how to check if number is prime python

stackoverflow.com › questions › 74773174 › checking-whether-a-number-is-prime

When you use a for loop, you are iterating the number of times that are in range(1, int(user_input)) and so your if/else gets executed over and over again, and in each iteration is printing something. Instead you only want to print when you are done.

To pull this off, you want to use a boolean True/False variable to track if the number is prime or not, and set it accordingly in your for loop.

user_input = input("pick a number")

is_prime = True
for i in range(2, int(user_input)):
    if int(user_input) % i == 0 :
        is_prime = False
        break #break out of the for loop since the number isn't prime

if is_prime:
   print("Your number is prime")
else:
   print("your number is not prime")

Some other small changes:

Removed condition and i != 1 as any iteration past 1 would fail this test and you want to check EVERY number in your iterations
Removed condition and i != int(user_input) since in your iterations you will never reach this number range() is not inclusive of the highest number.
Added a break once we determine that the number is not a prime so we don't waste iterations checking if it's even more not-prime.

Answer from JNevill on Stack Overflow

GeeksforGeeks

geeksforgeeks.org › python › python-program-to-check-whether-a-number-is-prime-or-not

Check Prime Number in Python - GeeksforGeeks

07:30

In the sympy module, we can test whether a given number 'n' is prime or not using sympy.isprime() function. For n < 264 the answer is definitive; larger n values have a small probability of actually being pseudoprimes.

Published October 11, 2025

Stack Overflow

stackoverflow.com › questions › 74773174 › checking-whether-a-number-is-prime

python - checking whether a number is prime - Stack Overflow

Top answer

1 of 4

4

When you use a for loop, you are iterating the number of times that are in range(1, int(user_input)) and so your if/else gets executed over and over again, and in each iteration is printing something. Instead you only want to print when you are done.

To pull this off, you want to use a boolean True/False variable to track if the number is prime or not, and set it accordingly in your for loop.

user_input = input("pick a number")

is_prime = True
for i in range(2, int(user_input)):
    if int(user_input) % i == 0 :
        is_prime = False
        break #break out of the for loop since the number isn't prime

if is_prime:
   print("Your number is prime")
else:
   print("your number is not prime")

Some other small changes:

Removed condition and i != 1 as any iteration past 1 would fail this test and you want to check EVERY number in your iterations
Removed condition and i != int(user_input) since in your iterations you will never reach this number range() is not inclusive of the highest number.
Added a break once we determine that the number is not a prime so we don't waste iterations checking if it's even more not-prime.

2 of 4

0

There are some issues with your code range should be 2 not 1. Add a break statemnt once condition satisfies.

user_input = input("pick a number")
user_input  = int(user_input)

for i in range(2, user_input)::  #for i in range(2, int(user_input /2)+1) also works
    if (user_input  % i) == 0:    
        print("Your number is prime")
        break
else:
    print("Your number is not a prime")

More simpler approch

user_input = int(input("pick a number"))
result = all(user_input % i for i in range(2, user_input))

if result:
     print("Your number is prime")
else:
     print("Your number not a is prime")

Discussions

Prime number or not?

One trick you can use is that you dont have to go over all numbers from 2 to N, you only need to check numbers from 2 to int(sqrt(N)) + 1. Reason is math, lets say N is 32, it doesnt matter if you check 4 * 8 or 8 * 4, you can kinda omit the second one as if 32 is divisible by 4 it is also divisible by 8 and you dont even need to check it. Essentially you dont need to check the second reversed pair. Question is, from what number you do not have to check, the answer is that you dont have to check from the point where the first number is better then the second one, because of reasons above. This point happens when X * X = N, so you only need to check from 2 to sqrt(N). More on reddit.com

r/learnpython

27

54

December 3, 2022

This is a question about a program to determine whether a number is a prime number

Could the following code be used to determine whether a number is prime? I’ve tested it and can’t find any exceptions. Could the following code be used to determine a prime number? def prime_checker(number): if numb… More on discuss.python.org

discuss.python.org

0

May 4, 2023

Prime number finding function

A function can only return once, so you're returning False as soon as you discover the number is even. More on reddit.com

r/learnpython

7

0

January 21, 2021

Finding Prime Numbers

Two easy ways to add elements to a list: The first method is to create an empty list, then append new elements to it. ceiling = int(input('Highest number to check? ')) primes = [] for number in range(1, ceiling + 1): if is_prime(number): primes.append(number) The second method is to use list comprehension. ceiling = int(input('Highest number to check? ')) primes = [number for number in range(1, ceiling + 1) if is_prime(number)] Your function to check if a number is prime can be set up in a number of ways. Here's one: def is_prime(number): if number < 2: prime = False else: hi_number = int(number ** 0.5 + 1) prime = all(number % i for i in range(2, hi_number)) return prime Good luck! Edit: oops in my prime check More on reddit.com

r/learnpython

14

9

July 9, 2019

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