Instead of the sum of prime numbers the output is the sum of all the numbers between the interval Here is my code:

x=int(input("lower limit"))
y=int(input("upper limit"))
s=0
for i in range(x, y 1):
    for j in range(2,int(i ** (0.5)) 1):
        if i % j == 0:
            break
    s  = i
print(s)

CodePudding user response：

You were adding i on every loop iteration(whether there are any factors other than 1 and itself). Instead, add i only when the factor is 1 and i:

x=int(input("lower limit"))
y=int(input("upper limit"))
s=0
for i in range(x, y 1):
    for j in range(2,int(i ** (0.5)) 1):
        if i % j == 0:
            break
    else:
        s  = i
print(s)

Output:

lower limit2
upper limit10
17

CodePudding user response：

We can use a variable found which we reset to 0 before each number tested.

x=int(input("lower limit"))
y=int(input("upper limit"))
s=0
for i in range(x, y 1):
    found=0
    for j in range(2,int(i ** (0.5)) 1):
        if i % j == 0:
           found=1
    if found==0:
        print("adding",i)
        s  = i
print(s)

Example run

lower limit2
upper limit7
adding 2
adding 3
adding 5
adding 7
17

CodePudding user response：

A possible problem with your code, making it harder to debug, is that it is not divided into functions.

Your basic operation is checking if a number is prime, so you should define a function is_prime checking if an integer is prime.

def is_prime(n):
    for j in range(2,int(i ** (0.5)) 1):
        if i % j == 0:
            return False
    return True

Once you have a function, you can test it and check if it works.

After that, you can use your building block and create a more readable and debuggable code:

x=int(input("lower limit"))
y=int(input("upper limit"))
s=0

for i in range(x, y 1):
    if is_prime(i):
        s  = i

print(s)

If can use external packages, there's an implementation for is_prime in the packages sympy and primePy - comparing your results to these implementations is also a good way to test your code.