Given two positive integers **A** and **B**, print to the console the first **A** non-negative numbers that are divisible by **B**

A number **X** is divisible by **B** if `X modulo B == 0`

**Example:**

Input:A= 5,B= 3Output:3 6 9 12 15Explanation:The first 5 positive integers that are divisible by 3 are 3, 6, 9, 12 and 15 1 modulo 3 = 1 => not divisible 2 modulo 3 = 2 => not divisible 3 modulo 3 = 0 => divisible 4 modulo 3 = 0 => not divisible 5 modulo 3 = 0 => not divisible 6 modulo 3 = 0 => divisible 7 modulo 3 = 0 => not divisible 8 modulo 3 = 0 => not divisible 9 modulo 3 = 0 => divisible 10 modulo 3 = 0 => not divisible 11 modulo 3 = 0 => not divisible 12 modulo 3 = 0 => divisible 13 modulo 3 = 0 => not divisible 14 modulo 3 = 0 => not divisible 15 modulo 3 = 0 => divisible

The core challenge of this problem is to find the first **A** non-negative numbers that are divisible by **B**. This problem is significant in various applications such as generating sequences, scheduling tasks, and more. A common pitfall is to iterate through all numbers and check divisibility, which is inefficient.

To solve this problem, we can use a straightforward approach:

- Start iterating from 1.
- Check if the current number is divisible by
**B**. - If it is, print the number and increment the counter.
- Stop when the counter reaches
**A**.

This approach ensures that we only print the required numbers and stop as soon as we have printed **A** numbers.

The naive solution involves iterating through all numbers and checking if each is divisible by **B**. This is not optimal because it may involve unnecessary checks.

An optimized solution involves directly calculating the multiples of **B**:

- Start from
**B**and keep adding**B**to get the next multiple. - Print each multiple until we have printed
**A**numbers.

This approach is more efficient as it directly generates the required multiples.

Here is a step-by-step breakdown of the optimized algorithm:

- Initialize a counter to 0.
- Start a loop from 1 to infinity.
- In each iteration, calculate the current multiple as
`currentMultiple = B * counter`

. - Print the current multiple.
- Increment the counter.
- Stop the loop when the counter reaches
**A**.

```
public class PrintMultiples {
public static void main(String[] args) {
int A = 5; // Number of multiples to print
int B = 3; // Divisor
// Counter to keep track of how many multiples have been printed
int count = 0;
// Start from 1 and keep checking for multiples of B
for (int i = 1; count < A; i++) {
if (i % B == 0) {
System.out.println(i);
count++;
}
}
}
}
```

The time complexity of the optimized solution is **O(A)** because we only iterate until we find **A** multiples. The space complexity is **O(1)** as we are not using any additional space.

Potential edge cases include:

**A**or**B**being very large.**B**being 1, which means every number is a multiple.

These cases are handled by the algorithm as it directly calculates the multiples.

To test the solution comprehensively, consider the following test cases:

- Simple cases like
`A = 5, B = 3`

. - Edge cases like
`A = 1, B = 1`

. - Large values like
`A = 1000, B = 100`

.

Use a testing framework like JUnit to automate the tests.

When approaching such problems:

- Understand the problem requirements clearly.
- Think of a straightforward solution first.
- Optimize the solution by reducing unnecessary computations.
- Practice similar problems to improve problem-solving skills.

In this blog post, we discussed how to print the first **A** non-negative numbers divisible by **B** efficiently. We explored both naive and optimized solutions, analyzed their complexities, and provided a detailed Java implementation. Understanding and solving such problems is crucial for developing strong problem-solving skills.

For further reading and practice, consider the following resources: