Letter division is a category of logic problem where you would take an ordinary math equation (in long form), and substitute all the numbers for letters, thereby in a direct sense masking the numeric values present that correctly enable the problem to work from start to completion. It is your task, through exploring, experimenting, and playing, to ascertain the numeric value of each letter (as many as 10, one for each numeric value 0-9).

Here we will be focusing on long division, something you learned (and perhaps last experienced, before becoming mindlessly addicted to pressing buttons on a calculator), in grade school. It entails a whole number (integer) division, involving aspects of multiplication, addition (through borrowing), and subtraction (primarily) to arrive at a quotient and a remainder.

Division is unique in that it produces two 'answers', each serving particular uses in various applications.

Here is an example (using numbers):

First up, we're going to divide 87654321 (the dividend) by 1224 (the divisor). Commonly, especially if punching into a calculator, we might express that equation as:

87654321/1224

Or in a language like C, assigning the quotient to the variable x (an integer):

    x = 87654321/1224;

But, we're not actually interested in the 'answer' (quotient or remainder); we are interested in the PROCESS. You know, the stuff the calculator does for you, which in order to perform this task and better explore the aspects of critical thinking, we need to take and encounter every step of the way:

          71613
     +---------
1224 | 87654321
      -8568
       ====
        1974
       -1224
        ====
         7503
        -7344
         ====
          1592
         -1224
          ====
           3681
          -3672
           ====
              9

Here we obtain the results (focusing on the quotient up top; as the remainder quite literally is what remains once we're done- we're specifically NOT delving into decimal points, but instead doing integer division, which as previously stated has MANY important applications in computing) through a step by step process of seeing how many times our divisor (1224) best and in the smallest fashion fits into some current value of the dividend (or intermediate result thereof).

For instance, seeking the smallest “best fit” of 1224 into 87654321, we find that 1224 fits best SEVEN times (1224 * 7 = 8568, which is the CLOSEST we can get to 8765… 1224 * 8 = 9792, which would be too big (and way too small for 87654). Clearly, we are seeking those values that best fit within a multiple of 0-9, staying away from double digits of multiplication (although, we COULD do it that way and still arrive at the same end result).

So: 8765-8568 = 197.

We have our first result, yet: there's still values in the dividend (87654321) remaining to process, specifically the 4321, so we take them one digit at a time.

The next available, unprocessed digit in 4321 is '4', so we 'drop that down' and append it to our previous result (197), giving us: 1974.

We now see how many times (via single digit multiplication), our divisor (1224) can fit into 1974. As it turns out, just once.

So: 1974-1224 = 750.

And we keep repeating the process until there are no more digits from the dividend to drop down; at which point, we are left with a remainder (in the above problem, the lone '9' at the very bottom; THAT is the remainder).

Clearly it is important to have a handle on and understanding of the basic long division process before attempting a letter division problem. So, be sure to try your hand at a few practice problems before proceeding.

Following will be a sample letter division problem, and a documented work through of it (and to be sure: the aim here is not merely to solve it, but to DOCUMENT HOW YOU SOLVED IT, you might want to keep notes as you go along to save you time and sanity).

Here goes:

            GLJK
      +---------
 KJKK | GLMBRVLR
       -VKOKL
        =====
         LJBGV
        -OKVKG
         =====
          JJGKL
         -LKBKV
          =====
           KVRMR
          -JKRKB
           =====
            VKMK
            
letters: BGJKLMOPRV

First off, note how this is NO DIFFERENT from the numeric problem above: just instead of numbers, which we've associated some concepts with, here we have letters (each letter maps to a unique number, 0-9). The trick will be to figure out which letter maps to which number.

So, let us begin.

One aim is to obtain the key to the puzzle, the mapping of the letters to numbers, so I will typically set up an answer key as follows:

Another thing I like to do is set up a more visual representation of what each letter COULD be. I do so in the following form:

B = { 0, 1, 2, 3, 4, 5, 6, 7, 8, 9 }
G = { 0, 1, 2, 3, 4, 5, 6, 7, 8, 9 }
J = { 0, 1, 2, 3, 4, 5, 6, 7, 8, 9 }
K = { 0, 1, 2, 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 }
O = { 0, 1, 2, 3, 4, 5, 6, 7, 8, 9 }
P = { 0, 1, 2, 3, 4, 5, 6, 7, 8, 9 }
R = { 0, 1, 2, 3, 4, 5, 6, 7, 8, 9 }
V = { 0, 1, 2, 3, 4, 5, 6, 7, 8, 9 }

Then, as I figure things out (either what certain are, but mostly, which ones they are NOT), I can mark it up accordingly.

Right from the start, we can already make some important connections; looking at EACH of the subtractions taking place, in the left-most position, we see an interesting phenomenon taking place- G-V=0, L-O=0, J-L=0, and K-J=0.

Now, since EACH letter is its own unique numeric value, subtracting one letter from another on its own won't result in a value of 0, but being borrowed from will.

That is: 7-6=1, but (7-1)-6=0. THAT is what is going on here.

So what we can infer from this, is some very important connections:

• V is one less than G (I'll write it as: V < G)
• O is one less than L (O < L)
• L is one less than J (L < J)
• J is one less than K (J < K)

Does that make sense? From looking at the puzzle, those four relations can be made.

Now, FURTHERMORE, some of those connections are thereby connected. Look at the 'L' and 'J' connections:

• O < L, but also: L < J
• L < J, but also: J < K

That implies a further connection, so we can chain them together:

• O < L < J < K

So from that initial observation and connection, we now have two disconnected relationships:

• V < G
• O < L < J < K

From what we've done so far, we do not know where V,G fall in respect to O,L,J,K. They might be less than, OR greater than. We won't know without further information.

Yet, even WITH this information, we can update our letter ranges:

• since V is less than G, we know V can NOT be 9.
• similarly, G can NOT be 0.
• O cannot be 9, 8, 7, because we know O is 3 less than K. So even though we don't know what K actually is, because K COULD be 9, we know what O, L, and J can NOT be.
• L cannot be 9 or 8
• J cannot be 9
• on the other side, K cannot be 0, 1, or 2
• J cannot be 0 or 1
• L cannot be 0.

So, if we update our range chart accordingly:

B = { 0, 1, 2, 3, 4, 5, 6, 7, 8    }
G = {    1, 2, 3, 4, 5, 6, 7, 8, 9 }
J = {       2, 3, 4, 5, 6, 7, 8,   }
K = {          3, 4, 5, 6, 7, 8, 9 }
L = {    1, 2, 3, 4, 5, 6, 7,      }
M = { 0, 1, 2, 3, 4, 5, 6, 7, 8, 9 }
O = { 0, 1, 2, 3, 4, 5, 6,         }
P = { 0, 1, 2, 3, 4, 5, 6, 7, 8, 9 }
R = { 0, 1, 2, 3, 4, 5, 6, 7, 8, 9 }
V = { 0, 1, 2, 3, 4, 5, 6, 7, 8, 9 }

Moving on, dealing with details of discovering those one-off relations, that tells us something about the NEXT subtractions: that they borrow (which means they are LESS THAN the thing being subtracted from them):

• L is less than K (which we actually know to be 2 less than K), so L - K needs to BORROW
• J is less than K (which we know is 1 less than K), so J - K needs to BORROW
• V is apparently also less than K (which we didn't previously know), so V - K needs to BORROW
• now knowing than V « K, we can connect our other relational fragment in (I use the double '«' to denote “less than” by an unknown amount, because while we know V is less than K, we don't know by how much).

So: V < G « O < L < J < K

This allows us some further whittling of our ranges:

• V cannot be 9, 8, 7, 6, or 5
• G cannot be 9, 8, 7, or 6
• O cannot be 0, or 1
• L cannot be 0, 1, or 2
• J cannot be 0, 1, 2, or 3
• K cannot be 0, 1, 2, 3, or 4

B = { 0, 1, 2, 3, 4, 5, 6, 7, 8    }
G = {    1, 2, 3, 4, 5,            }
J = {             4, 5, 6, 7, 8,   }
K = {                5, 6, 7, 8, 9 }
L = {          3, 4, 5, 6, 7,      }
M = { 0, 1, 2, 3, 4, 5, 6, 7, 8, 9 }
O = {       2, 3, 4, 5, 6,         }
P = { 0, 1, 2, 3, 4, 5, 6, 7, 8, 9 }
R = { 0, 1, 2, 3, 4, 5, 6, 7, 8, 9 }
V = { 0, 1, 2, 3, 4,               }

Already we can see that V and G are likely lower numbers, and O, L, J, and K are likely higher numbers.

What else do we have? Let's keep going:

We cannot instantly proceed to the next subtraction in as obvious a progression, as we'll need more information on the various letters involved.

However, looking at the puzzle, I'm interested in seeing if we can find any obvious examples of 0. You know, letter minus same letter sort of things. Because they will typically end up equalling 0 (or 9).

Why 9? Because of a borrow!

((5-1)+10)-5 = (4+10)-5 = 14 - 5 = 9

… that can be quite revealing too!

And it would appear we have one wonderful candidate in the bottom-most subtraction:

           KVRMR
          -JKRKB
           =====
            VKMK

Lookie there: R-R = K.

Usually, that would result in a 0. BUT, we also know that K can NOT be 0 (looking at our range table above).

So, that means it is being borrowed from, and it itself has to borrow, so we now also know that M is less than K (M « K).

And, as indicated above:

((R-1)+10)-R = 9!

We now know that K = 9!

That suddenly reveals a whole lot to us, due to our relational chains we've built. Let's update:

Also, with the new introduction of M being less than K:

B = { 0, 1, 2, 3, 4, 5,            }
G = {    1, 2, 3, 4, 5,            }
J = {                         8    }
K = {                            9 }
L = {                      7       }
M = { 0, 1, 2, 3, 4, 5,            }
O = {                   6          }
P = { 0, 1, 2, 3, 4, 5,            }
R = { 0, 1, 2, 3, 4, 5,            }
V = { 0, 1, 2, 3, 4,               }

And, our relational chains:

• V < G « O < L < J < K
• M « O < L < J < K

Because we don't yet know any relation of M compared to V or G, we have to keep them separate for now.

We also have a second disqualifier for K being 0… the ones place subtraction in that bottom-most subtraction:

R - B = K.

There's nothing further to the right that could borrow from this problem, so it can only exist in two states:

• R is greater than B
• R is less than B

Since we know that K is 9, there's NO OTHER pair of single digit numbers we can subtract to get 9, which tells us that:

• R is less than B (R « B)

Currently both R and B can be 0-5 (although now, B is 1-5, and R is 0-4). We'd need to find a combination where (R+10)-B is 9:

And from that, we're subtracting B, which is 1, 2, 3, 4, or 5. The answer has to be 9.

So:

10-1=9, 11-2=9, 12-3=9, 13-4=9, and 14-5=9

Hey, look at that… B is one greater than R (not just R « B, BUT: R < B)

Our relational chains:

• V < G « O < L < J < K
• M « O < L < J < K
• R < B « O < L < J < K

And our chart, of sorts:

B = {    1, 2, 3, 4, 5,            }
G = {    1, 2, 3, 4, 5,            }
J = {                         8    }
K = {                            9 }
L = {                      7       }
M = { 0, 1, 2, 3, 4, 5,            }
O = {                   6          }
P = { 0, 1, 2, 3, 4, 5,            }
R = { 0, 1, 2, 3, 4,               }
V = { 0, 1, 2, 3, 4,               }

If you look, the only letter we've not yet directly interacted with yet is 'P', although we already know enough about it (that it is 0-5, less than O, L, J, and K). And if you look closely, you'll notice that 'P' isn't even present in the letter division problem! So its identity will rely entirely on the proving of the other values.

Let's continue on:

M-K=M, BECAUSE we know M « K, AND BECAUSE we know the subtraction to the right is borrowing from it (because R < B), we have something like this: (M-1+10)-K=M

Can't really do much more with it at this point, but it is important to know to help us identify the borrows needing to happen.

Why don't we go ahead and find 0? If you look in the subtraction above the bottom one, we have another “letter minus same letter” scenario, and it doesn't equal K!

          JJGKL
         -LKBKV
          =====
           KVRM

We KNOW that V « L, so no borrow is happening there.

Therefore, K-K, or 9-9, equals 0. So R is 0!

… and B is 1! Because of our identified relationship.

Updating things!

Also, with the new introduction of M being less than K:

B = {    1                         }
G = {          3, 4, 5,            }
J = {                         8    }
K = {                            9 }
L = {                      7       }
M = {       2, 3, 4, 5,            }
O = {                   6          }
P = {       2, 3, 4, 5,            }
R = { 0                            }
V = {       2, 3, 4,               }

*NOTE: G is NOT 2, because G is greater than V (one greater, in fact), so we can similarly whittle that off.

Relational chains can look as follows now:

• R < B « V < G « O < L < J < K
• R < B « M « O < L < J < K
• R < B « P « O < L < J < K

Basically just down to V, G, P, and M.

And I think we have the means to find V: notice the second to last subtraction, the “LKBKV”. You know where we get that from? Multiplying the divisor (KJKK) by J (since it is the third subtraction taking place).

We KNOW the numeric values of K and J, in fact we know the values of L, K, and B. The only thing we don't know is 'V', and since V is in the one's place, that makes things super easy for us.

KJKK = 9899 J = 8

So: 9899 x 8 = 79192 = LKBKV!

V is 2!

Which means, because V < G, that G is 3!

Updating our records:

Also, with the new introduction of M being less than K:

B = {    1                         }
G = {          3                   }
J = {                         8    }
K = {                            9 }
L = {                      7       }
M = {             4, 5,            }
O = {                   6          }
P = {             4, 5,            }
R = { 0                            }
V = {       2                      }

Relational chains can look as follows now:

• R < B < V < G « M « O < L < J < K
• R < B < V < G « P « O < L < J < K

And then there were 2. We really just need to find M, or P, and we're done. And since there are no 'P' values in the puzzle, we need to target M. So let's look for some candidates:

Hey, how about this:

          JJGKL
         -LKBKV
          =====
           KVRM

One's place subtraction: L - V = M.

We KNOW L (7) is greater than V (2), so no borrow is happening.

L-V=M 7-2=5

M is 5. That means P is 4 by process of elimination.

Puzzle completed:

Also, with the new introduction of M being less than K:

B = {    1                         }
G = {          3                   }
J = {                         8    }
K = {                            9 }
L = {                      7       }
M = {                5             }
O = {                   6          }
P = {             4                }
R = { 0                            }
V = {       2                      }

Relational chains can look as follows now:

• R < B < V < G < P < M < O < L < J < K

I wasn't able to show it as well in text on the wiki, but I also made a point to mark up each subtraction to show whether a borrow occurred or not:

To be sure, there are likely MANY, MANY ways to arrive at these conclusions. What is important is being observant, performing little experiments, seeing if there can be any insights to have, even if whittling away knowing what things can NOT be.