=====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: IZNLI +--------- MZML | MNDMDILI -MZML ==== DNDDI -DLDDL ===== IMMML -ININN ===== INZNI - MZML ===== NMNM base: 6 letters: DILMNZ 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 6 the available counting digits are ONLY 0, 1, 2, 3, 4, and 5. 6, 7, 8, and 9 do NOT exist in base 6. Similarly, 5+1 = 10. Because 5 is the last counting digit available, when we exhaust the digits, we cycle to 0 and carry 1. However, 10 base 6 should be pronounced and thought of as "one zero base 6", and NOT "ten". Ten is a decimal (base 10) construct, and 10 base 10 does NOT equal 10 base 6. Lacking 6, 7, 8, and 9 means any memorized mathematical facts, especially with respect to carried/multi-place values, are no longer valid.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. ====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. Here, with this puzzle, one of my common starting strategies will not yield any results; the left-most subtractions do not reveal anything of value, but they do indicate something for the next layer of subtractions: the top letters are all greater than the bottom letters. ====Discovering I==== So, taking a look at that next layer of subtractions (from the left): * N-Z=D * Z << N * D << N * N-L=I * L << N * I << N * M-N=I * N << M * I << M * the bottom-most is an exception, for it tells us two things: * I-1 is zero. Zero plus one is one. I = 1. * this was also revealed in the quotient: when we multiply the divisor by I, we see the divisor as the subtrahend. * that the next subtraction in, N-M=N, that: * N << M (which we already knew from the previous subtraction) That gives us the following relational chains: Z << N << M D << N << M L << N << M I << N << M ====Revealing N and M==== This actually conveniently reveals N and M to us: * with only 6 total counting digits (0-5), N is demonstrated here to be none of 0, 1, 2, or 3; AND it is less than M (so not 5). That leaves one value: 4. * similarly, M cannot be 0, 1, 2, 3, or 4. That leaves only 5. Our chart: D = { 0 2, 3 } I = { 1 } L = { 0 2, 3 } M = { 5 } N = { 4 } Z = { 0 2, 3 } Our current key: 0 1 I 2 3 4 N 5 M ====Finding L==== Utilizing the divisor and dividend, we can use multiplication to also aid us in letter discovery. If we look at the divisor, we see it ends with L. Similarly, there's an L in the quotient. Multiplying MZML by L, we get the term ININN, the second to last subtrahend. That means, therefore, that L * L = N (again, looking at single digit ones places; there could be a carry, but that would impact the next round of multiplication). So, with L potentially being 0, 2, or 3, and N having been identified as 4, all we have to do is multiply each value of L by itself, looking for a 4 in the resultant one's place: * L = 0: 0 * 0 = 0 (nope) * L = 2: 2 * 2 = 4 (yes) * L = 3: 3 * 3 = 13 (nope) Only one of those multiplications resulted in a 4 in the one's place, and that is when L = 2. So L's value is now found. Our key: 0 1 I 2 L 3 4 N 5 M And our chart: D = { 0 3 } I = { 1 } L = { 2 } M = { 5 } N = { 4 } Z = { 0 3 } And some modified relational chains: Z << N << M D << N << M I << L << N << M Now we merely need to discover D or Z and we've got our key completed. ====Determining D==== If we look in our topmost subtraction, there are two transactions that nicely isolates D for us: * D - M = N * M - L = D We even know the situation regarding borrowing on both (D is less than M). For simplicity, let's go with the second subtraction, as no borrowing is needed, and being the rightmost subtraction, nothing is taking from us. M 5 -L -2 == --> == D 3 And there we have it: D = 3. As a result of elimination of possibilities, Z = 0. Our final relational chain: Z < I < L < D < N < M Our final chart: D = { 3 } I = { 1 } L = { 2 } M = { 5 } N = { 4 } Z = { 0 } And our final, completed key: 0 Z 1 I 2 L 3 D 4 N 5 M Solution is obtained! Onto verification: ====Verifying our key==== The best way to verify the puzzle with our key is to convert the dividend and divisor to its numeric equivalent, perform the division, and compare the resulting quotient and remainder against those found in the letterified puzzle: divisor: MZML --> 5052 dividend: MNDMDILI --> 54353121 And let's do some long division! +--------- 5052 | 54353121 5052 goes into 5435 exactly once: 1 +--------- 5052 | 54353121 -5052 ==== 3433 As 3433 is too small for 5052, it fits in 0 times, so we record that and drop down our next dividend digit (1), to get: 34331. 10 +--------- 5052 | 54353121 -5052 ==== 34331 5052 fits into 34331 plenty of times. Let's see how many: 5052 * 0 = 0 5052 * 1 = 5052 5052 * 2 = 14144 5052 * 3 = 23240 5052 * 4 = 32332 5052 * 5 = 41424 In the case of 34331, 41424 is too big, so 32332 is the closest match, so it goes in 4 times: 104 +--------- 5052 | 54353121 -5052 ==== 34331 -32332 ===== 15552 5052 best fits into 15552 two times: 1042 +--------- 5052 | 54353121 -5052 ==== 34331 -32332 ===== 15552 -14144 ===== 14041 5052 best fits into 14041 once: 10421 <-- quotient +--------- 5052 | 54353121 -5052 ==== 34331 -32332 ===== 15552 -14144 ===== 14041 - 5052 ===== 4545 <-- remainder Converting our quotient and remainder back to letters: * quotient: 10421 --> IZNLI * remainder: 4545 --> NMNM And comparing against the problem we were given: * quotient: IZNLI <-> IZNLI * remainder: NMNM <-> NMNM We have a match on both counts! Verification successful! This puzzle is now complete.