Write a program that prompts to enter a number and check whether the number is (i) prime or not
Answers
Answer:
#include <iostream>
using namespace std;
int main() {
int i, n;
bool isPrime = true;
cout << "Enter a positive integer: ";
cin >> n;
// 0 and 1 are not prime numbers
if (n == 0 || n == 1) {
isPrime = false;
}
else {
for (i = 2; i <= n / 2; ++i) {
if (n % i == 0) {
isPrime = false;
break;
}
}
}
if (isPrime)
cout << n << " is a prime number";
else
cout << n << " is not a prime number";
return 0;
}
Output
Enter a positive integer: 29
29 is a prime number.
This program takes a positive integer from the user and stores it in the variable n.
Notice that the boolean variable isPrime is initialized to true at the beginning of the program.
Since 0 and 1 are not prime numbers, we first check if the input number is one of those numbers or not. If the input number is either 0 or 1, then the value of isPrime is set to false.
Else, the initial value of isPrime is left unchanged and the for loop is executed, which checks whether the number entered by the user is perfectly divisible by i or not.
for (i = 2; i <= n / 2; ++i) {
if (n % i == 0) {
isPrime = false;
break;
}
}
The for loop runs from i == 2 to i <= n / 2 and increases the value of i by 1 with each iteration.
The loop terminates at i == n / 2 because we cannot find any factor for n beyond the number n / 2. So, any iterations beyond n / 2 is redundant.
If the number entered by the user is perfectly divisible by i, then isPrime is set to false and the number will not be a prime number.
But if the input number is not perfectly divisible by i throughout the entirety of the loop, then it means that the input number is only divisible by 1 and that number itself.
So, the given number is a prime number.
In the case of n == 2, the for loop fails to run and the value of isPrime remains true.
Explanation: