=====Walkthrough=====
Following will be  a walkthrough (the documentation in  a form acceptable
for submission, provided as  an example of both the form  I'd like to see
submissions in, but also providing illumination in the process of solving
a letter division puzzle) of the following puzzle:

            ???????
          +--------
    LIMNM | MEEFVLF
           -LIMNM
            =====
            TTVLFL
           -TBJBMW
            ======
             TLLVNF
            -TBJBMW
             ======
             ??????

    base: 11
    letters: BEFIJLMNTVW
    difficulty: medium

The  intent of  the  documentation  (and therefore  the  project you  are
performing) is  NOT merely  to SOLVE  the  problem,  but to  document the
step-by-step process of solving the problem. That is demonstrated here.

As this is a puzzle encoded in a base OTHER THAN 10, a few considerations
have to be kept in mind.

First, in base 11 the available counting digits  are 0, 1, 2, 3, 4, 5, 6,
7, 8, 9 AND A.

In addition  to the  familiar 0-9, the  letter A is  now a  number, whose
quantity is  equivalent to that  of the one  in decimal we  lovingly call
"ten".

As  8+1 =  9, 9+1  = A,  A+1=10.  Because A  is the  last counting  digit
available, when we exhaust the digits, we cycle to 0 and carry 1. 10 base
11 is the quantity we know of in base 10 as "eleven".

However, 10 base 11 should be pronounced and thought of as "one zero base
elevent", and  NOT "ten". Ten  is a decimal  (base 10) construct,  and 10
base 10 does NOT equal 10 base 11.

Having  an  extra number  (A)  means  any memorized  mathematical  facts,
especially  with respect  to  carried/multi-place values,  are no  longer
valid. The same  properties and concepts are there, but  the results will
be different (quantities remain the same, representations will differ).

Being on odd base,  that also means that all even and  odd numbers do not
end with an even or odd  digit (some certainly do, in alternating fashion
of runs of counts of 11).

====Getting underway====
Looking for anything to get us started,  let's look at a couple layers of
the left-most subtractions:

  * M-L=T
    * because the term being subtracted (LIMNM) is the same length as the
      divisor (indeed, exactly the same!), not only can we make the usual
      assertions of:
        * L << M
        * T << M
    * we can also state that while LIMNM is a smaller quantity than MEEFV
      that TTVLF is SMALLER than both MEEFV AND LIMNM. So, we can instead
      optimize our assertions to:
        * T << L << M

  * T-T=zero (no take)
  * T-B=T
    * B is 0 (no borrow, no take)

  * V-J=L (no borrow, unkrown presently on take)
    * J << V
    * L << V
      * and since we know something already about L, that most recent one
        can be further expanded:
        * T << L << V

Our initial relational chains:

    T << L << M
    T << L << V
    J << V

Our chart:

    B = { 0                               }
    E = {    1, 2, 3, 4, 5, 6, 7, 8, 9, A }
    F = {    1, 2, 3, 4, 5, 6, 7, 8, 9, A }
    I = {    1, 2, 3, 4, 5, 6, 7, 8, 9, A }
    J = {    1, 2, 3, 4, 5, 6, 7, 8, 9    }
    L = {       2, 3, 4, 5, 6, 7, 8       }
    M = {          3, 4, 5, 6, 7, 8, 9, A }
    N = {    1, 2, 3, 4, 5, 6, 7, 8, 9, A }
    T = {    1, 2, 3, 4, 5, 6, 7          }
    V = {             4, 5, 6, 7, 8, 9, A }
    W = {    1, 2, 3, 4, 5, 6, 7, 8, 9, A }

====Diving deeper====
In the middle row, we have an additional revealing subtraction by zero, a
further opportunity for gathering some clues. L-B-L, which clues us in to
the fact it is not borrowing, AND it is not being taken from.

This means that for the V-J=L subtraction, there's no borrow nor take.

Then, for the F-M=V subtraction, we now know there's no borrow, so:

  * M << F
  * V << F

This is a nice further expansion of our existing assertions:

    T << L << M << F
    T << L << V << F
         J << V << F

Our chart:

    B = { 0                               }
    E = {    1, 2, 3, 4, 5, 6, 7, 8, 9, A }
    F = {                   6, 7, 8, 9, A }
    I = {    1, 2, 3, 4, 5, 6, 7, 8, 9, A }
    J = {    1, 2, 3, 4, 5, 6, 7, 8       }
    L = {       2, 3, 4, 5, 6, 7          }
    M = {          3, 4, 5, 6, 7, 8, 9    }
    N = {    1, 2, 3, 4, 5, 6, 7, 8, 9, A }
    T = {    1, 2, 3, 4, 5, 6             }
    V = {             4, 5, 6, 7, 8, 9    }
    W = {    1, 2, 3, 4, 5, 6, 7, 8, 9, A }

====New clues from existing clues====
Looking for further clues, a look at the top row subtractions has us wuth
our eyes on both the rightmost subtractions of:

  * F-N=L
  * V-M=F

We know things about letters in both subtractions!

  * L << F, so: N << F (F does not borrow)
  * Both V and M are less than F, so:
    * V borrows
    * V << M

We can finally merge two similar assertions:

    T << L << V << M << F
         J << V << M << F
    N << F

Our chart:

    B = { 0                               }
    E = {    1, 2, 3, 4, 5, 6, 7, 8, 9, A }
    F = {                      7, 8, 9, A }
    I = {    1, 2, 3, 4, 5, 6, 7, 8, 9, A }
    J = {    1, 2, 3, 4, 5, 6, 7          }
    L = {       2, 3, 4, 5, 6, 7          }
    M = {                5, 6, 7, 8, 9    }
    N = {    1, 2, 3, 4, 5, 6, 7, 8, 9    }
    T = {    1, 2, 3, 4, 5, 6             }
    V = {             4, 5, 6, 7, 8       }
    W = {    1, 2, 3, 4, 5, 6, 7, 8, 9, A }

====Discovery through similarity====
There are two strikingly similar subtractions present in the puzzle, from
the top row E-M=V, and the second row F-M=V.

Notice how, despite the different top letter, the bottom two letters with
the subtraction are identical?

We could say that:

  * M+V = F
  * M+V = E

Clearly, two letters cannot have the same value, BUT, this is providing a
clue that:

  * E and F (or F and E) are next to each other.  We don't yet know which
    is the correct order.
  * that means one is taken from, and the other is not.

Now, in the case of E-M=V, we do happen to know something about that next
subtraction to its right, F-N=L. We know that F is larger than N or L.

Therefore:

  * In the case of E-M=V, E is not borrowing, nor is it being taken from.
  * In the case of F-M=V, F is not borrowing, but it is being taken from.
  * F-1 = E, so E < F

Our relational chains:

    T << L << V << M << E < F
         J << V << M << E < F
                   N << E < F

Our updated chart:

    B = { 0                               }
    E = {                      7, 8, 9    }
    F = {                         8, 9, A }
    I = {    1, 2, 3, 4, 5, 6, 7, 8, 9, A }
    J = {    1, 2, 3, 4, 5, 6             }
    L = {       2, 3, 4, 5, 6, 7, 8       }
    M = {                5, 6, 7, 8       }
    N = {    1, 2, 3, 4, 5, 6, 7, 8       }
    T = {    1, 2, 3, 4, 5                }
    V = {             4, 5, 6, 7          }
    W = {    1, 2, 3, 4, 5, 6, 7, 8, 9, A }

====Continuing to connect clues====
With our most recent discovery, we can now dive deeper into the rightmost
second row subtraction of L-W=N, since we now know it HAS to borrow:

  * L << W
  * L << N

Furthermore, since we know from the top row E-M=V that E does not borrow,
and from referencing our relational chain, we know that T << E, we can do
some work on the top row E-I=T subtraction too, namely:

  * E does not borrow (because E is bigger than T)
  * E does not get taken from.
  * I << E

Further updates to our relational chains:

    T << L << W
    T << L << V << M << E < F
         J << V << M << E < F
         T << L << N << E < F
                   I << E < F

Our updated chart:

    B = { 0                               }
    E = {                         8, 9    }
    F = {                            9, A }
    I = {    1, 2, 3, 4, 5, 6, 7, 8       }
    J = {    1, 2, 3, 4, 5, 6             }
    L = {       2, 3, 4                   }
    M = {                5, 6, 7, 8       }
    N = {          3, 4, 5, 6, 7, 8       }
    T = {    1, 2, 3                      }
    V = {             4, 5, 6, 7          }
    W = {          3, 4, 5, 6, 7, 8,    A }

Some things of note:

  * We were able  to whittle down the  possibilities for E and  F to just
    two values,  allowing for the elimination  of 9 as a  possibility for
    any other value.

  * Either F or W is the largest value, A.

====Eliminating some possibilities====
We are down to three candidates for 1: I, J, or T. Let's see if we cannot
rule any of them out.

In the top-row subtraction E-I=T, we know that E is larger. We have since
put those into our relational chain assertions, and we can use that now:

    T << L << V << M << E < F
    
                   I << E < F

Specifically, T can NOT be ONE less than E. At this point,  T is at least
FOUR values less than E.

So: I cannot be 1. For that matter, I cannot be 2 or 3 either.

On the flip side, with T currently reduced to 1, 2, or 3, we can cut down
on possibilities for I as well, since by our knowledge of T's range, I is
not able to be more than 3 less than E.

Also, if we explore what we currently know about J (V-J=L, where V is not
borrowing nor is being taken from):

    T << L << V << M << E < F

         J << V << M << E < F

We see that it is possible for J or L to be one less than V by what we do
know about them. But, notice how L is limited to just 2, 3, or 4. So, the
letter J is NOT able to be one less than V. Best it can be is 2 less than
V, and as great as 4 less than.

    V = { 4, 5, 6, 7 }
    J = { 1, 2, 3, 4, 5, 6 }
    L = { 2, 3, 4 }

From looking at these known limits to their ranges,  in no valid possible
combination can J be 6, right?

So, we were able to polish off a few possibilities:

    B = { 0                               }
    E = {                         8, 9    }
    F = {                            9, A }
    I = {                5, 6, 7, 8       }
    J = {    1, 2, 3, 4, 5                }
    L = {       2, 3, 4                   }
    M = {                5, 6, 7, 8       }
    N = {          3, 4, 5, 6, 7, 8       }
    T = {    1, 2, 3                      }
    V = {             4, 5, 6, 7          }
    W = {          3, 4, 5, 6, 7, 8,    A }

Still, J's range is too broad.

====Building some equation systems====
Calling upon another trick,  let us build some equations that may further
help us determine possibilities.

Please take note of the top row subtraction: V-M=F (no take)
and the second row subtraction: F-M=V (is being taken from)

Notice how for the letter pair F,  V, in each subtraction, one is on top,
and  in the  other,  the other  is  at the  top? We  can  use that,  with
knowledge of the  takes, to build the following  equation (basically, add
the two remaining letters together, and that equals the puzzle base minus
the number of takes):

  * (F, V) M + M = 10 - 1 = A

Hey, M is 5!

And because V << M, once we rule out the possibilities, V is 4!

    B = { 0                               }
    E = {                         8, 9    }
    F = {                            9, A }
    I = {                   6, 7, 8       }
    J = {    1, 2, 3                      }
    L = {       2, 3                      }
    M = {                5                }
    N = {          3,       6, 7, 8       }
    T = {    1, 2                         }
    V = {             4                   }
    W = {          3,       6, 7, 8,    A }

We can now go back and plug in additional values into equations, starting
with the no borrow, no take E-M=V subtraction in the top row:

  * E - M = V
  * E     = V + M
  * E     = 4 + 5
  * E     = 9

As E is exactly one less than F, F is A.

And in V-J=L (no borrow, no take), we can determine that J cannot be 3.

Our updated chart:

    B = { 0                               }
    E = {                            9    }
    F = {                               A }
    I = {                   6, 7, 8       }
    J = {    1, 2                         }
    L = {       2, 3                      }
    M = {                5                }
    N = {          3,       6, 7, 8       }
    T = {    1, 2                         }
    V = {             4                   }
    W = {          3,       6, 7, 8       }

Take note of how  J and T are limited in possibility  to only two values,
and in fact the SAME two values  (1 or 2). Therefore, nothing else can be
1 or 2, which inadvertently lets us solve for L (3):

    B = { 0                               }
    E = {                            9    }
    F = {                               A }
    I = {                   6, 7, 8       }
    J = {    1, 2                         }
    L = {          3                      }
    M = {                5                }
    N = {                   6, 7, 8       }
    T = {    1, 2                         }
    V = {             4                   }
    W = {                   6, 7, 8       }

... and  then plug back  into our  V-J=L subtraction, knowing  there's no
borrow, no take,  and knowing the values for  V and L, we now  have our J
(and then also, T!)

  * V - J = L
  * V     = L + J
  * V - L = J
  * 4 - 3 = 1
  * J is 1!
  * T is 2!

    B = { 0                               }
    E = {                            9    }
    F = {                               A }
    I = {                   6, 7, 8       }
    J = {    1                            }
    L = {          3                      }
    M = {                5                }
    N = {                   6, 7, 8       }
    T = {       2                         }
    V = {             4                   }
    W = {                   6, 7, 8       }

====Solving for the remaining letters====

Getting the  remaining letters (I, N,  W) is simply a  matter of plugging
into subtractions where we know enough about the other letters involved:

  * E-I=T (no borrow, no take)
  * E  =T+I
  * E-T=I
  * 9-2=I
  *   7=I

  * F-1-N=L (no borrow, but a take)
  * F-1  =L+N
  * F-1-L=N
  * A-1-3=N
  * 9  -3=N
  *     6=N

... and there we go:

    B = { 0                               }
    E = {                            9    }
    F = {                               A }
    I = {                      7          }
    J = {    1                            }
    L = {          3                      }
    M = {                5                }
    N = {                   6             }
    T = {       2                         }
    V = {             4                   }
    W = {                         8       }

The puzzle key:

    0    B
    1    J
    2    T
    3    L
    4    V
    5    M
    6    N
    7    I
    8    W
    9    E
    A    F

And there we go, puzzle solved. Now to verify:

====Determining the quotient====
With  the solve4  puzzles, the  verification step  is determining  of the
quotient  and remainder.  Here  we  will determine  the  quotient of  the
puzzle, through performing multiplication on the divisor:

    LIMNM x 1 (J) =  LIMNM
    LIMNM x 2 (T) =  IVBJF
    LIMNM x 3 (L) = JBBMWV
    LIMNM x 4 (V) = JLWBLE
    LIMNM x 5 (M) = JIVMFL
    LIMNM x 6 (N) = TBJBMW <-- hey, it matches a subtrahend (indeed, 2!)

So, all  we need to  do is  match our subtrahends,  in order from  top to
bottom, with the divisor factors here, and list them.

Our quotient is: JNN

====Determining the remainder====
With  the solve4  puzzles, the  verification step  is determining  of the
quotient  and remainder.  Here we  will  determine the  remainder of  the
puzzle, through performing math on the bottom row:

    TLLVNF
   -TBJBMW
    ======

Which can be split into 6 separate subtractions:

    T L L V N F
   -T B J B M W
    = = = = = =

And knowing the state of the individual borrows/takes:

  x T x L x L x V x N x F x
   -T   B   J   B   M   W
    =   =   =   =   =   =

In  this  case  there  are  no  borrows or  takes,  so  we  can  just  do
straightforward (non-borrowing) subtractions:

  x T x L x L x V x N x F x
   -T   B   J   B   M   W
    =   =   =   =   =   =
        L   T   V   J   T

So, our remainder is: LTVJT

With that, for verification on this puzzle, we merely need to provide:

JNN:LTVJT

This puzzle is now complete.