Another Rightmost Long Division

$\LARGE{ \begin{array}{rll} \phantom{0}\ \color{#D61F06}{\boxed{{6}}} && \\[-2pt] \color{#D61F06}{\boxed{\phantom0}\ \boxed{\phantom0}}\ \enclose{longdiv}{\boxed{\phantom0} \ \boxed{7}}\kern-.2ex \\[-2pt] \underline{\color{#D61F06}{\boxed{\phantom0} \ \boxed{8}}} && \\[-2pt] \boxed{9} \end{array} }$

Looking at the red highlighted numbers as shown above, we see that a 2-digit positive integer multiply by 6 gives another 2-digit integer. Let $x$ and $y$ denote these aforementioned integers respectively, then $6x = y< 100$ . So $6x < 100$ or equivalently $x < \dfrac{100}6 = 16 + \dfrac23$ . Thus $x=10,11,12,\ldots, 16$ only.

Since the multiplication of $x$ and 6 gives a last digit of 8, then by trial and error, of all the numbers $10,11,\ldots,16$ , only $13$ satisfy this constraint. hence $x=16$ and so $y = 6x =78$ . We can now fill in some some brackets in the long division:

$\LARGE{ \begin{array}{rll} \phantom{0}\ \boxed{{6}} && \\[-2pt] \boxed{1}\ \boxed{3}\ \enclose{longdiv}{\color{#3D99F6}{\boxed{\phantom0} \ \boxed{7}}}\kern-.2ex \\[-2pt] \underline{\color{#3D99F6}{\boxed{7} \ \boxed{8}}} && \\[-2pt] \color{#3D99F6}{\boxed{9}} \end{array} }$

Looking at the blue highlighted numbers. We know that $\boxed{\phantom0} 7 - 78 = 9$ , or equivalently $\boxed{\phantom0} 7 = 78 + 9 = 87$ , thus the remaining missing digit is 8. And now we can complete the entire long division:

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

The sum of all the missing digits is $1+3+8+7 = \boxed{19}$ .

It's 87/13

The answer is 1+3+7+8=19. The equation is 98/13 which gives remainder 9 and quotient=6