Python program to find Prime Numbers between 1 to100.

September 30, 2022

Find Prime number between 1 to100

Here, in this page we will discuss program to find Prime number between 1 to100 in python .A prime number is an positive integer that has no divisors  except one and itself or can only be exactly divided by the integers 1 and itself without leaving a remainder.

Prime number between 1 to 100

Methods Discussed in page

  • Method 1 :Basic checking prime by only checking first n
  • Method 2: Basic checking prime by only checking first n/2 divisors
  • Method 3: Checking prime by only checking first √n divisors
  • Method 4: Checking prime by only checking first √n divisors, but also skipping even iterations.
Find Prime Number between 1 to 100 in Python

Method 1

Python Code

Run 
def checkPrime(num):

    #  0, 1 and negative numbers are not prime
    if num < 2:
        return 0
    else:

        # no need to run loop till num-1 as for any number x the numbers in
        # the range(num/2 + 1, num) won't be divisible anyway
        # Example 36 won't be divisible by anything b/w 19-35

        x = num // 2
        for j in range(2, x + 1):
            if num % j == 0:
                return 0

    # the number would be prime if we reach here
    return 1


a, b = 1, 100
for i in range(a, b + 1):
    if checkPrime(i):
        print(i, end=" ")

Output

2 3 5 7 11 13 17 19 23 29 31 37 41 43 47 53 59 61 67 71 73 79 83 89 97

Method 2

Algorithm (Method 2)

  • Run a loop in the iteration of (i) b/w 1 and 100 bounds.
  • For each, i check if its prime or not using function checkPrime(i)
  • If i is prime print it else move to the next iteration

Python Code

Run 
def checkPrime(num):

    #  0, 1 and negative numbers are not prime
    if num < 2:
        return 0
    else:

        # no need to run loop till num-1 as for any number x the numbers in
        # the range(num/2 + 1, num) won't be divisible anyway
        # Example 36 won't be divisible by anything b/w 19-35
        x = num // 2
        for j in range(2, x + 1):
            if num % j == 0:
                return 0

    # the number would be prime if we reach here
    return 1


for i in range(1, 100 + 1):
    if checkPrime(i):
        print(i, end=" ")

Output

2 3 5 7 11 13 17 19 23 29 31 37 41 43 47 53 59 61 67 71 73 79 83 89 97

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Method 3

The outer logic remains the same. Only the method to check prime changes to make code more optimized. Has better time complexity of O(√N)

  • Run a loop in the iteration of (i) b/w 1 and 100 bounds.
  • For each, i check if its prime or not using function checkPrime(i)
  • If i is prime print it else move to next iteration

Python Code

Run 
from math import sqrt


def checkPrime(num):
    #  0, 1 and negative numbers are not prime
    if num < 2:
        return 0
    else:

        # A number n is not a prime, if it can be factored into two factors a & b:
        # n = a * b

        """Now a and b can't be both greater than the square root of n, since
        then the product a * b would be greater than sqrt(n) * sqrt(n) = n.
        So in any factorization of n, at least one of the factors must be
        smaller than the square root of n, and if we can't find any factors
        less than or equal to the square root, n must be a prime."""

        for j in range(2, int(sqrt(num))):
            if num % j == 0:
                return 0

    # the number would be prime if we reach here
    return 1


a, b = 1, 100
for i in range(a, b + 1):
    if checkPrime(i):
        print(i, end=" ")

Output

2 3 5 7 11 13 17 19 23 29 31 37 41 43 47 53 59 61 67 71 73 79 83 89 97

Method 4

The outer logic remains the same. Only the method to check prime changes to make code more optimized. Has same time complexity of O(√N).

But makes around half lesser checks

  • Run a loop in the iteration of (i) b/w 1 to 100 bounds.
  • For each, i check if its prime or not using function checkPrime(i)
  • If i is prime print it else move to next iteration

Python Code

Run 
from math import sqrt


def checkPrime(n):
    # 0 and 1 are not prime numbers
    # negative numbers are not prime
    if n <= 1:
        return 0
    # special case as 2 is the only even number that is prime
    elif n == 2:
        return 1

    # Check if n is a multiple of 2 thus all these won't be prime
    elif n % 2 == 0:
        return 0

    # If not, then just check the odds
    for i in range(3, int(sqrt(n)), 2):
        if n % i == 0:
            return 0
    return 1


a, b = 1, 100
for i in range(a, b + 1):
    if checkPrime(i):
        print(i, end=" ")

# This method is obviously faster as makes around half lesser comparison due skipping even iterations in the loop

Output

2 3 5 7 11 13 17 19 23 29 31 37 41 43 47 53 59 61 67 71 73 79 83 89 97

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5 comments on “Python program to find Prime Numbers between 1 to100.”


  • D

    n=int(input())
    for i in range(1,n):
    fact=0
    for j in range(2,i):
    if i%j==0:
    fact+=1
    if fact==0:
    print(i)


  • Arpita

    for i in range(1,100):
    c=0
    for j in range(1,i+1):
    if i%j==0:
    c=c+1
    if c==2:
    print(i,end=’ ‘)


  • KIRA

    More Easy:

    for i in range(2, 100):
    for j in range(2, i):
    if i % j == 0:
    break
    else:
    print(i)


  • Yashi

    start = int(input())
    end = int(input())

    for i in range(start, end+1):
    if i>1:
    for j in range(2,i):
    if(i % j==0):
    break
    else:
    print(i)


  • Rahul

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