# Python program to find Prime Numbers between 1 to100.

## 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. ### 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. ## 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`

## 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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## Getting Started

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

• KIRA

More Easy:

for i in range(2, 100):
for j in range(2, i):
if i % j == 0:
break
else:
print(i) 1
• 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) 0
• Rahul

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