Just Another 1-25 Logic Puzzle Walkthrough

Each whole number 01-25 occurs once in the puzzle.
Prime numbers are 2, 3, 5, 7, 11, 13, 17, 19, 23.
Type all one-digit numbers with a zero in front, e.g. 03. But 0 does not count as a digit except in 10 and 20.

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B5 tells you that B4 is 17.

Important to realize from B4 is that main diagonals are the only prime numbers. 9 cells are main diagonal cells and there are 9 prime numbers from 1-25. This means that all prime numbers are along the main diagonals and all non-prime numbers are not. So C3 has to be prime. It also has to be two-digit, according to B4. It also cannot be 13 or 19, according to E5. This leaves only 11 and 23. If C3 is 23, then 19 could not be adjacent to it. 19 also cannot be adjacent to 17, so if C3 is 23, 19 must go in A1 or E1, but E5 tells you that no number in Row 1 can contain a 9. Therefore C3 is 11.

Take a look at B5. Obviously, A5 might be 23. The other option is C2 – 8. You know that A5 is prime, and cannot be 13 or 19. So what’s left is 2, 3, 5, or 23. 23 is too big, because C2 would have to be 31. 5 doesn’t work for A5 because C2 would have to be 13, but C2 isn’t on the main diagonals, and is adjacent to 11. 3 doesn’t work because C2 would have to be 11, which is already in C3. 2 doesn’t work because then C2 would have to be 10. But Row 1’s multiple of 5 cannot be 5 itself because E5 says the greatest digit in Row 1 is 3. 5 is the only prime multiple of 5, so Row 1’s multiple of 5 must go in B1, C1, or D1. In any of those cells, C2 is ruled out as a multiple of 5 because of C3’s clue. So C2 cannot be 10, meaning that according to B5, A5 is E5 + 16, or 23.

A5 tells you that A4 is 24 or 25. It also tells you that C5 is 12, 18, or 24. This means, according to C3, that D5 must be the multiple of 5 in Row 5, and that means that C4, D4, and E4 cannot be a multiple of 5. So the multiple of 5 in Row 4 has to be A4. So A4 is 25.

According to E5, no number in Row 1 contains a 6 or 8. This means, according to A4, that the numbers containing a 6 or 8 must all be in Row 2. B2 and D2 must be prime, so cannot contain a 6 or 8. B5 is 6, so now you know that A2, C2, and E2 are some combination of 8, 16, and 18. Since 25 is A4, A3 and B3 cannot be a multiple of 5. According to C3 then, the multiple of 5 in Row 3 must be D3 or E3. Either way, D2 cannot be a multiple of 5, so B2 must be. And since B2 is also prime, it must be 5.

The only prime numbers remaining are 2, 3, 13, and 19. According to E5, A1 can only be 2 or 13. D4 can be 2, 3, or 19. What B2 tells you is that the two primes in that main diagonal are either 2 + 3 or 13 + 19. The other combinations make the whole diagonal fall in between 35 and 45. So one of the diagonals finishes with 2 + 3 and the other with 13 + 19. According to E5, 2 and 3 cannot be adjacent, so they must go in the A1-E5 diagonal. A1 cannot be 3, being adjacent to 5, so it is 2 and D4 is 3.

D2 cannot be 13, being adjacent to 11, so it is 19 and E1 is 13.

A4 told you that 18 is in Row 2. So A5 tells you that C5 is 12 or 24. Therefore, according to E1, the cell in question must be A3. That’s the only cell that shares a row/column/diagonal with A1, A4, A5, and C5.

Now you can narrow down B1. All the possibilities, according to E5, are 1, 2, 3, 10, 11, 12, 13, 20, 21, 22, 23. Of those, 2, 3, 11, 13, and 23 are already placed. 12, 20, 21, and 22 all contain a 2. B1 does not share a row/column/diagonal with A3, so it cannot contain a 2. B1 is also adjacent to 5, so it cannot be a multiple of 5, ruling out 10. This leaves only 1 for B1.

Since B2 is 5, the multiple of 5 in Row 1 has to be D1. According to E5 and what is left, it must be 10 or 20. But according to E1, it cannot be 20, so D1 is 10. D5, then, also not being able to be 20, is 15.

C1 can contain no digit greater than 3 (E5), so what’s left is 12, 20, 21, and 22. 20, of course, is the last multiple of 5, so it goes in Row 3. So C1 is 12 or 21 or 22. D5 says if 21 and 22 are in the same row, 11 is adjacent to 12. 21 and 22 in the same row would mean Row 3 because C4 and E4 do not share a row/column/diagonal with A3 and C5 is a multiple of 6. But if 21 and 22 are both in the same row, then C1, which only has three options at this point, has to be 12. And if C1 is 12, then 11 is not adjacent to 12. So according to D5, 21 and 22 are NOT in the same row, meaning one of them is C1. Now look at B1. Now that you know C1 is 21 or 22, the only place 12 could go in Column C is C5. 24 cannot go in C2 or C4, since it contains a 2, so 11 cannot share a column with 12. Since C5 at this point is either 12 or 24, now you know it is 24.

C5 refers to 19. C3 told you that a multiple of 5 still has to go in Row 3 but cannot be adjacent to 25. The only multiple of 5 left is 20, so D3 or E3 is 20. Either way, 19 is adjacent to 20, so according to C5, 19 cannot be adjacent to 18. You know from A4 that 18 is in Row 2, so 18 must be A2.

You just established that 19 is adjacent to 20. C2 and E2 are 8 and 16 or vice versa, so A2 must be referring to itself. So A3 and B3 must be greater than 9 (B4) but less than 18 (A2). 16 is C2 or E2 so all that’s left is 12 and 14. A3 must contain a 2 (E1), so A3 is 12 and B3 is 14.

What C5 said about odd numbers is false for some even number, according to B3. As you go through the options, you realize the only possibility is 20. 20 is D3 or E3, so it is adjacent to 19. And it also must be adjacent to 21, which could be C1 or D3/E3. C1 is not adjacent to D3/E3, so 20 and 21 are D3 and E3 or vice versa. This means that C1 is 22.

The only numbers yet to be placed that could satisfy C1’s clue are 10 and 4. But A1 says that if Column E contains 4, it also contains 16. E3 is 20 or 21, so if E4 is 4, then E2 is 16, not 8. 4 and 8 could both go in Column C, but that would create another instance, not another column; C1 would still not be satisfied. So 10 must be in the same column as 20, meaning 20 is D3.

According to E1, 21 must go in E3.

E3 tells you that 4 and 8 do, in fact, share a column. A1 says they can’t go in Column E. A4 says 8 and 16 are in Row 2, so C2 is 8 and C4 is 4.

9 and 16 remain. A4 tells you that 16 is E2 and so 9 is E4.

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