# Gapful Numbers

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

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

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 ` `using` `namespace` `std; ` ` `  `// Find the first digit ` `int` `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 digit ` `int` `lastDigit(``int` `n) ` `{ ` `    ``// return the last digit ` `    ``return` `(n % 10); ` `} ` ` `  `// A function to check Gapful numbers ` `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 ` `int` `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 approach ` `import` `math ` ` `  `# Find the first digit ` `def` `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 digit ` `def` `lastDigit(n): ` ` `  `    ``# return the last digit ` `    ``return` `(n ``%` `10``) ` ` `  `# A function to check Gapful numbers ` `def` `isGapful(n): ` ` `  `    ``concatenation ``=` `(firstDigit(n) ``*` `10``) ``+``\  ` `                     ``lastDigit(n) ` ` `  `    ``# Return true if n is gapful number ` `    ``return` `(n ``%` `concatenation) ` ` `  `# Driver Code ` `if` `__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 `

Output:

```Yes
```

Time Complexity: O(1)
Reference: https://oeis.org/A108343

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