def prime_finder(n):
prime =
for i in range(2,n+1):
if len(prime) == 0:
prime.append(i)
else:
count = 0
for j in prime:
if i%j != 0:
count+=1
else:
count-=1
if count == len(prime):
prime.append(i)
return prime
print(prime_finder(11))
def prime_finder(n):
# Write your code here
res = []
IsPrime = [True] * (n + 1)
IsPrime[0] = False
IsPrime[1] = False
d = 2
while d * d <= n:
if IsPrime[d]:
for i in range(d * d, n + 1, d):
IsPrime[i] = False
d += 1
for i in range(n + 1):
if IsPrime[i]:
res.append(i)
return res
print(prime_finder(13))
def prime_finder(n):
# Write your code here
primes = []
for num in range(2, n+1):
for i in range(2, num//2 +1):
if num % i == 0:
break
else:
primes.append(num)
return primes
print(prime_finder(11))
for this solution you have to make use of the uncommon for/else statement. if you just do a regular if/else statement within the inner loop you will get duplicate prime numbers in your list
def prime_finder(n):
# Write your code here
def isprimeultimate(n: int):
for x in range(2, int(n**0.5) + 1):
if n % x == 0:
return False
return True
return [int(x) for x in range(2,n+1) if isprimeultimate(x) == True]
print(prime_finder(11))
Here a nice oneliner solution for those of you who are a bit more advanced:
prime_finder = lambda n: [i for i in range(2, n + 1) if 0 not in [i % j for j in range(2, i)]]
def prime_finder(n):
prime = []
for numerator in range(2, n+1):
divisible = 0
for denominator in range(2, n+1):
if numerator == denominator: continue #skip the iteration if the number is itself
if numerator % denominator != 0: continue # skip the iteration if not divisible
divisible += 1
break # if the number is divisible in this iteration, we don't need to check the rest of denominator. This is not a prime number.
if divisible == 0: prime.append(numerator) #because we skip 1 and the number itself, divisible should be 0 for prime number
return prime
print(prime_finder(11))
def prime_finder(n:int):
return [i for i in range(2, n+1) if all([i % j != 0 for j in range(2, i)])]
def prime_finder(n):
Write your code here
list =
for number in range(1,n+1):
if number>1:
for prime in range(2,number):
if number%prime == 0:
break
else:
list.append(number)
return list
print(prime_finder(11))
Hi! So my solution is as follows:
def prime_finder(n):
# Initiate empty list
primes = []
# Looping through numbers smaller than n
for i in range(2,n+1):
isprime = True
# Checking if i is a prime
for k in range(2,i):
if i % k == 0:
isprime = False
break
# if it's a prime, add to the list
if isprime:
primes.append(i)
return primes
import math
def isPrime(n):
if n <= 3:
return n > 1
if (n % 6 != 1) and (n % 6 != 5):
return False
sqrt = math.sqrt(n)
i = 5
while(i <= sqrt):
if (n % i == 0) or (n % (i + 2) == 0):
return False
i += 6
return True
def prime_finder(n):
# Write your code here
prime_numbers = []
for i in range(1, n+1):
if isPrime(i):
prime_numbers.append(i)
return prime_numbers
print(prime_finder(1000000))
def is_prime(p):
for i in range(2, p):
if p%i == 0:
return False
return True
def prime_finder(n):
prime_list = []
for i in range(2, n+1):
if is_prime(i):
prime_list.append(i)
return prime_list
print(prime_finder(11))
def prime_finder(n):
prev_nums = [i for i in range(2, (n+1))]
prime_numbers = []
for num in prev_nums:
remainders = []
for i in range(2, 11):
if i != num:
remainders.append(num%i)
if 0 not in remainders:
prime_numbers.append(num)
return prime_numbers
print(prime_finder(110))
def prime_finder(n):
# Write your code here
prime = []
if n >= 2:
prime.append(2)
if n >= 3:
prime.append(3)
if n > 4:
for p in range(5, n + 1):
is_prime = True
for divisor in range(2, int(p/2)):
if p % divisor == 0:
is_prime = False
break
if is_prime:
prime.append(p)
return prime
print(prime_finder(11))
def prime_finder(n):
# Write your code here
l = []
for i in range(2,n+1):
if (i==2 or i==3 or i==5 or i==7):
l.append(i)
elif i%2 != 0 and i%3 != 0 and i%5 !=0 and i%7 != 0:
l.append(i)
return l
#print(prime_finder(100))
def prime_finder(n):
prime_numbers_before_n = list(range(2,n+1))
for i in range(1,n+1):
for j in range(1,n+1):
if i % j == 0 and j != 1 and i != j:
prime_numbers_before_n.remove(i)
break
return prime_numbers_before_n
print(prime_finder(11))
def prime_finder(n):
Write your code here
i = 0
num_list =
prime_list =
non_prime_list =
for num in range(1, n+1):
num_list.append(num)
for item in num_list:
position = num_list.index(item)
# preceeding_num = num_list[position-1]
preceeding_num_list = num_list[1:position]
if position==0:
pass
else:
for num in preceeding_num_list:
if item%num==0:
if item in non_prime_list:
pass
else:
non_prime_list.append(item)
num_list.remove(1)
for number in num_list:
if number in non_prime_list:
pass
else:
prime_list.append(number)
return(prime_list)
print(prime_finder(11))
#This was my solution for the prime number finder. What do you guys think? Constructive criticisms are welcomed. Have a nice day!
def prime_finder(n):
prime_number = []
for i in range(2,n+1):
for j in range(2,i):
if i % j == 0:
break
else:
prime_number.append(i)
return prime_number
print(prime_finder(21))
This is my solution. Any feedback is welcome. Thank you guys! Have a nice day!
def prime_finder(n):
num = 2
prime_nos=[]
counter=0
while num!=n+1:
for i in range(num):
if num%(i+1) == 0:
counter+=1
if counter<=2:
prime_nos.append(num)
counter=0
num+=1
return prime_nos
print(prime_finder(11))
import unittest
def prime_finder(n):
# mod 2 == 1
prime_list = []
is_prime = 1;
for i in range (2, n+1):
for j in range (2, 10):
if (i % j == 0) and (i != j) :
is_prime = 0
break
if (j == i):
break
if is_prime:
prime_list.append(i)
is_prime = 1
return prime_list
class TestPrime(unittest.TestCase):
def test_prime(self):
onehundred = [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]
self.assertEqual(prime_finder(100), onehundred)
unittest.main(exit=False)