highest palindrome with 3 digit numbers in python

Question

In problem 4 from http://projecteuler.net/ it says:

A palindromic number reads the same both ways. The largest palindrome made from the product of two 2-d

Answer 1

Tried making it more efficient, while keeping it legible:

def is_palindrome(num):
    return str(num) == str(num)[::-1]

def fn(n):
    max_palindrome = 1
    for x in range(n,1,-1):
        for y in range(n,x-1,-1):
            if is_palindrome(x*y) and x*y > max_palindrome:
                max_palindrome = x*y
            elif x * y < max_palindrome:
                break
    return max_palindrome

print fn(999)

Answer 2

Each time it doesnot have to start from 999 as it is already found earlier.Below is a simple method using string function to find largest palindrome using three digit number

def palindrome(y):
    z=str(y)
    w=z[::-1]
    if (w==z):
        return 0
    elif (w!=z):
        return 1        
h=[]
a=999
for i in range (999,0,-1):
    for j in range (a,0,-1):
    l=palindrome(i*j)
    if (l==0):
        h=h+[i*j]               
    a-=1
print h
max=h[0]

for i in range(0,len(h)):
    if (h[i] > max):
    max= h[i]
print "largest palindrome using multiple of three digit number=%d"%max  

Answer 3

This would more efficiently be written as:

from itertools import product

def is_palindrome(num):
    return str(num) == str(num)[::-1]

multiples = ( (a, b) for a, b in product(xrange(100,999), repeat=2) if is_palindrome(a*b) )
print max(multiples, key=lambda (a,b): a*b)
# (913, 993)

You'll find itertools and generators very useful if you're doing Euler in Python.

Answer 4

def div(n):
    for i in range(999,99,-1):
        if n%i == 0:
            x = n/i
            if x % 1 == 0:
                x = n//i
                if len(str(x)) == 3:
                    print(i)
                    return True
    return False


def palindrome():
    ans = []
    for x in range(100*100,999*999+1):
        s = str(x)
        s = int (s[::-1])
        if x - s == 0:
            ans.append(x)

    for x in range(len(ans)):
        y = ans.pop()
        if div(y):
            return y


print(palindrome())

Answer 5

Here is the function I made in python to check if the product of 3 digit number is a palindrome

Function:

def is_palindrome(x):
    i = 0
    result = True
    while i < int(len(str(x))/2):
        j = i+1
        if str(x)[i] == str(x)[-(j)]:
            result = True
        else:
            result = False
            break
        i = i + 1
    return result

---

Main:

---

max_pal = 0
for i in range (100,999):
    for j in range (100,999):
        x = i * j
        if (is_palindrome(x)):
            if x > max_pal:
                max_pal = x

print(max_pal)

Answer 6

Not the most efficient answer but I do like that it's compact enough to fit on one line.

print max(i*j for i in xrange(1,1000) for j in xrange(1,1000) if str(i*j) == str(i*j)[::-1])