Max Prime Palindrome in Python

Question

I'm trying to create a program that outputs the highest prime number than is a palindrome and is less than 1000. Expected output should be 929

Attempt 1

number = 1
prime = 1
maxNumber = 1000

while number > maxNumber:
    if str(number) == str(number)[::-1]:        #Inverts the string
        for number in range(number,maxNumber):  #Increments number until 1000
            for i in range(2, number):          #Checks current number and divides all lower
                if number % i == 0:
                    break
                else:
                    prime = number              
print(prime)

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Attempt 2

number = 3
prime = 3
maxNumber = 1000

while number < maxNumber:
    if str(number) == str(number)[::-1]:        #Inverts the string
        for i in range(2, number):
            if number % i == 0:
                break
            else:
                prime = number
        number+=1
print(prime)

---

Attempt 3, I followed the suggestions to separate the two functions to decrease processing time

for number in xrange(1000):
    if str(number) == str(number)[::-1]: and is_prime(number):
        prime = number
print(number)

#Method to find prime numbers
def is_prime(n):
    if n == 2 or n == 3: 
        return True
    elif n < 2 or n%2 == 0: 
        return False
    elif n < 9: 
        return True
    elif n%3 == 0: 
        return False
    r = int(n**0.5)
    f = 5
    while f <= r:
        if n%f == 0: 
            return False
        if n%(f+2) == 0: 
            return False
        f +=6
    return True   

---

Attempt 4: Receiving the error [name 'is_prime' is not defined]

for number in range(1000,3,-1):
    if str(number) == str(number)[::-1] and is_prime(number):      
        print(number)  
        break  

#Method to check if number is prime  
def is_prime(n):
    if n == 2 or n == 3: return True
    if n < 2 or n%2 == 0: return False
    if n < 9: return True
    if n%3 == 0: return False
    r = int(n**0.5)
    f = 5
    while f <= r:
        if n%f == 0: return False
        if n%(f+2) == 0: return False
        f +=6
    return True  

---

Final Solution: Thank you all for your help. This has been more helpful than I have ever expected.

#Method to check if number is prime  
def is_prime(n):
    if n == 2 or n == 3: return True
    if n < 2 or n%2 == 0: return False
    if n < 9: return True
    if n%3 == 0: return False
    r = int(n**0.5)
    f = 5
    while f <= r:
        if n%f == 0: return False
        if n%(f+2) == 0: return False
        f +=6
    return True  

#Checking for numbers that are palindromes and prime
for number in range(1000,3,-1):
    if str(number) == str(number)[::-1] and is_prime(number):
        print(number)  
        break

Answer 1

There are several issues with your code:

1. First version, your logic for the while loop is backward
2. Second version, your indentation is wrong. number+=1 is part of the if block above and the print is not part of the while loop.
3. The print at the end of the while loop is print the wrong value since it has dropped out of the loop at this point for the test while number < maxNumber:.

Try:

for n in xrange(1000):
    if str(n)==str(n)[::-1] and is_prime(n):
        print n

Which you can turn into a while loop easily:

n=0
while n<1000:
    if str(n)==str(n)[::-1] and is_prime(n):
        print n
    n+=1    

I suggest separating out the prime testing from the palindrome testing. Since testing a string for being a palindrome is fast in comparison to testing if a number is a prime (for larger numbers) test for a palindrome first.

There is a function for testing a number as being a prime here.

---

Based on your comments, I now see you are looking for the max and not all of the prime palindromes.

To get the max prime palindrome, you can just step backwards from the max value. Since you do not know if that max value is prime or not or even or not, you need to step by -1 (or write some additional code that confuses the concept):

for number in range(1000,3,-1):
    if str(number) == str(number)[::-1] and is_prime(number):      
        print(number)  
        break  

Which you can make 'Pythonic' by just using next and a generator:

>>> next(n for n in range(1000,3,-1) if str(n)==str(n)[::-1] and is_prime(n)) 
929

Or, use max with a list of the primes (way less efficient since you have to generate them all):

>>> max(n for n in range(1000) if str(n)==str(n)[::-1] and is_prime(n)) 
929