=====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:

                   MZN
           +----------
    RPPRMR | KMKMMNPZP
            -KSNNSRN
             =======
              RKPMMZZ
             -KNRSZRM
              =======
               RRRZMRP
              -RKNZSZZ
               =======
                 MSRMN

    base: 7
    letters: KMNPRSZ
    difficulty: easy

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 7 the available counting digits are ONLY 0, 1, 2, 3, 4, 5,
and 6.

7, 8, and 9 do NOT exist in base 7.

Similarly, 6+1 = 10. Because 6 is the last counting digit available, when
we exhaust the digits, we cycle to 0 and carry 1.

However, 10 base 7 should be pronounced  and thought of as "one zero base
7", and NOT "ten".  Ten is a decimal (base 10) construct,  and 10 base 10
does NOT equal 10 base 7.

Lacking 7,  8, and 9  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with respect to carried/multi-place values, are no longer valid.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.

Also,  being an  odd base,  means  that the  appearance of  even and  odd
numbers  may not  always  end with  their classical  even  or odd  value.
They'll  actually cycle  on  and off  every BASE  number  of digits.  For
instance, in base 7, we find that 6  (base 7) is even, but so is 11 (base
7). 10 base 7 is an odd number.

====Getting underway====
The beautiful  thing about  puzzles of  different bases,  is that  if you
stick to a  purely logical/relational approach, NOTHING  changes. Here we
will try and incorporate more math in our approach to a solution, so some
of the numerical properties of the base can come into play.

As  is  a common  starting  strategy,  we  look  for possible  clues  and
connections that we can later build upon. I find I like to start with the
left-most subtractions,  as this not  only can reveal some  initial range
comparisons, for example, looking top through bottom:

    K << R

What's more: K is exactly one less than R:

    K <  R

Not much to go on for that first pass, so let's move to the next ones in:

    from M-S=R: S << M (why? It doesn't need to borrow, so M is bigger)
    from M-S=R: R << M
    from K-N=R: K << N (why? Subtraction  from K  results in some  LARGER
    value? No, K is therefore less than N)

This allows the construction of some longer chains:

         S << M
    K <  R << M
    K <  R << N

Going another set inwards:

    K-N=K,  and the  last round's  K-N=R; we  already know  both have  to
    borrow, but  the difference  of one in  the result (K  and R  and one
    apart), we  have a scenario  where one is  being taken from,  and the
    other is not. Something to keep in mind as we dig deeper.

Okay, nothing new is revealing itself... let's update our chart:

    K = { 0, 1, 2, 3          }
    M = {          3, 4, 5, 6 }
    N = {       2, 3, 4, 5, 6 }
    P = { 0, 1, 2, 3, 4, 5, 6 }
    R = {    1, 2, 3, 4       }
    S = { 0, 1, 2, 3, 4, 5    }
    Z = { 0, 1, 2, 3, 4, 5, 6 }

So far  we know nothing about  P or Z.  Although we can make  some simple
eliminations by looking at the quotient and the subtrahend values.

The quotient: MZN

We see  that multiplying the  divisor (RPPRMR) by  M, we get  neither the
divisor nor nothing, but  some other values. This tells us  that M is not
0, and M is not 1.

Doing the same for Z, we end up with similar results. Therefore, Z is not
0 and not 1.

Finally, N has the similar results: not 0, not 1.

So, at  least we know a  little bit more about  Z (the same as  N at this
point).

Updated chart:

    K = { 0, 1, 2, 3          }
    M = {          3, 4, 5, 6 }
    N = {       2, 3, 4, 5, 6 }
    P = { 0, 1, 2, 3, 4, 5, 6 }
    R = {    1, 2, 3, 4       }
    S = { 0, 1, 2, 3, 4, 5    }
    Z = {       2, 3, 4, 5, 6 }

====Finding S (our zero value)====

There  is  another pattern  we  can  play off  of,  and  that is  in  the
bottom-most row: Z-Z=S.

The classic, which means we have two possibilities: S is zero, or S is 6.
In this  case, we luck  out, because we've  already determined that  S is
less  than  something (M),  so  S  cannot  be  6. Therefore,  there's  no
borrowing happening here, and nothing taken from us.

This realigns our relational chains as follows:

    S << K <  R << M
    S << K <  R << N

Which means the lower bounds for K, R, and N get bumped up by one.

So, we can update our key:

    0    S
    1
    2
    3
    4
    5
    6

And our chart:

    K = {    1, 2, 3          }
    M = {          3, 4, 5, 6 }
    N = {          3, 4, 5, 6 }
    P = {    1, 2, 3, 4, 5, 6 }
    R = {       2, 3, 4       }
    S = { 0                   }
    Z = {       2, 3, 4, 5, 6 }

Hanging  around that  neighborhood, look  one subtraction  to the  right:
M-S=R, actually  the EXACT  same subtraction  we saw in  the top  row: no
borrow was needed there, so no borrow is needed here.

However, now  that we know  that S  is 0, M-0=R?  It must be  being taken
from, and that means that M is  one greater than R (in R-Z=M, bottom row,
that means R is less than Z).

Relational chain:

    S << K <  R <  M << N
    S << K <  R <  M << Z

M, N, and Z can get whittled down a little:

  * M is no longer able to be 6 (and since it is one greater than R, if R
    can only be 2, 3, or 4; then M can only be 3, 4, or 5).
  * N is no longer able to be 0, 1, 2, or 3
  * Z cannot be 0, 1, 2, or 3.

The updated chart:

    K = {    1, 2, 3          }
    M = {          3, 4, 5    }
    N = {             4, 5, 6 }
    P = {    1, 2, 3, 4, 5, 6 }
    R = {       2, 3, 4       }
    S = { 0                   }
    Z = {             4, 5, 6 }

A further optimization:

  * if K is 1, 2, or 3, and R and M immediately follow:
    * if K = 1, R = 2, M = 3
    * if K = 2, R = 3, M = 4
    * if K = 3, R = 4, M = 5

In all scenarios,  3 is used. So nothing  outside of K, R, or M  can be 3
(such as P):

    K = {    1, 2, 3          }
    M = {          3, 4, 5    }
    N = {             4, 5, 6 }
    P = {    1, 2     4, 5, 6 }
    R = {       2, 3, 4       }
    S = { 0                   }
    Z = {             4, 5, 6 }

====Finding K (our one value), along with R and M====
P or  K is  one (we're  down to just  two choices),  so could  we perhaps
eliminate one or the other as the one value? Let's see.

Scanning  the problem  for subtractions  that utilize  K, or  P (but  not
together), I see the following two subtractions:

  * top row:    K - N = K
  * middle row: K - N = R

We know that  K << N, so  it has to borrow  from its left. But  in one of
these cases, it is being taken from, where in the other, it is not.

  * top row:    K - N = K is followed by: M - N = P
    * we know that M << N, so it has to borrow.
  * middle row: K - N = R is followed by: P - R = R
    * since this isn't needing to borrow, we now know that R << P

Because  P is  greater than  R,  it cannot  be  less than  R. This  would
eliminate 1 and 2 as possible values for P, giving us one remaining value
to be the match for one: K.

Hurrah. And  because K,  R, and  M are  all together,  we now  know those
values as well! Updating our chart:

    K = {    1                }
    M = {          3          }
    N = {             4, 5, 6 }
    P = {             4, 5, 6 }
    R = {       2             }
    S = { 0                   }
    Z = {             4, 5, 6 }

Our updated key:

    0    S
    1    K
    2    R
    3    M
    4
    5
    6

====Finding P====
We  also  now  have  the  means   of  finding  P,  due  to  our  previous
investigations. For the subtraction: P-R=R, we have determined:

  * as P is greater than R, it doesn't borrow.
  * the  subtraction to  the right  is the  familiar M-S=R,  and as  M is
    greater than S, there is also no borrow generated, so we're not being
    taken from.
  * This means, by rewriting the subtraction to an addition: R+R=P
    * since R is 2: 2+2=4, therefore P is 4.

Updating the key:

    0    S
    1    K
    2    R
    3    M
    4    P
    5
    6

Updating the chart:

    K = {    1                }
    M = {          3          }
    N = {                5, 6 }
    P = {             4       }
    R = {       2             }
    S = { 0                   }
    Z = {                5, 6 }

====Finding N and Z====
Now we  are down to the  final two: N  and Z. Finding one  will instantly
reveal the  other. So, are  there any good  candidates in the  problem to
work with?

  * M - N = P
 
We know the values for  M and P, and we know that M <<  N, so it needs to
borrow, but  we also know  we are not  being taken from  (S << M,  in the
subtraction to our right).

So:

    M+10       3+10         13
    -  N       -  N       -  N
    ====  -->  ====  -->  ====
       P          4          4

Remember, this isn't decimal: N is NOT 9. But at the same time, is it the
maximum counting digit available to us in this base (6)?

What IS 4+6 in base 7?

Let's count it:

  * 4  + 1 = 5  (1)
  * 5  + 1 = 6  (2)
  * 6  + 1 = 10 (3)
  * 10 + 1 = 11 (4)
  * 11 + 1 = 12 (5)
  * 12 + 1 = 13 (6)

Aha! N is 6.

Which means Z is 5.

Updated chart:

    K = {    1                }
    M = {          3          }
    N = {                   6 }
    P = {             4       }
    R = {       2             }
    S = { 0                   }
    Z = {                5    }

Updated relational chain:

    S <  K <  R <  M <  P <  Z <  N

Updated key:

    0    S
    1    K
    2    R
    3    M
    4    P
    5    Z
    6    N

And we've got a solution!