Largest palindrome product - euler project

Question

I was trying to solve project Euler problem 4 which is:

A palindromic number reads the same both ways. The largest palindrome made from the product of tw

Answer 1

Here is the Python code for the Project_Euler-4 problem.

We have to find the largest palindrome number which is a product of two three digit numbers

import math
def isPal(x):
    x=str(x)
    t=x[::-1]
    if(x==t):
        return True
    else:
        return False
max=0
for i in range(999,99,-1):
    for j in range(999,99,-1):
        if(isPal(i*j)):
            if((i*j)>max):
                max=(i*j)
print(max)

The answer will be 906609

Answer 2

The for loop runs for i from 998001 down to 100000. Nowhere in your program are you checking that i can actually be the product of two 3-digit numbers.

Answer 3

Well I am seeing a lot of things wrong here.

• First of all you are using multiplication of 2 highest 3 digit numbers and then decrementing it to find palindrome. What you need to do according to question is to use variables having highest 3 digit no.s and then decrement them to check there resultant product.
• Second for checking if the no. is palindrome you used an array to store it then used a loop to check it, I find it incorrect, as you could simply store the resultant no. in another integer variable by using the simple approach.(reverseNum * 10 + (num % 10) )

And I am seeing a correct code already posted by a user (ROMANIA)

Answer 4

Here is a solution that doesn't iterate through all the 6-digit numbers:

public static boolean isPalindrome(int nr) {
    int rev = 0;                    // the reversed number
    int x = nr;                     // store the default value (it will be changed)
    while (x > 0) {
        rev = 10 * rev + x % 10;
        x /= 10;
    }
    return nr == rev;               // returns true if the number is palindrome
}

public static void main(String[] args) {

    int max = -1;

    for ( int i = 999 ; i >= 100 ; i--) {
        for (int j = 999 ; j >= 100 ; j-- ) {
            int p = i * j;
            if ( max < p && isPalindrome(p) ) {
                max = p;
            }
        }
    }
    System.out.println(max > -1? max : "No palindrome found");
}

Edit:

An improved solution for the main method ( according to Peter Schuetze ) could be:

public static void main(String[] args) {

    int max = -1;

    for ( int i = 999 ; i >= 100 ; i--) {
        if ( max >= i*999 ) { 
            break;
        }
        for (int j = 999 ; j >= i ; j-- ) {             
            int p = i * j;
            if ( max < p && isPalindrome(p) ) {
                max = p;
            }
        }
    }       
    System.out.println(max > -1? max : "No palindrome found");
}

For this particular example, the time is about 2 times better, but if you have bigger numbers, the improvement will be more significant.

Output:

906609

Answer 5

You are decrementing i sequentially from 999*999 to 100 *100. It does not necessarily mean that the first palindrome you are finding is a product of two 3 digit numbers.

The palindrome 997799 has 11 and 90709 as prime factors which is not a product of two 3 digit numbers.

Answer 6

Here is the code in c++

#include <iostream>
using namespace std;
int reverse(int a){
int reverse=0;
for(int i=0;i<6;i++){
    reverse = reverse*10+a%10;
    a/=10;
}
return reverse;
}
int main(){
int a=999,max=1,rev;;
int b=999;
for(a=999;a>100;a--){
    for(b=999;b>100;b--){
        int p = a*b;
         rev = reverse(p);
        if (p==rev) {
            if(max<rev){
                max = rev;
            }
        }
    }
}
cout<<"\n"<<max<<"\n";
return 0;
  }