# Clean slate

$\LARGE{ \require{enclose} \begin{array}{rlll} \phantom{0}\ \phantom{0} \ \phantom{0} \ \boxed{\phantom{0}} && \\[-3pt] \boxed{\phantom0} \ \boxed{\phantom0}\ \enclose{longdiv}{\boxed{\phantom0} \ \boxed{\phantom0} \ \boxed{\phantom{0}} } && \\[-3pt] \underline{\boxed{\phantom0}\ \boxed{\phantom0} \ \boxed{\phantom0}} && \\[-3pt] \boxed{\phantom0} \ \boxed{\phantom{0}} && \\[-2pt] \end{array} }$

The above shows a long division, where each box represents a single-digit integer between 0 and 9 inclusive. Fill the boxes in such a way that

• Every digit 0-9 must occur at least once in the boxes.
• The leading box (leftmost box) of the 5 sections (1.divisor, 2.dividend, 3.remainder, 4.quotient, 5.the section just below the dividend) must have a non-zero digit.

Given that there are 11 unique solutions, find the sum of remainders of all the 11 solutions.

Details and Assumptions:

• As an explicit example, the long division below is one such solution, and the remainder in this case is 18.

$\LARGE{ \require{enclose} \begin{array}{rlll} \phantom{0}\ \phantom{0} \ \phantom{0} \ \boxed{7} && \\[-3pt] \boxed{5} \ \boxed{6}\ \enclose{longdiv}{\boxed{4} \ \boxed{1} \ \boxed{0} } && \\[-3pt] \underline{\boxed{3}\ \boxed{9} \ \boxed{2}} && \\[-3pt] \boxed{1} \ \boxed{8} && \\[-2pt] \end{array} }$

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