<– Back to Lively Letters Logic Puzzle II

  1. A1 tells you “C4 is not a letter or a number”, which means it has to be an X

  2. C4 tells us that “All letters are in cells that have BOTH a prime cell number and a cell letter whose position in the alphabet is also a prime number. This means that the letter can only be in cells that have both a cell number of either, 2, 3 or 5 (1 and 4 are not prime numbers) and a cell letter of either B, C or E. Therefore all other cells are either a number or an X. A2 is one of these other cells since it does have a cell letter of B, C or E. We are also told that A2 is not a number. Therefore A2 is an X.

  3. A2 tells us that “There are no numbers in the corners, but not all corners are Xs”. Since C4 tells us that A5 and E1 can’t be letters, they have to be Xs. This also means that E5 is a letter since they are not all Xs.

  4. A2 also tells us that “If D2 is not an X, then B2 has no letters in its row and no letters in its column”. So if B2 had no letters in its row and no letters in its column, that would only leave 4 cells as possibilities for letters based on the criteria laid out in the C4 clue: C3, C5, E3 and E5. However, since we know that there are 5 letters (A-E) this is not possible. Therefore D2 has to be an X.

  5. D2 tells us that “D4 is the only number in this row” (row D). So, since C4 tells us that there are no letters in this row, we know that all of the cells in the row other than D4 are Xs (D1, D3, and D5)

  6. A5 tells us that “C5 is a number that is either its cell letter’s position in the alphabet or its cell number”. This tells us that C5 is either a 3 or a 5, meanwhile E1 tells us that “No numbers that are in columns left of a number are greater than that number”. This tells us that C5 can’t be a 3 because if it were, there would have to be a number greater than 3 in a leftward column (and there is not enough room in column 5 for 3, 4 and 5 since we already know that E5 is not a number) Therefore C5 is 5.

  7. D3 tells us that “Every column has exactly 3 Xs”. This tells us that B5 is an X since it is the only place that the third X in column 5 can be (again, E5 is a letter).

  8. D3 also tells us that B1 and C1 have to be numbers since there are already 3 Xs in the column and the cells don’t match the C4 requirement for letters.  Then B5 tells us “There are 2 numbers in this row” (row B). This has accounted for the last of the 5 numbers. We know now that B1, C1, D4 and C5 and somewhere from B2-B4 are the 5 numbers. Therefore all other cells that don’t meet C4’s criteria for letters now have to be Xs. So A3, A4, and E4 are Xs.

  9. E4 tells us that “There are two rows with only one X.” Since we already have multiple Xs for rows, A, D and E, we know this means rows B and C have to be these rows with only one X. This tells us C2 and C3 have to be letters, since we already know they can’t be numbers. As for row B, B4 does not meet the C4 criteria, which means B4 has to be the last number (and B2 and B3 have to be the last letters). So now that all numbers and letters are accounted for, all other cells have to be an X. The only cells left are E2 and E3.

  10. E3 tells us that “Three numbers are the same as their cell number and/or their cell letter’s position in the alphabet”. We already know that 5 is one of these numbers since it has a cell number of 5. 4 has to be one of these numbers since it has to be column 4 (because of the clue in cell E1 and the fact that E5 is a letter). 3 can’t be one of these numbers since, because of the clue in E1 combined with where the letters have to go based on the clue in C4, it also has to be in column 4. Therefore 1 or 2 has to be the other one.  This means B1 has to be 1 and C1 has to be 2, because if it were reversed, they would both satisfy the condition and there would be four numbers that do so instead of three.

  11. Since A3 tells us that “B and C are adjacent”, D5 tells us that “A is adjacent to B”, and B1 tells us that “B is NOT diagonally adjacent to C”, then A, B, and C must be in the block of cells that includes B2, B3, C2 and C3. (Since no letters can be in row A.) D1’s condition states that “All letters except for one are both different from their cell letter and have a position in the alphabet that is different from their cell number”. B1’s clue guarantees that is either B or C that is the single exception for this D1 condition. This is because the only possibility that would have both B and C as “different from their cell letter AND have a position in the alphabet that is different from their cell number” is with B and C diagonally adjacent (B2 being C and C3 being B). So, all of this adds up to telling us that E5 can’t be E (because there is only one letter that is the exception for the D1 condition). And since we’ve already said, A B, and C are in one of the other four cells, the only letter left is D. So E5 is D.

  12. C1 tells us that “A is not adjacent to 4” and E5 tells us “E is not adjacent to 4”. This means that cell C3 is neither A nor E since cell C3 has to be adjacent to 4. It also cannot be C because if it were then both C and B would be exceptions for the D1 condition, but we know that only one of them is. Therefore, C3 has to be B.

  13. C3 tells us that “E and 4 are in the same row”. This means B4 has to be 4 because it is impossible for D4 to share a row with E. That also makes D4 have to be 3, as it is the last number. Then we know that “E is not adjacent to 4”, so E has to be in cell B2 in order for it to share the same row as 4. Then we also know that “A is not adjacent to 4”, so A has to be in the only non-adjacent cell left: C2. This leaves us with the last cell having to be C (also note that this satisfies the D1 condition).