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Trying to find the largest prime factor but I don't know what is wrong. Please give me advice [closed]

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I don't know what is wrong. I am trying to find largest prime factor
import math
def getfactor(num):
factors = []
for i in range(2,int((math.sqrt(num)))+1):
if num % i == 0:
factors.append(i)
return factors
def determineprime(num):
factor = []
for i in range(2,num + 1):
if num % i == 0:
factor.append(i)
if len(factor) == 1:
return True
else:
return False
factors = getfactor(600851475143)
primes = []
print(factors)
for i in factors:
determineprime(i)
if i:
primes.append(i)
print(primes[-1])

Change:
determineprime(i)
if i:
primes.append(i)
to:
if determineprime(i):
primes.append(i)
The original version was calling determineprime(i), discarding the result, and always appending i to primes. The corrected version will only append i to primes if determineprime(i) returns true.
Update: As Mark Tolonen pointed out, getfactor will return an empty list if its argument is prime. In that case, the number itself should be returned as the sole element of the list. It can just check to see if the list is empty before returning, and if so, append the number itself, since it must be prime in that case.

Trying to code a function with one numeric input. If the number <= 0, return -1. If the number > 0, divide that integer over-and-over by 2 [closed]

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I am new to Python and coding. This is what I have so far, but can't get it to work.
def halve_to_2( num ):
while True:
num = num /2
if num < 2:
else:
if num <=0:
print ("-1")
return num
k = int(input("Enter a number"))
print(halve_to_2(k))

def halve_to_2( num ):
while True:
if num <= 0:
print("-1")
return -1
else:
num = num / 2
k = int(input("Enter a number"))
print(halve_to_2(k))

def divisibleByTwo( num) :
if num <=0 :
return - 1
else :
divisibleByTwo (num/2)
Its called recursion. The break point is the return - 1 for this recursive loop

def divisibleByTwo( num) :
print ( num)
if int(num) > 0 :
divisibleByTwo (num/2)
return num/2
This one print the values that fall within integers not scientific decimals

Why we use count variable & and how its work? [closed]

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#program of prime number from 2 to 100
for i in range (2,100):
count = 0
for j in range(1, i+1):
if i % j = 0:
count = count + 1
if count == 2:
print(i)

First of all, your code seems to not work due to a syntax error(might be a typo).
if conditions in python, requires double equals (==).
if i % j == 0:
In the question you posted, the count variable acts as a flag or counter variable.
A prime number is a number which can only be perfectly divided, by 1 or the number itself. (eg: 11 can be divided by 1 and 11 only, therefore a prime number)
So the counter variable i.e. count holds the number of times that a number(2-200) gets perfectly divided(i.e. (%) modulo operation results in value 0). If the value of count is 2 then we can confirm that it is a prime number since a prime number can only be divided with 1 and the number itself (hence count=2). And if the count == 2 after all the possible divisions happening in the j loop (2nd for loop), we can conclude that it is a prime number.
for i in range (2,200):
count = 0
for j in range(1, i+1):
if i % j == 0:
count += 1
if count == 2:
print(i)

count variable is being used to ensure if the particular number i is being divisible by any other number other than 1. Because if it does, then it is not a prime number. The moment it becomes 2, it means that number is divisible by 1 and some other number less than i proving it be a non-prime number.

Run this to get primes from 100, you can change the 100 to whatever you want
for num in range(100):
if num > 1:
for i in range(2,num):
if (num % i) == 0:
print(num,"is not a prime number")
print(i,"times",num//i,"is",num)
break
else:
print(num,"is a prime number")
else:
print(num,"is not a prime number")

Count simple numbers in diapasone python [closed]

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Task is : input two numbers that is a diapasone . Then count a quantity of simple numbers into this diapasone.
Please correct my python code below
import math
count = (int, input().split())
for i in count:
n = int(input())
for j in range(2, int(math.sqrt(n)) + 1):
if n % j == 0:
break
print(count)

If you want to count the prime numbers in the closed interval [j, k] (which is given with input()), then you could use
import math
interval = tuple(map(int, input().split()))
count = 0
for n in range(max(2, interval[0]), interval[1]+1):
for j in range(2, int(math.sqrt(n)) + 1):
if n % j == 0:
break
else:
count += 1
print(count)
Explanation:
use map to apply int on each string
use count as a counter variable
use range as an iterator over integers in given interval
start prime number tests with 2, even if lower bound is smaller
count up only if no divisor is found (else clause of for loop)

I do not understand foor loop [closed]

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Why inside for loop is written upper + 1 and not just upper. If I change upper+1 with upper and run the code, I will have the same result!
for num in range(lower,**upper + 1)**
# Python program to display all the prime numbers within an interval
lower = int(input("Enter lower range: "))
upper = int(input("Enter upper range: "))
print("Prime numbers between",lower,"and",upper,"are:")
for num in range(lower,upper + 1):
# prime numbers are greater than 1
if num > 1:
for i in range(2,num):
if (num % i) == 0:
break
else:
print(num)

It's because range(x,y) stops at y-1. For example:
for i in range(2,5):
print(i)
will give you:
2
3
4
So if you want to check upper as well, you should add 1, so upper+1. Your code returns the same results since upper is not prime, so even if upper is checked (with upper+1 in range), upper will not be printed.