Blanks In Division

I got to admit that I got inspired by Agnishom's other problem recently .

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Let me color these boxes first and I'll show the relationship between each of these boxes:

$\LARGE{ \begin{array}{rll} \phantom{0}\ \color{#D61F06}{\boxed{\phantom{0}}} && \\[-2pt] \boxed{{8}}\ \enclose{longdiv}{\color{#3D99F6}{\boxed{\phantom0}\ \boxed{\phantom{0}}}}\kern-.2ex \\[-2pt] \underline{\color{#20A900}{\boxed{7} \ \boxed{\phantom{0}}}} && \\[-2pt] \boxed{{6}} \end{array} }$

We have
[1] : $\Large 8 \; \times \; \color{#D61F06}{\boxed{\phantom0} } \; = \; \color{#20A900}{ \boxed7 \boxed{\phantom0} }$ , and
[2] : $\Large \color{#3D99F6}{\boxed{\phantom0}\boxed{\phantom0} } \; - \; \color{#20A900}{ \boxed7 \boxed{\phantom0} } \; = \; \boxed6$ .

Let us solve [1] first: The only 2-digit multiple of 8 that begins with 7 is $72 = 8\times9$ . So the digit that represents the red square is 9, and the digit that represents the green square is 2, $\Large 8 \; \times \; \color{#D61F06}{\boxed9 } \; = \; \color{#20A900}{ \boxed7 \boxed{2} }$ .

And for [2] : $\Large \color{#3D99F6}{\boxed{\phantom0}\boxed{\phantom0} } \; - \; \color{#20A900}{ \boxed7 \boxed{2} } \; = \; \boxed6$ . Equivalently, we have $\large \color{#3D99F6}{\boxed{\phantom0}\boxed{\phantom0} } = 72 + 6 = 78$ .

We can now complete the long division, $\LARGE{ \begin{array}{rll} \phantom{0}\ \color{#D61F06}{\boxed{9}} && \\[-2pt] \boxed{{8}}\ \enclose{longdiv}{\color{#3D99F6}{\boxed{7}\ \boxed{8}}}\kern-.2ex \\[-2pt] \underline{\color{#20A900}{\boxed{7} \ \boxed{2}}} && \\[-2pt] \boxed{{6}} \end{array} }$

Thus, the only unused digit in the options is $\boxed6$ .