I am not allergic to long division

I used logical deductions to get the answer. First let us assign the unknown digits as follows:

$\quad \quad \quad \quad \quad DEAF\\BAC ) GHIJAKL\\ \quad \quad MNAA\\ \quad \quad \quad \quad OPQA\\\quad \quad \quad \quad \quad RSA\\\quad \quad \quad \quad \quad TUVK\\\quad \quad \quad \quad \quad WAXY\\\quad \quad \quad \quad \quad \quad T \Delta \Theta L \\\quad \quad \quad \quad \quad \quad T \Delta \Theta L\\\\\quad \quad \quad \quad \quad \quad \quad \quad 0$

We note that $BAC \times D = MNAA$ and $AC \times D = ?AA$ . Assume $A=0$ , then $C \ne 0$ , $D \ne 0$ and to get the last digit to be $A$ or $0$ . then $0C \times D = 2 \times 5$ , $5 \times 2$ , $4 \times 5$ , $5 \times 4$ , $6 \times 5$ , $5 \times 6$ , $8 \times 5$ , or $5 \times 8$ , But it can noted that all the products above don't give second last digit to be $0$ , therefore, $A \ne 0$ . Do the same elimination process for $A =1,2,3,4,5,6,7,8,9$ . And it is found that the possible cases are $AC \times D=37 \times 9 = 333, 43 \times 8 = 344, 48 \times 3 = 144$ or $84 \times 7 = 588$ .

Now we look at $AC \times E = ?SA$ of $BAC \times E = RSA$ for all cases of $AC$ . For $AC=37$ , the only possible case for the last digit of the product $37 \times E$ to be $3$ is $E=9$ , but $37 \times 9 = 333$ which means that $S=3=A$ and it is unacceptable. Similarly $43 \times 8 = 344$ and $48 \times 3 = 144$ are unacceptable. Next is $48 \times 8 = 384$ which means that $E=8$ and that $B48 \times 8 \le 1184 \ne R84$ , again unacceptable. Left $84 \times 2 = 168$ which means $A=8$ , $C=4$ , $E=2$ and $S=6$ . For $BAC \times E = RSA$ , the possible $B=1,2,3,4$ .

Now consider $BAC \times A = WAXY$ ; the only possible case is $484 \times 8 = 3872$ . So $B=4$ , $W=3$ , $X=7$ and $Y=2$ .

From $BAC \times D = MNAA$ ; the only possible case is $484 \times 7 = 3388$ . So $D=7$ and $M=N=3$ .

From $BAC \times F = Z \Delta \Theta L$ for all $F=0,1,2,3,4,5,6,7,9$ and checking for the validity of $TUVK - WAXY = Z \Delta \Theta$ , it is found that it is only possible when $F=9$ . Therefore, the divisor $BAC=484$ , the quotient $DEAF=7289$ and the dividend is $7289 \times 484 = 3527876$ and the sum of divisor, dividend and quotient is $484+3527876+7289=\boxed{3535649}$ .

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Or, this for a programming solution

Solution can be found through a series of logical deductions. My full solution can be found at this link: https://beyondmathsolutions.wordpress.com/2014/08/22/soln-feynmans-division-problem/

(Anyone know how to insert hyperlinks in a solution?)