Gapful Number is a number N of at least 3 digits such that it is divisible by the concatenation of it’s first and last digit.
Few Gapful Numbers are:
100, 105, 108, 110, 120, 121, 130, 132, 135, 140,…
Check if N is a Gapful Number
Given an integer N, the task is to check whether N is a Gapful Number or not. If N is a Gapful Number then print “Yes” else print “No”.
Examples:
Input: N = 108
Output: Yes
Explanation:
108 is divisible by 18
Input: N = 112
Output: No
Approach: The idea is to create a number(say num) using the first and last digits of the given numbers and check whether N is divisible by num or not. If N is divisible by num then it is a Gapful Number and print “Yes”, else print “No”.
Below is the implementation of the above approach:
C++
// C++ program for the above approach#include <bits/stdc++.h>using namespace std;// Find the first digitint firstDigit(int n){ // Find total number of digits - 1 int digits = (int)log10(n); // Find first digit n = (int)(n / pow(10, digits)); // Return first digit return n;}// Find the last digitint lastDigit(int n){ // return the last digit return (n % 10);}// A function to check Gapful numbersbool isGapful(int n){ int first_dig = firstDigit(n); int last_dig = lastDigit(n); int concatenation = first_dig * 10 + last_dig; // Return true if n is gapful number return (n % concatenation == 0);}// Driver Codeint main(){ // Given Number int n = 108; // Function Call if (isGapful(n)) cout << "Yes"; else cout << "No"; return 0;} |
Java
// Java program for the above approach class GFG{ // Find the first digit static int firstDigit(int n) { // Find total number of digits - 1 int digits = (int)(Math.log(n) / Math.log(10)); // Find first digit n = (int)(n / Math.pow(10, digits)); // Return first digit return n; } // Find the last digit static int lastDigit(int n) { // Return the last digit return (n % 10); } // A function to check Gapful numbers static boolean isGapful(int n) { int first_dig = firstDigit(n); int last_dig = lastDigit(n); int concatenation = first_dig * 10 + last_dig; // Return true if n is gapful number return (n % concatenation == 0); } // Driver code public static void main(String[] args) { // Given number int n = 108; // Function call if (isGapful(n)) System.out.print("Yes"); else System.out.print("No");} } // This code is contributed by Pratima Pandey |
Python3
# Python3 program for the above approachimport math# Find the first digitdef firstDigit(n): # Find total number of digits - 1 digits = math.log10(n) # Find first digit n = (n / math.pow(10, digits)) # Return first digit return n# Find the last digitdef lastDigit(n): # return the last digit return (n % 10)# A function to check Gapful numbersdef isGapful(n): concatenation = (firstDigit(n) * 10) +\ lastDigit(n) # Return true if n is gapful number return (n % concatenation)# Driver Codeif __name__=='__main__': # Given Number n = 108 # Function Call if (isGapful(n)): print("Yes") else: print("No")# This code is contributed by Ritik Bansal |
C#
// C# program for the above approach using System;class GFG{ // Find the first digit static int firstDigit(int n) { // Find total number of digits - 1 int digits = (int)(Math.Log(n) / Math.Log(10)); // Find first digit n = (int)(n / Math.Pow(10, digits)); // Return first digit return n; } // Find the last digit static int lastDigit(int n) { // Return the last digit return (n % 10); } // A function to check Gapful numbers static bool isGapful(int n) { int first_dig = firstDigit(n); int last_dig = lastDigit(n); int concatenation = first_dig * 10 + last_dig; // Return true if n is gapful number return (n % concatenation == 0); } // Driver code public static void Main() { // Given number int n = 108; // Function call if (isGapful(n)) Console.Write("Yes"); else Console.Write("No");} } // This code is contributed by Code_Mech |
Javascript
<script>// Javascript program for the above approach // Find the first digit function firstDigit( n) { // Find total number of digits - 1 let digits = parseInt( (Math.log(n) / Math.log(10))); // Find first digit n = parseInt( (n / Math.pow(10, digits))); // Return first digit return n; } // Find the last digit function lastDigit( n) { // Return the last digit return (n % 10); } // A function to check Gapful numbers function isGapful( n) { let first_dig = firstDigit(n); let last_dig = lastDigit(n); let concatenation = first_dig * 10 + last_dig; // Return true if n is gapful number return (n % concatenation == 0); } // Driver code // Given number let n = 108; // Function call if (isGapful(n)) document.write("Yes"); else document.write("No");// This code is contributed by aashish1995 </script> |
Output:
Yes
Time Complexity: O(1)
Reference: https://oeis.org/A108343
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