This puzzle was initially solved in the puzzl-y-tastic channel on discord. The transcript was taken from the daily backup, with discord-specific formatting removed. XPW +------- XQJJ | TQYLJL -QYWX ==== XWLJJ -XWGPE ===== TTXL -TJPL ==== WG base: 10 letters: EGJLPQTWXY Lets do a strafing run along the left side (remember, leftmost subtraction top values are NOT borrowing): Q << T Second row, we’ve got two in a row that cancel out (X-X=nothing, W-W=nothing), so there’s no borrows, no takes there Next one up is therefore L-G=T, which isn’t borrowing: G << L Bottom row we’ve got a cancellation leading the way (T-T=nothing), which then leaves us with T-J=nothing, a hopefully familiar pattern (T isn’t borrowing) We know that in order for this math to work, the T is being taken from So: T-1-J=nothing Therefore: J < T (J is exactly one less than T) We have some common letters in our assertions, so let’s do some combining: Q << J < T << L X << J < T << L G << L Applying these assertions to our range table: E = { 0 1 2 3 4 5 6 7 8 9 } G = { 0 1 2 3 4 5 6 7 8 } J = { 2 3 4 5 6 7 } L = { 5 6 7 8 9 } P = { 0 1 2 3 4 5 6 7 8 9 } Q = { 0 1 2 3 4 5 6 } T = { 3 4 5 6 7 8 } W = { 0 1 2 3 4 5 6 7 8 9 } X = { 0 1 2 3 4 5 6 } Refreshing puzzle display: XPW +------- XQJJ | TQYLJL -QYWX ==== XWLJJ -XWGPE ===== TTXL -TJPL ==== WG base: 10 letters: EGJLPQTWXY Q << J < T << L X << J < T << L G << L Low hanging fruit- bottom right: L-L=G. We know from exploring in class that G is either 0 or the largest possible value in the base (9). We also know, that since this subtraction is on the far right end of a term, it isn’t being taken from, so we’re not borrowing. Therefore, G is 0. Looking elsewhere in the puzzle for other G’s... there’s one we’ve already visited: L-G=T We know that L isn’t borrowing, but at the time we hadn’t explored whether the L is being taken from. Turns out, it IS being taken from (as L-zero equalling something other than L would infer) So, not only is T somewhat less than L, T is in fact *exactly one less than* L T < L Updating: XPW +------- XQJJ | TQYLJL -QYWX ==== XWLJJ -XWGPE ===== TTXL -TJPL ==== WG base: 10 letters: EGJLPQTWXY Q << J < T < L X << J < T < L Now that we know T is exactly one less than L, we can resolve differences between the ranges of J/T and L Updating range table: E = { 1 2 3 4 5 6 7 8 9 } G = { 0 } J = { 3 4 5 6 7 } L = { 5 6 7 8 9 } P = { 1 2 3 4 5 6 7 8 9 } Q = { 1 2 3 4 5 6 } T = { 4 5 6 7 8 } W = { 1 2 3 4 5 6 7 8 9 } X = { 1 2 3 4 5 6 } Continuing on, we’ve got some more candidates to investigate: Second row, 2nd from right: J-P=T Here, we know that J < T (J is exactly one less than T) Which actually narrows down P considerably. To two possible values (one if it isn’t being taken from: 9, one if it is being taken from: 8) Then, looking at the next subtraction to the right: J-E=X, we know that X is somewhat less than J, therefore J doesn’t need to borrow Which means: P is 9 E is somewhat less than J Updating: XPW +------- XQJJ | TQYLJL -QYWX ==== XWLJJ -XWGPE ===== TTXL -TJPL ==== WG base: 10 letters: EGJLPQTWXY E << J < T < L Q << J < T < L X << J < T < L Updating range table: E = { 1 2 3 4 5 } G = { 0 } J = { 4 5 6 } L = { 6 7 8 } P = { 9 } Q = { 1 2 3 4 5 } T = { 5 6 7 } W = { 1 2 3 4 5 6 7 8 } X = { 1 2 3 4 5 } A further optimization: because of J < T < L, and the fact we have them locking into a subrange of 3 a piece No matter what J, T, and L are, ONE of them will be 6 (4 5 6, 5 6 7, 6 7 8) updating range table: E = { 1 2 3 4 5 } G = { 0 } J = { 4 5 6 } L = { 6 7 8 } P = { 9 } Q = { 1 2 3 4 5 } T = { 5 6 7 } W = { 1 2 3 4 5 7 8 } X = { 1 2 3 4 5 } Bottom row, 2nd from right: X-P=W We know the X is borrowing, we know the X is NOT being taken from We know P is 9 Highest digit trick: X < W (X is exactly one less than W) Updating: XPW +------- XQJJ | TQYLJL -QYWX ==== XWLJJ -XWGPE ===== TTXL -TJPL ==== WG base: 10 letters: EGJLPQTWXY E << J < T < L Q << J < T < L X < W << J < T < L Updating range table: E = { 1 2 3 4 5 } G = { 0 } J = { 5 6 } L = { 7 8 } P = { 9 } Q = { 1 2 3 4 5 } T = { 6 7 } W = { 2 3 4 5 } X = { 1 2 3 4 } Further J T L optimization: now that we’ve got them down to 2 values a piece, our possible solutions are: 5 6 7 or 6 7 8. Notice how 6 and 7 (or the middle term, T) are used regardless of scenario. So, nothing else can be 6 or 7 (really just knocks the 7 off from Y) Top row, rightmost: L-X=J (L is bigger than both X and J- it doesn’t borrow) We know the relationship of L to J—- L is exactly TWO greater than J Therefore, we know what X is: 2 And, because W is one greater than X, W is 3 Updating range table: E = { 1 4 5 } G = { 0 } J = { 5 6 } L = { 7 8 } P = { 9 } Q = { 1 4 5 } T = { 6 7 } W = { 3 } X = { 2 } Second row, rightmost (again): J-E=X Because we know J is bigger than X, and we now know that X is 2... E is exactly 2 less than J Updating: XPW +------- XQJJ | TQYLJL -QYWX ==== XWLJJ -XWGPE ===== TTXL -TJPL ==== WG base: 10 letters: EGJLPQTWXY E < ? < J < T < L Q << J < T < L X < W << J < T < L Looking at what we know about E and what we know about J, and how they have to relate… we actually have a solution for E and J. The only value E can be is 4, making J- 6. Updating range table: E = { 4 } G = { 0 } J = { 6 } L = { 8 } P = { 9 } Q = { 1 5 } T = { 7 } W = { 3 } X = { 2 } So, we’re basically down to Q and Y. Figure out one, we instantly have the other There are two candidates. Both on the top row. I’m going to do Y-W=L, because we know L is 8 Y-W=L Y has to be less than L (because we know what 8 (L) and 9 (P) are. And we know Y is NOT being taken from So: Y is exactly 2 less than W (even if we didn’t do this 2nd largest digit track- Y << W, and that would give us the identity of Y regardless) Y is 1, Q is 5 Updating range table: E = { 4 } G = { 0 } J = { 6 } L = { 8 } P = { 9 } Q = { 5 } T = { 7 } W = { 3 } X = { 2 } We can verify some things... Q-Y=W. If this is all correct, Q is not borrowing, and Q IS being taken from. Y is 1. So: Q-1-1=W 5-1-1=3 and... W is 3 Looks like things are checking out.