Python Program to Print all Prime Numbers in an Interval - Complete Tutorial

[@wscubetech]

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[@saurabhsingh5236]

Understanding the else in a for loop: • The else block after a for loop executes only if the loop is not terminated by a break statement. • In this case, the else block is executed if no divisors are found for the current num. Step-by-Step Execution: 1. Outer loop (for num in range...): • Iterates through numbers between Lower and upper + 1. 2. Inner loop (for i in range(2, num)): • Checks divisors of num starting from 2 up to num - 1. • If a divisor is found (num % i == 0), the break statement terminates the loop. 3. Else block: • Executes only if the inner for loop completes without a break (i.e., num has no divisors and is therefore prime).

[@vaibhavv475]

More explanation is needed as to how the value continues to be printed......

[@PythonProjectSolver]

why 2 is consider as a prime number? because when 2 going to second for loop it is range of (2,2) than it check 2%2 that is equal to 0 than that is not a prime number. the solution is : find_p = range(1, 101) for p in find_p: if p > 1: for x in range(2, int(p**0.5) + 1): if p % x == 0: break else: print(p)

[@venkatagovindpiradi1397]

I'm also thinking the same😢

[@sushantchougale6818]

Thank you very much Madam 😊

[@Picture_of_truth]

start=int(input("enter starting no:")) end=int(input("enter ending no:")) num=0 for num in range(2,end): for i in range(2,num): if num%i==0: break; else: print(num)

[@shaikhkashif9973]

Thanks

[@velocxty2477]

Great explanation ma’am!⚡️

[@durgaprasadbandari968]

great explanation

[@bhavyasachula7006]

thanks a lott

[@__ayu]

THANK YOU...

[@Neerajahir02.ahir.]

Nice

[@agrimgupta0805]

we can make it more optimized?

[@ashsaha5750]

Code is not working... Expexted indented block saying... irritating

[@mohammadajharuddeen8933]

Reach native ka full course upload kro please

[@Abdul-tu6mq]

Sprrrrrrrrr maam

[@dawood3578]

mam I have a question. If I try to print else: statement with a tab space, right under second if, it prints the prime number more times than one. But when I write it like you have, behind the if: statement, it prints correctly. Isn't it necessary to use else: directly under if: statement. Please let me know. Thanks!

[@ROHITH_PRO_51]

Same doubt bro

[@nayansharma224]

same doubt

[@saurabhsingh5236]

@@nayansharma224 Understanding the else in a for loop: • The else block after a for loop executes only if the loop is not terminated by a break statement. • In this case, the else block is executed if no divisors are found for the current num. Step-by-Step Execution: 1. Outer loop (for num in range...): • Iterates through numbers between Lower and upper + 1. 2. Inner loop (for i in range(2, num)): • Checks divisors of num starting from 2 up to num - 1. • If a divisor is found (num % i == 0), the break statement terminates the loop. 3. Else block: • Executes only if the inner for loop completes without a break (i.e., num has no divisors and is therefore prime).

[@ArtiKumari-rb6lj]

Ma'am prime number repeated in output.

[@Harshdhillon]

Don’t print them in for loop break out from it

[@KritarthShuklaA]

Apna indentation check kro

[@PannaLal-y7o]

Same problem

[@a.dashish2844]

Your second If and else statment must in the block first range function

[@a.dashish2844]

x = int(input(" x: ")) y = int(input(" y: ")) for i in range(x, y+1): if i > 1: # Ensuring we only consider numbers greater than 1 for m in range(2, i): if i % m == 0: break else: print(i) # Only prints if no divisor was found

[@ammarahmad7036]

It's true

[@mekalachinnikrishna8372]

Mam we are using num here but we are not defined num then how your output will come

[@Nannu-l4i]

Cz num koi variable ni h jisme values store krvai ho , simply "i" ko replace Kia h "num" se ,which means items in that range of numbers

[@swastikpanda5670]

this is my code yet i am not getting the correct output: a=int(input("a: ")) # b=int(input("b: ")) # num=0 # for num in range (a,b+1): # if num>1: # for i in range(2,num): # if num%i==0: # break # else: # print(num)

[@Ram-yw5ky]

Num_s = int(input("Enter no =")) end_s = int(input("Enter no =")) for num in range(Num_s,end_s + 1): if num > 1: for i in range(2,num): if num%i == 0: break else: print(num) break Mam mei code run kr rha fir bhi 9 prime kyu show kr rha

[@saiishwarya812]

Because you are writing else statement inside the for In this case while checking for first iteration only it will show as prime

[@kshitijsahdev4480]

"Herro everyone" 😂

[@MuhammadAwais-yt6je]

Unreal engine 5 ka b cours layaa

[@ichchakushwaha3209]

hey i printed "num+1 " at the place of "num", and not getting the "2" in output as a prime number l=int(input("lower limit: ")) u=int(input("upperr limit: ")) for num in range(l,u+1): if num>1: for i in range(2,num+1): if num%i==0: break else: print(num) break this is my code

[@GamingGalaxyFF]

remove the last break command then it will work😊

[@AryanHuria]

Hey you are doing in second loop num+1 , then it's wrong suppose from the first loop you took 99 , and the second loop gets divided by 99 , if you will write num+1 , it will get divided by 99 too and therefore program won't return a single prime number.

[@shashankpandey5187]

check indentation once(your else statement)

[@AmaanKhan-cb4rn]

❤❤❤

[@skcreates343]

Mai es video par pahli baar aaya hu or Python learn karna chahta hu. Kya hum Python me koi app devlop kar sakte hai ? Kya hum Python se online tool bana sakte hai ? Can you guide me god bless you 🙏🙏🙏🌸🌸

[@TilakRaj-qo6ic]

Python is becoming a general purpose language slowly so you can do anything you want

[@Cyclopentasi1_oxane]

i%j==0. Ie... 2%2==0.... thats right? Then how 2 printed in prime

[@kishorem5461]

Same doubt

[@zulfiquarshaikh3461]

@wscubetech pls explain if 2%2 ==0 , then how 2 is getting printed in output

[@PythonProjectSolver]

this is correct code : find_p = range(0, 100) for p in find_p: if p > 1: for x in range(2, int(p**0.5) + 1): if p % x == 0: break else: print(p)

[@PythonProjectSolver]

this is correct code : find_p = range(0, 100) for p in find_p: if p > 1: for x in range(2, int(p**0.5) + 1): if p % x == 0: break else: print(p)@@kishorem5461

[@PythonProjectSolver]

@@zulfiquarshaikh3461 this is correct code : find_p = range(0, 100) for p in find_p: if p > 1: for x in range(2, int(p**0.5) + 1): if p % x == 0: break else: print(p)

[@ZenBeats01]

❤️❤️

[@muzamilrana8854]

1st comment😍

[@sakshikadu7722]

i kya h

[@Nannu-l4i]

Items in range(0, upto to given num)