=====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: ??????? +-------- HPCHC | NHEYMCN -CWEYE ===== CIIYC -HPCHC ===== HIOHIN - PRERE ====== ?????? base: 12 letters: CEHIJMNOPRWY difficulty: medium missing: J 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 12 the available counting digits are 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, A, and B. * In addition to the familiar 0-9, the letters A and B are now numbers, whose quantities are equivalent to that of the ones in decimal we lovingly call "ten" and "eleven". * As 8+1 = 9, 9+1 = A, A+1=B. B+1=10. Because B is the last counting digit available, when we exhaust the digits, we cycle to 0 and carry 1. 10 base 12 is the quantity we know of in base 10 as "twelve". * However, 10 base 12 should be pronounced and thought of as "one zero base twelve", and NOT "ten". Ten is a decimal (base 10) construct, and 10 base 10 does NOT equal 10 base 12. * Having extra numbers (A, B) than what we are acclimated to with base 10 means any memorized mathematical facts, especially with respect to carried/multi-place values, are no longer valid.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 even base, that also means that even numbers will still be even. ====Getting underway==== Looking for anything to get us started, let's look at a couple layers of the left-most subtractions: * (top row, leftmost): N-C=nothing * nothing is zero; since we cannot have two letters with the same value, something is happening to make both letters be identical; in our case, while the N is not borrowing (it is on the left, it cannot borrow), N however is being taken from. And, N-1 is the same as C. So: C is exactly one less than N * C < N * (second row, leftmost): C-H=H * being on the left, C does not borrow. It MIGHT be taken from, or it may not. H is somewhat less than C (H << C), and H+H is not greater than "twelve". * since we also know something else about C, we can combine both the assertions into a longer chain: H << C < N * (third row, leftmost): H-nothing=nothing * as nothing is zero, and H isn't zero, yet H becomes zero from being taken from, this lone dangling letter is a dead giveaway for the value of one. H is 1. * as H+H is C (but C may or may not be taken from). C is 2 or 3. * With C down to exactly two values. And N being one greater than C, we can eliminate any other letter being 3 (if C is 2, N is 3; if C is 3, N is 4... 3 is occupied in either scenario). Our current relational chains: H << C < N Our chart: C = { 2, 3 } E = { 0, 2, 4, 5, 6, 7, 8, 9, A, B } H = { 1 } I = { 0, 2, 4, 5, 6, 7, 8, 9, A, B } J = { 0, 2, 4, 5, 6, 7, 8, 9, A, B } M = { 0, 2, 4, 5, 6, 7, 8, 9, A, B } N = { 3, 4 } O = { 0, 2, 4, 5, 6, 7, 8, 9, A, B } P = { 0, 2, 4, 5, 6, 7, 8, 9, A, B } R = { 0, 2, 4, 5, 6, 7, 8, 9, A, B } W = { 0, 2, 4, 5, 6, 7, 8, 9, A, B } Y = { 0, 2, 4, 5, 6, 7, 8, 9, A, B } ====Diving deeper==== Taking a second pass, let's see what can be gleaned from the subtractions that are second in from the left on each row (as best as can currently be gleaned): * (top row, 2nd from left): H-W=C * we know from the subtraction to its left that the H here needs to borrow. Therefore: * H << W * H << C (which we already knew) * (second row, 2nd from left): I-P=I * while we don't yet know if I needs to borrow (or is being taken from), we DO know that P is either 0 or our largest counting symbol, B. We know this because we have two of the same letter involved (I, I). So: P = { 0, B } * (third row, 2nd from left): I-P=? * here we can actually solve for P, because we know the I in this case is borrowing. Therefore, P is GREATER than something. As zero is greater than nothing, P is our highest symbol, B. Our current relational chains: H << C < N H << W Our chart: C = { 2, 3 } E = { 0, 2, 4, 5, 6, 7, 8, 9, A } H = { 1 } I = { 0, 2, 4, 5, 6, 7, 8, 9, A } J = { 0, 2, 4, 5, 6, 7, 8, 9, A } M = { 0, 2, 4, 5, 6, 7, 8, 9, A } N = { 3, 4 } O = { 0, 2, 4, 5, 6, 7, 8, 9, A } P = { B } R = { 0, 2, 4, 5, 6, 7, 8, 9, A } W = { 2, 4, 5, 6, 7, 8, 9, A } Y = { 0, 2, 4, 5, 6, 7, 8, 9, A } ====Round three, discovering zero==== On our attempted third pass, we see what there is to discover: * (top row, 3rd from left): E-E=I * that double letter pattern appears again, letting us know that I can be either 0 or B, our largest symbol. * BUT, since we already know what B is, I is therefore automatically determined to be 0, by process of elimination. * I = 0. * This also means that E is NOT borrowing, nor is it being taken from * (2nd row, 3rd from left): I-C=O * now that we know I is 0, we know it needs to borrow. * (3rd row, 3rd from left): O-R * at present nothing can be determined Our current relational chains (unchanged from the prior round): I < H << C < N << P I < H << W << P Our chart: C = { 2, 3 } E = { 2, 4, 5, 6, 7, 8, 9, A } H = { 1 } I = { 0 } J = { 2, 4, 5, 6, 7, 8, 9, A } M = { 2, 4, 5, 6, 7, 8, 9, A } N = { 3, 4 } O = { 2, 4, 5, 6, 7, 8, 9, A } P = { B } R = { 2, 4, 5, 6, 7, 8, 9, A } W = { 2, 4, 5, 6, 7, 8, 9, A } Y = { 2, 4, 5, 6, 7, 8, 9, A } ====Connecting some dots==== Now that we have some further clues, especially in the form of identified borrows/takes, and known letters, we can solve for some additional values: * (2nd row, left-most): C-H=H * before, we had C narrowed down to 2 or 3, depending on whether the C was being taken from or not. Now that P has been identified as B, in I-P=I, we know I has to borrow. So: C is being taken from. * Therefore: C is 3 (N by extension is 4). * (1st row, 2nd from left): H-W=C * we knew H, we knew approximately what C was, and now we know that H is borrowing, but NOT being taken from. So, now knowing H and C, we can solve for W: * H+base-W=C * 1+10-W=3 * 11-W=3 * 11-3-W=0 * 11-3=W * A=W (W is A). Our current relational chains: I < H << C < N << W < P Our chart: C = { 3 } E = { 2, 5, 6, 7, 8, 9 } H = { 1 } I = { 0 } J = { 2, 5, 6, 7, 8, 9 } M = { 2, 5, 6, 7, 8, 9 } N = { 4 } O = { 2, 5, 6, 7, 8, 9 } P = { B } R = { 2, 5, 6, 7, 8, 9 } W = { A } Y = { 2, 5, 6, 7, 8, 9 } ====Quick grabs==== Glancing around the puzzle, we can solve for two additional letters: * (2nd row, 2nd from right): Y-H=H hey, notice the similarity to C-H=H. * The C is taken from, so we know here the Y isn't (and the Y is borrowing, because of math- H+H) * Y therefore is one less than C, or 2. * (2nd row, 3rd from right): I-C=O * we know I and C, we know I borrows, and is NOT being taken from. * I+base-C=O * 0+10-3=O * 10-3=O * 9=O (O is 9). * (top row, rightmost): M-E=Y * we know M doesn't borrow (Y-Y=I indicates that) * we know Y << M, and E << M. * as Y is 2, E is exactly TWO less than M. Our current relational chains: I < H < Y < C < N << E < ? < M << O < W < P Our chart: C = { 3 } E = { 5, 6 } H = { 1 } I = { 0 } J = { 5, 6, 7, 8 } M = { 7, 8 } N = { 4 } O = { 9 } P = { B } R = { 5, 6, 7, 8 } W = { A } Y = { 2 } ====Solving for E==== At this point, our vectors for deriving further relational clues starts to wane: while E has a connection with M (two less), we still don't know anything about J or R, and indeed, with J entirely missing, we would need to solve for E, M, and R. But, at present, we have no other helpful clues to derive relationally. So what can we do? Multiplication! Notice how the divisor ends in C (which we know is 3), and two of the subtrahend terms end in E (which is 5 or 6)? We can see what possibilities there are in the multiplication table: 1 2 3 4 5 6 7 8 9 A B x--------------------------------- 3 | 3 6 9 10 13 16 19 20 23 26 29 ^ ^^ ^^ So, we need to multiply C (3) times something to get a values that ends in a 5 or a 6 (our two whittled down possibilities for E). And look: we get no 5's! So, E is 6! And, because M is two greater than E, M is 8! Our current relational chains: I < H < Y < C < N < ? < E < ? < M < O < W < P Our chart: C = { 3 } E = { 6 } H = { 1 } I = { 0 } J = { 5, 7 } M = { 8 } N = { 4 } O = { 9 } P = { B } R = { 5, 7 } W = { A } Y = { 2 } ====Solving for R==== Now, we're down to J and R. We have no context on J, save for whatever everything else isn't. And the ONLY lead we have on R is in the bottom-most subtrahend, the second letter from the right. So, we'll need to dive in deeper with multiplication: Because there were THREE possible values when multiplied by 3 to equal 6, we need to test each of them until we rule them out. So: C: 3 3 3 H: 1 1 1 x2 x6 xA -> x2 x6 xA == == == == == == E: 6 16 26 2 6 A +0 +1 +2 <-- handle the carry from round 1 == == == 2 7 10 As we are looking for R, which can be 5 or 7, notice how only ONE of those results suits us: multiplying by 6, gives us 7. R is 7. J, by process of elimination, is 5. And to look ahead to our verification: multiplying by 2 in this second stage (and addressing the carry), gives us our Y. Our current relational chains: I < H < Y < C < N < J < E < R < M < O < W < P Our key: IHYCNJERMOWP Our chart: C = { 3 } E = { 6 } H = { 1 } I = { 0 } J = { 5 } M = { 8 } N = { 4 } O = { 9 } P = { B } R = { 7 } W = { A } Y = { 2 } ====Determining our quotient==== Part of the joy of the solve4 puzzling experience is needing to ascertain both the quotient and remainder as part of the completion of the puzzle. Due to the steps we needed to take to solve all our letters, we basically have the quotient figured out: HPCHC x ? = CWEYE (what did we multiply above to get our Y: 2) HPCHC x ? = HPCHC (this is simple, 1: H) HPCHC x ? = PRERE (this would be the 6 to get us our R value, so E) Our quotient therefore is: YHE ====Determining our remainder==== Finally, let us get our remainder through performing a set of subtractions, on that last term: xH>IxO>H>I>N - P R E R E = = = = = = H H E N W Our remainder is: HHENW Then, for verification, our string would be: YHE:HHENW This puzzle is now complete.