Simple cryptographic puzzle

Question

I'm looking for a way to solve this crypt arithmetic problem of:

ROBERT +  GERALD =  DONALD

and potentially others as well, where each letter represents a digit.

How would you go about solving this by hand and how does that relate to solving it programmatically?

Thank you in advance

Answer 1

You can actually work this out as a sum:

  robert
+ gerald
  ------
= donald

and use basic mathematical knowledge.

For example, there's no carry possible in the right column (6) and we have T + D = D. That means T must be zero.

Similarly, for column 5, there's no carry from column 6 and R + L = L means R is zero as well, and no carry to column 4.

Same with column 4, E + A = A so E is zero.

So we now have:

  0ob000
+ g00ald
  ------
= donald

From there, we can infer from columns 3 and 1 that b==n and g==d and they (along with o/a/l/d) can be any value since every digit is being added to zero so there is no chance of carry anywhere. So let's just make them all one:

  011000
+ 100111
  ------
= 111111

In fact, you could make them all zero and end up with 000000 + 000000 = 000000.

But that's hardly programming related, so let's make it so:

#include <stdio.h>
int main (void) {
 int robert, gerald, donald;
 for (int r = 0; r < 10; r++) {
  for (int o = 0; o < 10; o++) {
   for (int b = 0; b < 10; b++) {
    for (int e = 0; e < 10; e++) {
     for (int t = 0; t < 10; t++) {
      for (int g = 0; g < 10; g++) {
       for (int a = 0; a < 10; a++) {
        for (int l = 0; l < 10; l++) {
         for (int d = 0; d < 10; d++) {
          for (int n = 0; n < 10; n++) {
           robert = r * 100000 + o * 10000 + b * 1000 + e * 100 + r * 10 + t;
           gerald = g * 100000 + e * 10000 + r * 1000 + a * 100 + l * 10 + d;
           donald = d * 100000 + o * 10000 + n * 1000 + a * 100 + l * 10 + d;
           if (robert + gerald == donald) {
            printf ("  %06d\n", robert);
            printf ("+ %06d\n", gerald);
            printf ("  ------\n");
            printf ("= %06d\n", donald);
            printf ("........\n");
           }
          }
         }
        }
       }
      }
     }
    }
   }
  }
 }
 return 0;
}

That will give you a whole host of solutions.

And, before you complain that you cannot have repeated digits, there is no solution if that's the case, since mathematically both T and R must be zero, as shown in the original reasoning above. And you can prove this empirically with:

#include <stdio.h>
int main (void) {
 int robert, gerald, donald;
 for (int r = 0; r < 10; r++) {
  for (int o = 0; o < 10; o++) {
   if (o==r) continue;
   for (int b = 0; b < 10; b++) {
    if ((b==r) || (b==o)) continue;
    for (int e = 0; e < 10; e++) {
     if ((e==r) || (e==o) || (e==b)) continue;
     for (int t = 0; t < 10; t++) {
      if ((t==r) || (t==o) || (t==b) || (t==e)) continue;
      for (int g = 0; g < 10; g++) {
       if ((g==r) || (g==o) || (g==b) || (g==e) || (g==t)) continue;
       for (int a = 0; a < 10; a++) {
        if ((a==r) || (a==o) || (a==b) || (a==e) || (a==t) || (a==g)) continue;
        for (int l = 0; l < 10; l++) {
         if ((l==r) || (l==o) || (l==b) || (l==e) || (l==t) || (l==g) || (l==a)) continue;
         for (int d = 0; d < 10; d++) {
          if ((d==r) || (d==o) || (d==b) || (d==e) || (d==t) || (d==g) || (d==a) || (d==l)) continue;
          for (int n = 0; n < 10; n++) {
           if ((n==r) || (n==o) || (n==b) || (n==e) || (n==t) || (n==g) || (n==a) || (n==l) || (n==d)) continue;
           robert = r * 100000 + o * 10000 + b * 1000 + e * 100 + r * 10 + t;
           gerald = g * 100000 + e * 10000 + r * 1000 + a * 100 + l * 10 + d;
           donald = d * 100000 + o * 10000 + n * 1000 + a * 100 + l * 10 + d;
           if (robert + gerald == donald) {
            printf ("  %06d\n", robert);
            printf ("+ %06d\n", gerald);
            printf ("  ------\n");
            printf ("= %06d\n", donald);
            printf ("........\n");
           }
          }
         }
        }
       }
      }
     }
    }
   }
  }
 }
 return 0;
}

which outputs no solutions.

Now DONALD + GERALD = ROBERT, that's a different matter but you can solve that simply by modifying the code above slightly, making the if statement into:

if (donald + gerald == robert) {
 printf ("  %06d\n", donald);
 printf ("+ %06d\n", gerald);
 printf ("  ------\n");
 printf ("= %06d\n", robert);
 printf ("........\n");
}

and you get the single solution:

  526485
+ 197485
  ------
= 723970