The 7-Eleven problem (idea)

You can also write a small program to solve this a bit more efficiently, by pre-computing all the divisors of 711000000 that are less than 711, and then only trying those options. Something like this would work:

#include <stdio.h>

int main() {
  unsigned int A, B, C, D;
  unsigned int Aind, Bind, Cind;
  unsigned int divs [61] =
    {1, 2, 3, 4, 5, 6, 8, 9, 10, 12, 15, 16, 18, 20, 24, 25, 30, 32, 36,
    40, 45, 48, 50, 60, 64, 72, 75, 79, 80, 90, 96, 100, 120, 125, 144,
    150, 158, 160, 180, 192, 200, 225, 237, 240, 250, 288, 300, 316, 320,
    360, 375, 395, 400, 450, 474, 480, 500, 576, 600, 625, 632};
  for(Aind=0, A=divs[0]; Aind<61; Aind++, A=divs[Aind])
    for(Bind=0, B=divs[0]; A+B<=709 && Bind<=Aind; Bind++, B=divs[Bind])
      for(Cind=0, C=divs[0]; A+B+C<=710 && Cind<=Bind; Cind++, C=divs[Cind]) {
        D = 711 - (A+B+C);
        if(D<=C) {
          if(A*B*C*D == 711000000) {
            printf("A: %u, B: %u, C:%u, D: %u\n", A, B, C, D);
          }
        }
      }
  return(1);
}

This approach finds the result in something like 350 inner loops. If you are interested in a non-brute-force method to solve this problem, you can take a look at the detailed solution in section 2.2 of this paper.