Determine all possible positive integers, two or more digits long, like ABC...XYZ, such that ABC...XY0Z is a multiple of it.

Note: digits may be repeated.

//If Brute-Force is the only approach, following C++

//program should help.

int i=10;    //starting with a positive non-zero two digit number
int lastdigit=i mod 10;   //last digit of number i
int divident= (i-lastdigit) * 10 + lastdigit;

//the bigger number (ABC...XY0Z)

while (true)
{
 if (lastdigit mod i == 0)
     cout<<i<<";"<<lastdigit<<"\n";

 i++;
 lastdigit=i mod 10;
 divident= (i-lastdigit) * 10 + lastdigit;
}

Posted by elementofsurprize on 2007-01-04 05:29:44