time for another “let’s solve” letter division funtime!

        UUC
    +------
GLM | YGJYY
     -HFY
      ===
       JGY
      -HFY
       ===
       GJTY
      -GYUL
       ====
         YL

base: 10
letters: CFGHJLMTUY

Standard initial attack run: hit up the left-most sides of each term, since we know there is NO borrow happening there:

top row: H << Y (in fact, H < Y because the Y is being taken from)

second row: H < Y << J
and: G << J

third row: G-G=nothing

we can now prove the Y < J (because the J is being taken from)

rewriting some rules to reflect recent discoveries:

          U U C
    +-- - - - -
GLM | Y>G J Y Y
     -H F Y
      = = =
        J G Y
      - H F Y
        = = =
        GxJ>T Y
      - G Y U L
        = = = =
            Y L

base: 10
letters: CFGHJLMTUY

G << H < Y < J

Range table:

C = { 0 1 2 3 4 5 6 7 8 9 }
F = { 0 1 2 3 4 5 6 7 8 9 }
G = { 0 1 2 3 4 5 6       }
H = {   1 2 3 4 5 6 7     }
J = {       3 4 5 6 7 8 9 }
L = { 0 1 2 3 4 5 6 7 8 9 }
M = { 0 1 2 3 4 5 6 7 8 9 }
T = { 0 1 2 3 4 5 6 7 8 9 }
U = { 0 1 2 3 4 5 6 7 8 9 }
Y = {     2 3 4 5 6 7 8   }

top row, second from left: G-F=J

because we know the G borrows:

G << F

we already know G << J

third row, third from left: T-U=Y

because T borrows:

T << U
T << Y

          U U C
    +-- - - - -
GLM | Y>G J Y Y
     -H F Y
      = = =
        J>G Y
      - H F Y
        = = =
        GxJ>T Y
      - G Y U L
        = = = =
            Y L

base: 10
letters: CFGHJLMTUY

T << U
T << H < Y < J
G << H < Y < J
G << F

Range table:

C = { 0 1 2 3 4 5 6 7 8 9 }
F = {   1 2 3 4 5 6 7 8 9 }
G = { 0 1 2 3 4 5         }
H = {     2 3 4 5 6 7     }
J = {         4 5 6 7 8 9 }
L = { 0 1 2 3 4 5 6 7 8 9 }
M = { 0 1 2 3 4 5 6 7 8 9 }
T = { 0 1 2 3 4 5 6 7     }
U = {   1 2 3 4 5 6 7 8 9 }
Y = {       3 4 5 6 7 8   }

next pass:

top row right-most (we know we’re definitely not being taken from): J-Y=G

hey, our existing assertions covers this. J is bigger, therefore J does not borrow.

second row, right-most: Y-Y=T

hey, another pattern. T is EITHER the smallest (0) or the largest (9), depending on the state of the identical borrow/take

because this is on the far right edge of the term, we know we’re not being taken from. Because the borrow has to match the take in this pattern, Y is NOT borrowing.

T is 0

          U U C
    +-- - - - -
GLM | Y>GxJ Y Y
     -H F Y
      = = =
        J>GxY
      - H F Y
        = = =
        GxJ>T Y
      - G Y U L
        = = = =
            Y L

base: 10
letters: CFGHJLMTUY

G << H < Y < J
G << F

Range table:

C = {   1 2 3 4 5 6 7 8 9 }
F = {     2 3 4 5 6 7 8 9 }
G = {   1 2 3 4 5         }
H = {     2 3 4 5 6 7     }
J = {         4 5 6 7 8 9 }
L = {   1 2 3 4 5 6 7 8 9 }
M = {   1 2 3 4 5 6 7 8 9 }
T = { 0                   }
U = {   1 2 3 4 5 6 7 8 9 }
Y = {       3 4 5 6 7 8   }

trying to finish up our borrow/take marking, we only have one left to determine (bottom row, to the right)

some interesting things there, although I don’t know if we can yet determine that remaining state

we can make some other determinations, however:

for T-U=Y:

because T is 0, it has to borrow

because we do NOT yet know whether a take happens, we have two solution states for this subtraction (T is 10, T is 9)

solving for each:

T is 10 (borrow, no take): 10-U=Y

>Tx: 10
-Y : 3 4 5 6 7 8
 U : 7 6 5 4 3 2


>T>: 9
-Y : 3 4 5 6 7 8
 U : 6 5 4 3 2 1

just from this, we can eliminate 8 and 9 as possibilities for U

to really get more out of this, we need to add in H and J, since they have known relationships with Y (exactly one less than / one greater than):

>Tx: 10
 J : 4 5 6 7 8 9
-Y : 3 4 5 6 7 8
 H : 2 3 4 5 6 7
 U : 7 6 5 4 3 2
         X


>T>: 9
 J : 4 5 6 7 8 9 
-Y : 3 4 5 6 7 8
 H : 2 3 4 5 6 7
 U : 6 5 4 3 2 1
       X X

notice how the case of H Y J being 4 5 6 is eliminated in both states: because of that we can eliminate that possibility from the range table

regarding U, its possibility of being 5 is eliminated in both states, so we can eliminate U as being 5 from the range table

Range table:

C = {   1 2 3 4 5 6 7 8 9 }
F = {     2 3 4 5 6 7 8 9 }
G = {   1 2 3 4 5         }
H = {     2 3   5 6 7     }
J = {         4 5   7 8 9 }
L = {   1 2 3 4 5 6 7 8 9 }
M = {   1 2 3 4 5 6 7 8 9 }
T = { 0                   }
U = {   1 2 3 4   6 7 8 9 }
Y = {       3 4   6 7 8   }

furthermore, the right-most subtraction on the bottom row: Y-L=L

we can infer some stuff about L (even if we do not know if the Y needs to borrow)

we have a classic doubling pattern here, so let’s put that to use:

   no borrow | borrow
Y    4 6 8   | 14 16 18
L    2 3 4   | 7  8  9

bringing in H and J:

   no borrow | borrow
J    5 7 9   |  5  7  9
Y    4 6 8   | 14 16 18
H    3 5 7   |  3  5  7
L    2 3 4   |  7  8  9
                      X

so: three things we can glean:

1. HYJ are now down to exactly 3 values each (from 5), due to Y being guaranteed to be an even number (the lack of take, and the even base guarantees this in the doubling pattern, regardless of borrow)

2. we now have L down to 5 values

3. in the case of Y borrowing, one of the three possibilities has been eliminated, as both J and L cannot both be 9

Range table:

C = {   1 2 3 4 5 6 7 8 9 }
F = {     2 3 4 5 6 7 8 9 }
G = {   1 2 3 4 5         }
H = {       3   5   7     }
J = {           5   7   9 }
L = {     2 3 4     7 8   }
M = {   1 2 3 4 5 6 7 8 9 }
T = { 0                   }
U = {   1 2 3 4   6 7 8 9 }
Y = {         4   6   8   }

at this point, we still don’t have that last borrow/take state, and we’re out of other new relationships to draw upon. So let’s try a different strategy: magnitude comparisons!

          U U C
    +-- - - - -
GLM | Y>GxJ Y Y
     -H F Y
      = = =
        J>GxY
      - H F Y
        = = =
        GxJ>T Y
      - G Y U L
        = = = =
            Y L

base: 10
letters: CFGHJLMTUY

G << H < Y < J
G << F

notice how the divisor (GLM) is 3 digits?

GLM x 1 = GLM

multiply GLM x anything higher than 1 (2-9), if the result is still a 3 digit value, THAT value is a bigger number than divisor (they all are, but since we don’t know the identities of the letters yet, we can only derive relationship values from similar magnitudes)

with this, we’ve got: GLM << HFY (G << H, which we already know)

we also have: GJT << GLM (the difference from the second term subtraction- the difference MUST be smaller than the divisor, otherwise we’d be doing the division wrong)

so here: G == G, nothing valuable there. So we proceed to the next value:

J << L

ah! That’s actually QUITE useful!

if J is less than L, that implies we can go with the borrowing column (for Y-L=L) in this table:

   no borrow | borrow
J    5 7 9   |  5  7  9
Y    4 6 8   | 14 16 18
H    3 5 7   |  3  5  7
L    2 3 4   |  7  8  9
     X X X            X

so: HYJ and L are now all down to 2 values a piece

Range table:

C = {   1 2 3 4 5 6 7 8 9 }
F = {     2 3 4 5 6 7 8 9 }
G = {   1 2 3 4 5         }
H = {       3   5         }
J = {           5   7     }
L = {               7 8   }
M = {   1 2 3 4 5 6 7 8 9 }
T = { 0                   }
U = {   1 2 3 4   6 7 8 9 }
Y = {         4   6       }

furthermore: because we’ve got HYJ and L down to 2 values a piece, and locked in relationships between all: we can eliminate any other letter being 5 or 7

Range table:

C = {   1 2 3 4   6   8 9 }
F = {     2 3 4   6   8 9 }
G = {   1 2 3 4           }
H = {       3   5         }
J = {           5   7     }
L = {               7 8   }
M = {   1 2 3 4   6   8 9 }
T = { 0                   }
U = {   1 2 3 4   6   8 9 }
Y = {         4   6       }

          U U C
    +-- - - - -
GLM | Y>GxJ Y Y
     -H F Y
      = = =
        J>GxY
      - H F Y
        = = =
        GxJ>T>Y
      - G Y U L
        = = = =
            Y L

base: 10
letters: CFGHJLMTUY

G << H < Y < J
G << F

updating that last borrow state, we can now better complete that T-U=Y equation (last row, second from right)

>T>: 9
 J : 5 7 
-Y : 4 6
 H : 3 5
 U : 5 3

now we have U whittled down more, AND we can eliminate anything other than H and U as being 3

Range table:

C = {   1 2   4   6   8 9 }
F = {     2   4   6   8 9 }
G = {   1 2   4           }
H = {       3   5         }
J = {           5   7     }
L = {               7 8   }
M = {   1 2   4   6   8 9 }
T = { 0                   }
U = {       3             } !!!
Y = {         4   6       }

and actually, because we previously eliminated the possibility of U being 5: by process of elimination, U is 3

trace back through our recent state tables, solving for H Y J and L:

Range table:

C = {   1 2   4         9 }
F = {     2   4         9 }
G = {   1 2   4           }
H = {           5         }
J = {               7     }
L = {                 8   }
M = {   1 2   4         9 }
T = { 0                   }
U = {       3             }
Y = {             6       }

          U U C
    +-- - - - -
GLM | Y>GxJ Y Y
     -H F Y
      = = =
        J>GxY
      - H F Y
        = = =
        GxJ>T>Y
      - G Y U L
        = = = =
            Y L

base: 10
letters: CFGHJLMTUY

U << H < Y < J < L
G << H < Y < J < L
M << H < Y < J < L
G << F

picking on a convenient subtraction we know a lot about, top row, right-most: J-Y=G

we know that J is exactly one bigger than Y.

therefore: G is 1

Range table:

C = {     2   4         9 }
F = {     2   4         9 }
G = {   1                 }
H = {           5         }
J = {               7     }
L = {                 8   }
M = {     2   4         9 }
T = { 0                   }
U = {       3             }
Y = {             6       }

F is our next likely target: the G-F=J subtraction

 G: 11
-J: 7
 =====

Range table:

C = {     2             9 }
F = {         4           }
G = {   1                 }
H = {           5         }
J = {               7     }
L = {                 8   }
M = {     2             9 }
T = { 0                   }
U = {       3             }
Y = {             6       }

now down to just C and M, whichever way we get one of them, so let’s go:

          U U C
    +-- - - - -
GLM | Y>GxJ Y Y
     -H F Y
      = = =
        J>GxY
      - H F Y
        = = =
        GxJ>T>Y
      - G Y U L
        = = = =
            Y L

base: 10
letters: CFGHJLMTUY

U << H < Y < J < L
G << H < Y < J < L
M << H < Y < J < L
G << F

M x U = Y
M x 3 = 6
M is 2 or 9
2 x 3 = 6
9 x 3 = 27

M cannot be 9, M is 2

Range table:

C = {                   9 }
F = {         4           }
G = {   1                 }
H = {           5         }
J = {               7     }
L = {                 8   }
M = {     2               }
T = { 0                   }
U = {       3             }
Y = {             6       }

key: TGMUFHYJLC

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what do you think about how awesome letter divisions are, and how they are immeasurably enhancing your life, and the lives of those around you? Be sure to leave a comment in the comment section below!