pyPrimeChecker - Revised

# Check If A Number Is Prime
# by Mike Kerry - Revisited July 2021 - acclivity2@gmail.com
import time

def checkPrime(number):
    if number < 4:
        return number > 1               # 2 and 3 return True, 0, 1 and all negative numbers return False
    if number % 6 not in (1, 5):        # Any prime > 3, divided by 6 gives a remainder of 1 or 5
        return False
    # Compute 1 plus the square root of the number to be tested. This will be the largest divisor we need to consider
    sqrt = 1 + int(number ** 0.5)

    # Test the remainder for each division of divisors up to the square root
    # We start at 5 and only need to go up to the square root of number, and only in steps of 2
    for divisor in range(5, sqrt, 2):
        if not number % divisor:
            return False
    return True


aa = [1, 2, 3, 4, 5, 6, 7, 8, 9, 100, 101, 120, 121, 127, 131, 163, 12345, 54321, 987654323]
for a in aa:
    print(str(a).ljust(12), checkPrime(a))


# Find the one-millionth prime number. Measure the execution time
st = time.time()
a, b = -1, 0
while b < 1000000:
    a += 1
    b += checkPrime(a)

print(a, (time.time() - st))

# RESULTS
# 1            False
# 2            True
# 3            True
# 4            False
# 5            True
# 6            False
# 7            True
# 8            False
# 9            False
# 100          False
# 101          True
# 120          False
# 121          False
# 131          True
# 127          True
# 163          True
# 12345        False
# 54321        False
# 987654323    True

# 15485863     this is the one-millionth prime number, computed in 284 seconds (i7-2670QM processor)