# How to find Prime Factors till a limit

In this article, let's try to find a solution for the coding problem - Print all the possible prime factors for all the numbers till the given limit.

Let’s assume we are given a number n. we need to print all the numbers till the given number n, such that for every number if there exists a prime factor, we need to print those prime factors instead of that number in the set.

For example, if we are given the number as 10. we need to print the output as: 1 2 3 2 5 2 3 7 2 3 2 5

The input is an upper limit, until which we will move and print all the prime factors for every number we have visited.

The output for the program is as below –

``````1
2
3
2 <- for 4, since 4 = 2 x 2
5
2 3 <- for 6, since 6 = 2 x 3
7
2 <- for 8, since 8 = 2 x 4
3 <- for 9, since 9 = 3 x 3
2 5 <- for 10, since 10 = 2 x 5``````

## How do we solve this?

Let’s try to approach it like this:

1. Maintain a list of primes we have encountered through out the loop, say primes.
2. For every element in the series till the limit n, say x
3. maintain a flag isPrime to check if there was a factor or not, set to true
4. Loop from 2 till the number x, with a counter i
5. Check if there is any i that is a factor of x
6. if there is this i in the primes, print the number i
7. unset the flag isPrime, since we know that i is a factor of x
8. End of the loop
9. if the counter i is equal to x and isPrime is still set to true, add the current number x to the set of primes we’re maintaining.
10. The same process is repeated for all the numbers till the limit n

## Code in C#

``````public class PrimeFactorBuzz
{
HashSet<int> primes = new HashSet<int>();

public void DoPrimeFactors(int n)
{
for (int i = 1; i <= n; i++)
{
PrimeFactor(i);
}
}

public void PrimeFactor(int number)
{
bool isPrime = true;
int up = number;

int i = 2;
while (i < up)
{
if (number % i == 0)
{
if (primes.Contains(i))
{
Console.Write(\$"{i}"); isPrime = false;
}
}
i++;
}
if (number == 1) Console.Write(\$"{number}");

if (i == up && isPrime)
{ 