Write a program to accept a number and using ternary operator, check whether it is a buzz number or not. A
buzz number either ends with 7 or is divisible by 7. For example: numbers like 49, 47, 63,427 are buzz
numbers. But numbers like 100, 25 and 62 are not
Answers
Explanation:
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.
Recommended: Please try your approach on {IDE} first, before moving on to the solution.
// 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";
}
Output:
Buzz Number
Not a Buzz Number
Time complexity : O(1) as there are atomic statements and no loops.
import java. util. *
class Buzz
{
public static void main (string args[])
{
Scanner sc= new Scanner (System. in);
int x;
System.out.println ("Enter a number");
x=sc. nextInt ();
if(x%==0||x/10==7)
System.out.println (x+"is Buzz number");
else
System.out.println (x+"is not Buzz");
}
}
A Buzz number is a number which is divisible by 7 or whose last digit is 7.