</script> |
Output:
10080
Time complexity: O(logN)
Auxiliary Space: O(1)
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