Feynman Long division

$\LARGE{ \begin{array}{rll} \phantom{0}\ \mathrm{7} \ \mathrm{8} \ \mathrm{0} \ \mathrm{9} && \\[-1pt] \mathrm{1} \ \mathrm{2} \ \mathrm{7} \ \enclose{longdiv}{ \mathrm{9} \ \mathrm{9} \ \mathrm{1} \ \mathrm{7} \ \mathrm{5} \ \mathrm{0} }\kern-.2ex \\[-1pt] \underline{ \mathrm{8} \ \mathrm{8} \ \mathrm{9} \ \phantom 0 \ \phantom 0 \ \phantom 0 } \\[-1pt] { \mathrm{1} \ \mathrm{0} \ \mathrm{2} \ \mathrm{7} \ \phantom0 \ \phantom0 \ }\kern-.2ex \\[-1pt] \underline{ \mathrm{1} \ \mathrm{0} \ \mathrm{1} \ \mathrm{6} \ \phantom 0 \ \phantom 0 } && \\[-1pt] { \mathrm{1} \ \mathrm{1} \ \mathrm{5} \ \mathrm{0} }\kern-.2ex \\[-1pt] \underline{ \mathrm{1} \ \mathrm{1} \ \mathrm{4} \ \mathrm{3} } && \\[-1pt] \mathrm{7} \end{array} }$

Since $7$ times the divisor, plus a $3$ -digit number, is still a $3$ -digit number, the divisor is no greater than $128$ . The fourth digit of the dividend must be $7$ . The third digit of the quotient must be $0$ (since two digits are brought down at the last stage). Since the second and the fourth digits of the quotient multiply the divisor to give a $4$ -digit number, both the second and the fourth digits of the quotient are each either $8$ or $9$ . Thus the quotient is one of $7808,7809,7908,7909$ . In the first three cases (since $8$ times the divisor must be a $4$ -digit number), the divisor must be at least $125$ . If the quotient is $7909$ , the divisor must be at least $112$ .

It is now possible to check the possible divisors and quotients and, using the fact that the remainder is $7$ , retrieve the possible dividends. Only three scenarios give a dividend whose fourth digit is $7$ , namely $991750 = 127 \times 7809 + 7 \hspace{2cm} 893724 = 113 \times 7909 + 7 \hspace{2cm} 980723 = 124 \times 7909 + 7$ The second and third scenarios yield divisions which omit the digit $5$ (the second omits $6$ as well). On the other hand, the first scenario is just fine. From this we obtain the answer $\boxed{991750}$ .