Mistakes do give rise to problems

The equation can be written as $HULK\cdot DAD = BOB\cdot 9999.$ We rule out $D = 9$ because that would give $HULK$ and $BOB$ equal last digits. The fact that the quotient is less than 1 shows $1\leq B < D$ . Thus $D \geq 2$ and $B \leq 7$ .

Since $\text{gcd}(DAD,BOB) = 1$ , these have no common factors. Therefore $9999 = n\cdot DAD$ and $HULK = n\cdot BOB$ , for a certain factor $n | 9999$ . From the numbers of digits of (\DAD) it is clear that $4 \leq n < 50$ .

Since 9999 is divisible by 101, $DAD = 101\cdot D$ . This implies that $A = 0$ and $n\cdot D = 99$ . This means that $D = 3, n = 33$ : $HULK = 33\cdot BOB.$ Since $B < D$ , $B$ must be equal to 1 or 2. But $B = 1$ would make $K = 3 = D$ , which is impossible. Therefore $B = 2$ .

For $O$ we naturally rule out $0, 2, 3$ . Also, since $D \not| BOB$ we have $BOB \equiv 2B + O \not\equiv 0$ mod 3. Therefore $O \not= 5, 8$ . This leaves five possibilities.

1
2
3
4
5

BOB = 212  HULK = 6996
BOB = 242  HULK = 7986 *
BOB = 262  HULK = 8646
BOB = 272  HULK = 8976
BOB = 292  HULK = 9636

Only the second option has no overlap of digit values. Thus the solution of the cryptogram is $\frac{242}{303} = 0.798679867968\dots$ The solution of the question is $2 + 4 + 3 + 0 + 7 + 9 + 8 + 6 = \boxed{39}$ .

(11^3)*6/9999=242/303

=0.79867986.....

Gives

Answer=39