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i think it should be 9999*9901 = 9900 0099

since we're looking for the largest palindrome no, it's natural to take 9999
as one of the factors.

mul by 9999 is easy, ie.: x * 9999 = x * 10000 - x

hence we need to find the largest 4-digit no, say, 'abcd', s.t.,

abcd 0000 - abcd = mnpq qpnm
from this equation one can derive that it is 9901

Irrelevant.

You can write a program to find the largest palindromic binary representation that results from multiplying numbers together without converting to string. But you can't find the largest arabic numeral palindrome without converting to string because arabic numeral are chars.

#include <stdio.h>
#include <stdbool.h>
#include <math.h>

int main (int argc, char *argv[]){
bool found = false;
  for (int i = 998; i >= 100; i--)
  {
    char j[7];
    sprintf(j,"%d",i);
    j[3]= j[2];
    j[4]= j[1];
    j[5]= j[0];
    int x =atoi(j);
    int limit = sqrt((float) x);
    for (int z = 999; z >= limit; z--)
    {
      if (x%z==0){
        printf("%d",x);
        found = true;
        break;
      }
    }
    if (found) break;
  }
}

http://stackoverflow.com/questions/3250957/finding-the-largest-palindrome-of-the-product-of-two-three-digit-numbers-problem

{
void LargestPalindrome()
{
unsigned int Res = 0;
unsigned int A = MyPow(10,7) + 1;
unsigned int B = MyPow(10,6) + MyPow(10,1);
unsigned int C = MyPow(10,5) + MyPow(10,2);
unsigned int D = MyPow(10,4) + MyPow(10,3);


for(unsigned int i = 9999; i >= 1000; i--)
{
for(unsigned int a = 9; a >= 1; a--)
{
for(int b = 9; b >= 0; b--)
{
for(int c = 9; c >= 0; c--)
{
for(int d = 9; d >= 0; d--)
{
Res = A*a + B*b + C*c + D*d;

if(Res % i == 0 && Res / i <= 9999)
{
cout << i << " " << Res << " " << Res / i << endl;
return;
}
}
}
}
}
}
}
}

As noted elsewhere, a 6-digit base-10 palindrome abccba is a multiple of 11. This is true because a*100001 + b*010010 + c*001100 is equal to 11*a*9091 + 11*b*910 + 11*c*100. So, in our inner loop we can decrease n by steps of 11 if m is not a multiple of 11.

bool ispalin (long num){
   long newnum=0;
   long tmp;
   int digit;
   while(num) {
       digit = num%10;
       num = num/10;
       newnum = newnum + digit*10;
    }
    if (num == newnum)
      return 1;
    else
      return 0;

}

long findlargestpalin () {
    res = 0;
    for (i = 9999 ; i > 1001 ; i--) {
        for (j = i ; j > 1000; j = j - 11) {
             if (j%11!=0) {
                 div = j/11;
                 j = div * j;
             }
             prod = i*j;
             if (ispalin(prod) && prod > res)
                 res = prod;
        }
    }
    return res;
}