# Given a number n, print all primes smaller than or equal to n. It is also given that n is a small number.
# For example, if n is 10, the output should be “2, 3, 5, 7”. If n is 20, the output should be “2, 3, 5, 7, 11, 13, 17, 19”.


# Python program to print all primes smaller than or equal to
# n using Sieve of Eratosthenes

def SieveOfEratosthenes(n):
	
	# Create a boolean array "prime[0..n]" and initialize
	# all entries it as true. A value in prime[i] will
	# finally be false if i is Not a prime, else true.
	prime = [True for i in range(n + 1)]
	p = 2
	while (p * p <= n):
		
		# If prime[p] is not changed, then it is a prime
		if (prime[p] == True):
			
			# Update all multiples of p
			for i in range(p * 2, n + 1, p):
				prime[i] = False
		p += 1
	prime[0]= False
	prime[1]= False
	# Print all prime numbers
	for p in range(n + 1):
		if prime[p]:
			print (p)

# driver program
if __name__=='__main__':
    n = 30
    print("Following are the prime numbers smaller")
    print("than or equal to ", n)
    print("than or equal to ", n)
    SieveOfEratosthenes(n)

prime_lists=[] # a list to store the prime numbers

def prime(n): # define prime numbers
    if n <= 1:
        return False
    # divide n by 2... up to n-1
    for i in range(2, n):
        if n % i == 0:  # the remainder should'nt be a 0
            return False
    else:
        prime_lists.append(n)
        return True



for n in range(30,1000):  # calling function and passing starting point =30 coz we need primes >30
    prime(n)

check=0 # a var to limit the output  to 10 only
for n in prime_lists:
    for x in prime_lists:
        val=  n *x
        if (val > 1000 ):
            check=check +1
        if (check <10) :
            print("the num is:", val , "=",n , "* ", x )
        break

https://codecatch.net/post/06c9f5b7-1e00-40dc-b436-b8cccc4b69be

import pandas as pd



x = pd.read_excel(FILE_NAME)

print(x)

from collections import Counter

def find_parity_outliers(nums):
  return [
    x for x in nums
    if x % 2 != Counter([n % 2 for n in nums]).most_common()[0][0]
  ]

find_parity_outliers([1, 2, 3, 4, 6]) # [1, 3]

from typing import Optional

from datetime import datetime

def convert_timestamp_string_to_epoch(timestamp: str) -> Optional[int]:
    epoch_time = None
    time_obj = datetime.strptime(timestamp, "%Y-%m-%d %H:%M:%S.%f")
    epoch_time = int((time_obj - datetime(1970, 1, 1)).total_seconds() * 1000)
    return epoch_time


print(int(convert_timestamp_string_to_epoch("2025-08-01 13:11:47.171")))
#above outputs 1754053907171.0

#how to I remove the .0 ?