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Program to check whether the given number is Buzz Number or not in C++
Given with a number ‘n’ and the task is to determine whether the given positive integer is a buzz number or not and display the result as an output.
What is Buzz Number?
For being a buzz number there are two conditions either of which must be true −
Number should end with digit 7 e.g. 27, 657, etc.
Number should be divisible by 7 e.g 63, 49, etc.
Input
number: 49
Output
it’s a buzz number
Explanation − since the number is divisible by 7 so it’s a buzz number
Input
number: 29
Output
it’s not a buzz number
Explanation − since the number is neither divisible by 7 nor end with digit 7 so it’s not a buzz number
Approach used in the given program is as follows
Input the number to check for the condition
Check whether the number is ending with digit 7 or divisible by 7
If the condition holds true print its a buzz number
If the condition doesn’t holds true print its not a buzz number
Algorithm
Start Step 1→ declare function to check if a number is a buzz number of not bool isBuzz(int num) return (num % 10 == 7 || num % 7 == 0) Step 2→ In main() Declare int num = 67 IF (isBuzz(num)) Print "its a buzz Number\n" End Else Print "its not a buzz Number\n" End Stop
Example
#include <cmath>
#include <iostream>
using namespace std;
// function to check if its a buzz number
bool isBuzz(int num){
return (num % 10 == 7 || num % 7 == 0);
}
int main(){
int num = 67;
if (isBuzz(num))
cout << "its a buzz Number\n";
else
cout << "its not a buzz Number\n";
}Output
If run the above code it will generate the following output −
its a buzz Number
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