Cryto-Division

I tried to narrow down the possibilities a lot before crunching numbers!

The equation can be written as $HULK\cdot DAD = BOB\cdot 9999.$ Since $101 | 9999$ is a prime factor, $HULK$ or $DAD$ should be a multiple of 101. But it can't be $HULK$ , for four-digit multiples of 101 are of the form $XY\cdot 101 = XYXY$ , with repeating digits. Therefore $DAD$ is a multiple of 101. It follows immediately that $DAD = 101\cdot D$ , and $A = 0$ . The equation becomes $HULK\cdot D = BOB\cdot 99.$ From the fact that $BOB/DAD < 1$ it follows that $B \leq D$ . In fact, since $O > A = 0$ , we know that $B < D$ . This immediately rules out $D = 1$ and $B = 9$ .

Further analysis of prime factors shows that $11|HULK$ . The quotient $HULK/11 = ??K$ ends in the same digit as $HULK$ . This gives $??K \cdot D = BOB \cdot 9$ ; because $K \not= B$ it follows that $D \not= 9$ and $B \leq 8$ .

It is also easy to see that $K, D, B \not= 5$ : if $K$ or $D$ were equal to 5, then $B$ should be 0 or 5, which is not allowed. And $B$ can only be 5 if either $K$ or $D$ is 5.

We also see that $D$ should be a divisor of $BOB\cdot 9$ . This means that $D|BOB$ , unless $D = 6$ , in which case we only require that $BOB$ is even. (Why is $D = 3$ no exception? We are told that the gcd of $BOB$ and $DAD = 101\cdot D$ is not 1, so these numbers have a common factor. It can't be 101, because $O \not= 0$ ; if $D = 3$ , the only possible common factor is 3, so that in this case also $D | BOB$ .)

This condition of divisibility can be worked out in more detail.

• $D = \text{even}$ : Then $B$ must also be even. Combined with $B < D$ , we rule out $D = 2$ and $B = 8$ . Moreover, for $D = 4$ we have $BOB = 101B + 10O \equiv B + 2O \equiv 0$ mod 4, showing that $B$ is a multiple of four if and only if $O$ is even; for $D = 8$ this condition can be sharpened.

• $D = 3$ : We have $BOB \equiv 2B + O \equiv 0$ mod 3.

• $D = 7$ : We have $BOB = 101B + 10O \equiv 3B + 3O \equiv 3(B+O) \equiv 0$ mod 7, so that $B + O = 7 \ \text{or}\ 14$ .

With this information, the possible values for $B, O, D$ are severely limited! I generated all 29 remaining possibilities by hand, and used a calculator to find $HULK = BOB\cdot 99/D$ :

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D   BOB HULK
3   141 4653                        
3   171 5643                        
3   252 8316
3   282 9306
4   212 5247
4   232 5742
4   252 6237
4   272 6732
4   292 7227
6   212 3498
6   232 3828
6   252 4158
6   272 4488
6   292 4818
6   424 6996
6   484 7986
7   161 2277
7   252 3564
7   343 4851
7   434 6138
7   616 8712
7   686 9702
8   232 2871
8   272 3366
8   424 5247
8   464 5742
8   616 7623
8   656 8118
8   696 8613

The only case where there is no overlap in digits within $HULK$ or between $HULK$ and $D, BOB$ is the combination $D = 6$ , $BOB = 212$ , and $HULK = 3498$ . The total of these digits, together with $A = 0$ , is $\boxed{33}$ .

Extra: The condition that gcd $(BOB,DAD) \not= 1$ is essential. From the discussion above, it played a role when $D = 3$ . Indeed, if we allow $D = 3$ but $3\not|BOB$ , we find an additional solution $D = 3$ , $BOB = 242$ and $HULK = 7986$ .

Here's the short to type, brute force BASH in Python!

It uses a bit of reduction by Math, that $0.HULKHULK... = \dfrac{HULK}{9999}$ .

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from fractions import gcd
from itertools import permutations

for i in permutations('0123456789'):
    k=''.join(i)
    [B,O,D,A,H,U,L,K]=list(k)[:8]
    if B!='0' and D!='0':
        if int(B+O+B)*9999==int(H+U+L+K)*int(D+A+D) and gcd(int(B+O+B),int(D+A+D))!=1:
            print int(B)+int(O)+int(D)+int(A)+int(H)+int(U)+int(L)+int(K)

It prints the answer $\boxed{33}$ .

And I feel so happy, people haven't forgotten Mistakes Give Rise To Problems ,

because Julian Poon has posted Mistakes do give rise to Problems

x = 0.HULKHULKHULK...

10000 x = HULK.HULKHULKHULK...

9999 x = HULK

x = $\frac{HULK}{9999}$ = $\frac{BOB}{DAD}$ {Not visualizing simplification allowed.}

With B, D, H from 1 to 9, and O, U, L, K and also A from 0 to 9, where among all alphabets are not equal, AND 1009899 B + 99990 O = (1000 H + 100 U + 10 L + K)(101 D + 10 A):

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212 606 0.349834983498350
242 303 0.798679867986799

$\frac{212}{606} = \frac{106}{303}$ but not applicable in the context for verification.

gcd (212, 606) $\neq$ 1 wanted means $\frac{212}{606} = 0.349834983498...$ wanted, although quite unusual.

2 + 1 + 6 + 0 + 3 + 4 + 9 + 8 = 33

Answer: $\boxed{33}$

I used a Lua program where it prints a list of factors which shows a recurring 4-digit to compute this and i tried the first result with unique digits: BOB = 212 DAD = 606 0.HULK = 0.3498... Then computed for bodahulk(w/c i forgot to include in my program): 2+1+6+0+3+4+9+8=33