I need to find the sum of all the prime numbers under 2,000,000. This code takes way too long

Question

##Problem 10
primes=[]
sumofprimes=0
for number in range(1,2000000):
##This goes through all the potential primes.
    p=0
    for tester in range(2,number):
        if ((number%tester)==0):
            break
        else:
##every time the number is not divisible by something it adds 1 to p and when everything up to the number is not divisible by anything then it is a prime.
            p=p+1
        if(p==(number-2)):

the "number minus 2" is because lets say you have the prime "3" it will only test "2" so p will equal 1 or (3-2).

            sumofprimes=sumofprimes+number
print(sumofprimes)

score 0 · Answer 1 · answered Apr 27 '18 at 07:30

You could upgrade your second loop using this theorem : Fermat's little theorem

Basically, if sqrt(p) is prime, p is. So you can create a loop like this (not tested):

for tester in range(3,sqrt(number), 2):
    if ((number%tester)==0):
        break

(no need to past through multiple of 2)

I need to find the sum of all the prime numbers under 2,000,000. This code takes way too long

the "number minus 2" is because lets say you have the prime "3" it will only test "2" so p will equal 1 or (3-2).

1 Answers1