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-digit numbers is 9009 = 91 * 99.

Find the largest palindrome made from the product of two 3-digit numbers.

I have this code here

def isPalindrome(num):
    return str(num) == str(num)[::-1]
def largest(bot, top):
    for x in range(top, bot, -1):
        for y in range(top,bot, -1):
            if isPalindrome(x*y):
                return x*y
print largest(100,999)

It should find the largest palindrome, it spits out 580085 which I believe to be correct, but project euler doesn't think so, do I have something wrong here?

When I revered the for loop I didn't think it through, I removed the thing that checks for the biggest, silly me. Heres the working code

def isPalindrome(num):
    return str(num) == str(num)[::-1]
def largest(bot, top):
    z = 0
    for x in range(top, bot, -1):
        for y in range(top,bot, -1):
            if isPalindrome(x*y):
                if x*y > z:
                    z = x*y
    return z
print largest(100,999)

it spits out 906609

http://forum.projecteuler.net/viewforum.php?f=50&sid=96f5f001e9fdeae0f2af2e049d872f9e and more specifically, http://forum.projecteuler.net/viewtopic.php?f=50&t=1989 — Andbdrew, Oct 01 '12 at 13:57

score 10 · Accepted Answer · answered Oct 01 '12 at 13:40

10

Iterating in reverse doesn't find the largest x*y, it finds the palindrome with the largest x. There's a larger answer than 580085; it has a smaller x but a larger y.

answered Oct 01 '12 at 13:40

John Kugelman

349,597
67
533
578

Hmm, lets change this around a bit, brb. – FabianCook Oct 01 '12 at 13:41
Agreed. Rather than returning as soon as you find a palindrome, you need to test every combination and keep track of the largest. – japreiss Oct 01 '12 at 13:41

score 5 · Answer 2 · answered Oct 01 '12 at 13:48

5

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.

answered Oct 01 '12 at 13:48

Jon Clements

138,671
33
247
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Im only using python for this because it is an interpreted language, otherwise I would use java – FabianCook Oct 01 '12 at 13:50
@SmartLemon fair enough - Haskell's very useful as well though ;) – Jon Clements Oct 01 '12 at 13:50
@SmartLemon For these kind of things - definitely - look at the haskell solutions on the Euler answer board for the ones you've already solved - you'll find code snippet (I think there's also a website which has the code written for the first 'n' many problems in Haskell as well) – Jon Clements Oct 01 '12 at 13:54

score 2 · Answer 3 · answered Jul 13 '14 at 09:27

2

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])

answered Jul 13 '14 at 09:27

user3833942

125
1
6

score 1 · Answer 4 · answered Sep 26 '13 at 08:31

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)

CooLee · Answer 5 · 2015-01-21T15:04:25.240

0

Here I added two 'break' to improve the speed of this program.

def is_palindrome(num):
    return str(num) == str(num)[::-1]
def max_palindrome(n):
    max_palindrome = 1
    for i in range(10**n-1,10**(n-1)-1,-1):
        for j in range(10**n-1,i-1,-1):
            if is_palindrome(i*j) and i*j > max_palindrome:
                max_palindrome = i * j
                break
            elif i*j < max_palindrome:
                break
    return max_palindrome
n=int(raw_input())
print max_palindrome(n)

edited Jan 21 '15 at 15:04

answered Jan 20 '15 at 15:01

CooLee

101
2

1

Comments to clarify your code are highly appreciated – olyv Jan 20 '15 at 15:13

score 0 · Answer 6 · answered Mar 01 '15 at 03:30

Simple:

def is_pallindrome(n):
    s = str(n)
    for n in xrange(1, len(s)/2 + 1):
        if s[n-1] != s[-n]:
            return False    
    return True

largest = 0
for j in xrange(100, 1000):
    for k in xrange(j, 1000):
        if is_pallindrome(j*k):
            if (j*k) > largest: largest = j*k
print largest

score 0 · Answer 7 · answered Aug 01 '16 at 12:27

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

score 0 · Answer 8 · answered Jul 03 '17 at 23:28

Here is my code to solve this problem.

lst = []
for i in range(100,1000): 
    for n in range(2,i) : 
        lst.append (i* n) 
        lst.append(i*i)

lst2=[]
for i in lst: 
    if str(i) == str(i)[::-1]:
        lst2.append(i)
print max(lst2)

Abdul Hasib Uddin · Answer 9 · 2020-09-10T14:32:10.100

0

580085 = 995 X 583, where 906609 = 993 X 913. Found it only by applying brute-forcing from top to bottom!

edited Sep 10 '20 at 14:32

answered Oct 28 '17 at 17:41

Abdul Hasib Uddin

11
2

Irmak Aydeniz · Answer 10 · 2017-12-21T13:59:36.713

0

Here is my Python code:

max_pal = 0
for i in range(100,999):
    for j in range(100,999):
      mult = i * j 
      if str(mult) == str(mult)[::-1]: #Check if the number is palindrome
          if mult > max_pal: 
              max_pal = mult
print (max_pal)

edited Dec 21 '17 at 13:59

answered Dec 18 '17 at 12:39

Irmak Aydeniz

21
4

It would be useful adding a few lines noting/explaining your approach. – Yannis Dec 18 '17 at 12:56

score 0 · Answer 11 · answered Aug 02 '19 at 22:29

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())

Ronald Dela Peña · Answer 12 · 2020-09-21T15:47:43.907

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)

score 0 · Answer 13 · answered Feb 19 '21 at 10:12

Here is my solution for that:

lst1 = [x for x in range(1000)]
palindrome = []

def reverse(x):
    a = str(x)[::-1]
    return int(a)

x = 0
while x < len(lst1):
    for y in range(1000):
        z = lst1[x] * y
        if z == reverse(z):
            palindrome.append(z)
    x += 1

duppal = set(palindrome)
sortpal = sorted(duppal)
total = sortpal[-1]
print(sortpal)
print('Largest palindrome: ' + str(total))

Rudney · Answer 14 · 2014-04-26T22:10:10.113

-1

ReThink: efficiency and performance

def palindrome(n):    

    maxNumberWithNDigits = int('9' * n) #find the max number with n digits

    product = maxNumberWithNDigits * maxNumberWithNDigits 

    #Since we are looking the max, stop on the first match

    while True:        
        if str(product) == str(product)[::-1]: break;

        product-=1

    return product

start=time.time()
palindrome(3)
end=time.time()-start

palindrome...: 997799, 0.000138998031616 secs

edited Apr 26 '14 at 22:10

answered Apr 26 '14 at 20:58

Rudney

1
1

2

The palindrome 997799 is not a product of two 3 digit numbers, it has 11 and 90709 as prime factors. – rakvium Dec 01 '16 at 15:19

highest palindrome with 3 digit numbers in python

14 Answers14

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

Function:

Linked