Question

Use phyton

TheSieve of Eratosthonesis an algorithm to find all the prime numbers between 1 andsome integerN. It can be implemented with nestedforloops:(a) Make a list of all the integers from 2 throughN.(b) Cross off all the multiples of 2 (except for 2 itself). The smallest number that remains(after 2) is 3.(c) Cross off all the multiples of 3 (except for 3 itself). The smallest number that remainsis 5.(d) Cross off all the multiples of 5 (except for 5 itself). The smallest number that remainsis 7.(e)···(f) Repeat looking for the smallest numberpthat remains and crossing off all of its multiples(except forpitself) until you reach√N.(g) All of the numbers that remain are primeUse the Sieve to print all the prime numbers under 1000.

Answer 1

import math

if __name__=="__main__":
    n=int(input("Enter the value of n: "))
    #creating a list of numbers from 2 to n
    primes=[i for i in range(2,n+1)]
  
    #start with number 2
    p=2
    #when p less than square root of n
    while (p<=math.sqrt(n)):
        #if the value current value in list is not None
        if(primes[p-2]!=None):
            #make all the mulitples as None
            for i in range(p*p,n+1,p):
                primes[i-2]=None
        p=p+1
  
    #print all values in the list which is not prime  
    for i in primes:
        if(i!=None):
            print(i)

############################ PGM END ###################################

OUTPUT
#########

Enter the value of n: 20 2 7 13 17 19