Print series of prime numbers in python

Question

I am trying to learn Python programming, and I'm pretty new at this.

I was having issues in printing a series of prime numbers from one to hundred. I can't figure our what's wrong with my code.

Here's what I wrote; it prints all the odd numbers instead of primes:

for num in range(1,101):
    for i in range(2,num):
        if (num%i==0):
            break
        else:
            print(num)
            break

Answer 1

def prime_number(a):
    yes=[]
    for i in range (2,100):
        if (i==2 or i==3 or i==5 or i==7) or (i%2!=0 and i%3!=0 and i%5!=0 and i%7!=0 and i%(i**(float(0.5)))!=0):
            yes=yes+[i]
    print (yes)

Answer 2

Adding to the accepted answer, further optimization can be achieved by using a list to store primes and printing them after generation.

import math
Primes_Upto = 101
Primes = [2]
for num in range(3,Primes_Upto,2):
    if all(num%i!=0 for i in Primes):
       Primes.append(num)
for i in Primes:
    print i

Answer 3

Instead of trial division, a better approach, invented by the Greek mathematician Eratosthenes over two thousand years ago, is to sieve by repeatedly casting out multiples of primes.

Begin by making a list of all numbers from 2 to the maximum desired prime n. Then repeatedly take the smallest uncrossed number and cross out all of its multiples; the numbers that remain uncrossed are prime.

For example, consider the numbers less than 30. Initially, 2 is identified as prime, then 4, 6, 8, 10, 12, 14, 16, 18, 20, 22, 24, 26, 28 and 30 are crossed out. Next 3 is identified as prime, then 6, 9, 12, 15, 18, 21, 24, 27 and 30 are crossed out. The next prime is 5, so 10, 15, 20, 25 and 30 are crossed out. And so on. The numbers that remain are prime: 2, 3, 5, 7, 11, 13, 17, 19, 23, and 29.

def primes(n):
  sieve = [True] * (n+1)
  for p in range(2, n+1):
    if (sieve[p]):
      print p
      for i in range(p, n+1, p):
        sieve[i] = False

An optimized version of the sieve handles 2 separately and sieves only odd numbers. Also, since all composites less than the square of the current prime are crossed out by smaller primes, the inner loop can start at p^2 instead of p and the outer loop can stop at the square root of n. I'll leave the optimized version for you to work on.

Answer 4

a=int(input('enter the lower no.'))
b=int(input('enter the higher no.'))
print("Prime numbers between",a,"and",b,"are:")
for num in range(a,b):

    if num>1:
        for i in range(2,num):
            if (num%i)==0:
                break
        else:
            print(num)

Answer 5

# computes first n prime numbers
def primes(n=1):
    from math import sqrt
    count = 1
    plist = [2]
    c = 3
    if n <= 0 :
        return "Error : integer n not >= 0"
    while (count <= n - 1):    # n - 1 since 2 is already in plist
        pivot = int(sqrt(c))
        for i in plist:
            if i > pivot :    # check for primae factors 'till sqrt c
                count+= 1
                plist.append(c)
                break
            elif c % i == 0 :
                break    # not prime, no need to iterate anymore
            else :
                continue 
        c += 2    # skipping even numbers              
    return plist

Answer 6

min=int(input("min:"))
max=int(input("max:"))
for num in range(min,max):
    for x in range(2,num):
        if(num%x==0 and num!=1):
            break
        else:
            print(num,"is prime")
            break