Program to check whether the given number is Buzz Number or not

A number is said to be Buzz Number if it ends with 7 OR is divisible by 7.
The task is to check whether the given number is buzz number or not.

Examples:

Input : 63
Output : Buzz Number
Explanation: 63 is divisible by 7, one
of the condition is satisfied.

Input : 72
Output : Not a Buzz Number
Explanation: 72 % 7 != 0, 72 is neither 
divisible by 7 nor it ends with 7 so 
it is not a Buzz Number.

C++

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// C++ program to check whether the
// given number is Buzz Number or not.
#include <cmath>
#include <iostream>
using namespace std;
  
// function to check BUzz number.
bool isBuzz(int num)
{
    // checking if the number
    // ends with 7 and is divisible by 7
    return (num % 10 == 7 || num % 7 == 0);
}
  
// Driver method
int main(void)
{
    int i = 67, j = 19;
    if (isBuzz(i))
        cout << "Buzz Number\n";
    else
        cout << "Not a Buzz Number\n";
    if (isBuzz(j))
        cout << "Buzz Number\n";
    else
        cout << "Not a Buzz Number\n";
}

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Java

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// Java program to check whether the
// given number is Buzz number or not
import java.io.*;
import java.util.*;
class GFG {
  
    // function to check BUzz number.
    static boolean isBuzz(int num)
    {
        // checking if the number
        // ends with 7 and is divisible by 7
        return (num % 10 == 7 || num % 7 == 0);
    }
  
    // Driver method
    public static void main(String args[])
    {
        int i = 67, j = 19;
        if (isBuzz(i))
            System.out.println("Buzz Number");
        else
            System.out.println("Not a Buzz Number");
        if (isBuzz(j))
            System.out.println("Buzz Number");
        else
            System.out.println("Not a Buzz Number");
    }
}
// This code is contributed by Nikita Tiwari.

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Python

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# Python program to check whether the
# given number is Buzz Number or not.
  
# function to check BUzz number.
def isBuzz(num) :
      
    # checking if the number 
    # ends with 7 and is divisible by 7 
    return (num % 10 == 7 or num % 7 == 0
  
# Driver method
i = 67
j = 19
if (isBuzz(i)) :
    print "Buzz Number"
else :
    print "Not a Buzz Number"
if (isBuzz(j)) :
    print "Buzz Number"
else :
    print "Not a Buzz Number"    

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PHP

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<?php
// PHP code to check Buzz Number
  
// Function for Buzz Number
function isBuzz($num)
{
    // checking if the number
    // ends with 7 and is divisible by 7
    return($num % 10 == 7 || $num % 7 == 0);    
}
  
// Driver Code
$i = 67;
$j = 43;
  
if(isBuzz($i))
print("Buzz Number\n");
else
print("Not Buzz Number\n");
      
if(isBuzz($j))
print("Buzz Number\n");
else
print("Not Buzz Number\n");
  
// This code is contributed by akash7981
?>

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C#

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// C# program to check whether the
// given number is Buzz number or not
using System;
  
class GFG {
  
    // function to check BUzz number.
    static bool isBuzz(int num)
    {
          
        // checking if the number
        // ends with 7 and is 
        // divisible by 7
        return (num % 10 == 7 || num % 7 == 0);
    }
  
    // Driver method
    public static void Main()
    {
        int i = 67, j = 19;
          
        if (isBuzz(i))
            Console.WriteLine("Buzz Number");
        else
            Console.Write("Not a Buzz Number");
              
        if (isBuzz(j))
            Console.Write("Buzz Number");
        else
            Console.Write("Not a Buzz Number");
    }
}
  
// This code is contributed by Nitin mittal.

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Output:

Buzz Number
Not a Buzz Number

Time complexity : O(1) as there are atomic statements and no loops.

This article is contributed by Nikita Tiwari. If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

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Improved By : nitin mittal



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