﻿ Java - Accepts an integer and count the factors of the number

# Java: Accepts an integer and count the factors of the number

## Java Basic: Exercise-57 with Solution

Write a Java program to accept an integer and count the factors of the number.

Sample Solution:

Java Code:

``````import java.util.*;

public class Exercise57 {
public static void main(String[] args) {
// Create a Scanner object for user input
Scanner in = new Scanner(System.in);
System.out.print("Input an integer: ");

// Read an integer from the user
int x = in.nextInt();

// Call the result method and print the result
System.out.println(result(x));
}

// Define a method to calculate the number of divisors for a given integer
public static int result(int num) {
int ctr = 0;

// Iterate from 1 to the square root of the input number
for (int i = 1; i <= (int) Math.sqrt(num); i++) {
// Check if 'i' is a divisor, and if it's not a perfect square
if (num % i == 0 && i * i != num) {
ctr += 2;  // Increase the count by 2
} else if (i * i == num) {
ctr++;  // If 'i' is a perfect square, increase the count by 1
}
}
return ctr;  // Return the total count of divisors
}
}
```
```

Sample Output:

```Input an integer: 25
3
```

Pictorial Presentation:

Flowchart:

Java Code Editor:

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