How do I solve this using Python? I’m struggling 😭

Question

A function isPrime () is given to you that given as an integer as an argument returns True if it is prime and False otherwise.

def isPrime (number) :
if number > 1:
for x in range (2, number) :
if number % X == 0:
return False
return True
else:
return False

Copy this code in ur code file. Write a function addPrimes that sums all prime numbers between 1 and x (where, x > 1). You need to use this function isPrime to check if a given number is prime.

Answer 1

Right that is because it is running through range (1, X) when it shall be (1, X+1)

Try:

def isPrime (number) :
if number > 1:
for x in range (2, number) :
if number % x == 0:
return False
return True
else:
return False

def addPrime(x):
sum = 0
for i in range(1, x+1):
if isPrime(i):
sum+=i;
return sum

print(addPrime(5))

Answer 2

Alright here is an addPrime function that uses isPrime:

def addPrime(x):
sum = 0
for i in range(1, x):
if isPrime(i):
sum+=i;
return sum

However I'm almost certain your isPrime function is incorrect.

Answer 3

It works now. Thank you very much !

Answer 4

Here is a good tutorial on how to use nested functions:

https://realpython.com/inner-functions-what-are-they-good-for/

Answer 5

Coding Cat , oh yeah right you are just skipping through all even numbers... It will be way faster

Answer 6

Olivia , khm.... The same way as you remember how to walk)

Answer 7

Check this code
https://code.sololearn.com/cJxJf9Ki5652/?ref=app

Answer 8

Your attempt?

it's really not that hard you just make a loop that goes through every number from 1 to X and check on whether it is prime using isPrime() function. (if it is prime add it to the sum)

Answer 9

Aleksei Radchenkov I’m new to programming. This is what i did:

def isPrime(number):
x= isPrime(number)
if number > 1 :
for x in range(2,number):
if number % x == 0:
return False
return True
else:
return False
if x is True:
total= 0
addPrimes= total + x
return addPrimes

Answer 10

Actually your isPrime function will work but it is far not the most efficient approach:

Here is a better code:

from math import sqrt
def isPrime (n) :
prime_flag = 0
if(n > 1):
for i in range(2, int(sqrt(n)) + 1):
if (n % i == 0):
prime_flag = 1
break
if prime_flag == 0:
return True
else:
return False

else:
return False

def addPrime(x):
sum = 0
for i in range(1, x):
if isPrime(i):
sum+=i;
return sum

print(addPrime(7))

Answer 11

Ok. Thank you
I was having trouble defining a function in another function.
Idk how to do that as yet

Answer 12

I run the code against some test cases and i think something is missing from the code. So these are the answers that I’m supposed to get when call addPrime:
>>>addPrime(5)
10
>>>addPrime(200)
4227
>>>addPrime(3)
5

I only got the correct ans for addPrime(200)

Answer 13

Aleksei Radchenkov maybe you are interested in this:
It works w/o a flag and needs only the half of steps in the for loop.

def isPrime (n) :
if(n <= 1):
return False
if (n == 2):
return True
if (n%2 == 0):
return False
for i in range(3, int(sqrt(n)),2):
if (n % i == 0):
return False
return True

Answer 14

Aleksei Radchenkov How can the code be written so that the isPrime function is in the addPrime function?

Answer 15

Trizzy , something like:

def addPrime(x):
def isPrime (number) :
if number > 1:
for x in range (2, number) :
if number % x == 0:
return False
return True
else:
return False
sum = 0
for i in range(1, x+1):
if isPrime(i):
sum+=i;
return sum

print(addPrime(5))

Answer 16

Ooh ok. Nice

Answer 17

You should use range(2,int(number**0.5)+1) instead using range(2,number)
Because if that num is not prime , there are probably 1 or more pair, each pair has a num <= square root of num and a num >= square root of num
Ex :
num is 10
10 % 2 == 0
2 < 10 ** 0.5
=> It's not prime