A ‘Scending Numbers Logic Puzzle Walkthrough

Prime numbers: 2, 3, 5, 7, 11, 13, 17, 19, 23

Perfect squares (and their square roots): 1(1), 4(2), 9(3), 16(4), 25(5)

Remember to enter every answer as a 2-digit value, e.g. 01.

(Back to puzzle)


1.) You start with all the main rules of the puzzle. Each row is in ascending or descending order, with a prime number in the middle, starting or ending with a perfect square. Perfect squares are not adjacent to their square roots, and no number is diagonally adjacent to a consecutive number, e.g. D1 cannot be 1.
Look at Row 4. It is obviously in descending order so only 5 numbers can possibly go in B4, C4, D4. C4 must be a prime number, so it must be 5 or 7. It cannot be 5 because then nothing would fit in D4. So C4 is 07.

2.) B4 is now either 8 or 9. According to E4, each row starts or ends with a perfect square. There are five rows and five perfect squares, so in effect, all perfect squares must be in Column A or E. This means B4 cannot be 9, so it is 08.

3.) E5 must be a prime number, according to C4. It also must be between 5 and 12, according to B4, so it must either be 5 or 11, since 7 is already used. If E5 is 5, then D4 would have to be 6 because you know D4 must be 5 or 6. But if E5 is 5 and D4 is 6, E2’s rule is violated. So E5 cannot be 5; it must be 11.

4.) Since E5 is 11, A5 must be a perfect square. The ones remaining are 1, 9, and 16. It clearly cannot be 9 because then the row would be ascending and 9 would be separated from 11 by 4 spaces. You’d violate C3’s rule. So A5 is 1 or 16. You also know that C5 is prime. If A5 is 1, then Row 5 is ascending, so C5 could be 3, 5, or 7. 7 is already used. If C5 is 3, B5 would have to be 2, but 2 is already used, so C5 cannot be 3. If C5 is 5, again, D4 would have to be 6, and again E2’s rule would be violated. So C5 cannot be 5. Nothing works for A5 to be 1, so it must be 16.

5.) The only prime number greater than 11 and less than 16 is 13. So C5 is 13.

6.) Obviously D5 has to be 12.

6.) C5 tells you that 17 is not in Column C. This means the only prime numbers left for C1 and C2 are 3, 5, and 23. The only way for 3 to work in Column C is if that row begins 1-2 or ends 2-1. But 2 is in E2, so 3 cannot be in Column C. So C1 and C2 are 5 and 23 or vice versa. Similar to 3, if you put 23 in Column C, the column must start or end with 25. Since 25 is already used, you know that 23 must go in C2.

7.) So B2 is 24.

7.) The only other prime to go in C1 is 05.

8.) Now you know that D4 must be 6.

8.) The only remaining perfect squares are 1 and 9. If 9 is in Row 1, B1 or D1 would need to be 6 or 7 or 8, but all of those are already used. Therefore 1 must go in A1 or E1 and 9 must go in A3 or E3. According to E2, A3 cannot be 9. Therefore E3 is 09.

9.) Row 1 must begin 1-3-5 or end 5-3-1, since 2 and 4 are used. According to E2, D1 cannot be 3. Therefore Row 1 begins 1-3-5. A1 is 01 and A2 is 03.

10.) The A1-E5 diagonal sums to 61. Right now the A5-E1 diagonal sums to 43. Since D2 is part of one of the main diagonals, what E3 is saying is that either
61 + D2 = 43 + D2 + E1 OR
61 = 43 + D2 + D2 + E1. The smallest remaining numbers are 14 and 15, so the second equation certainly cannot be true because the A5-E1 diagonal would sum to at least 86, which is much greater than 61. So it must be the first equation:
61 + D2 = 43 + D2 + E1. Subtract D2 from both sides and you get 61 = 43 + E1. In other words, since D2 will be added to both diagonals, you can ignore it. E1 must be 18.

11.) What’s left at this point are 14, 15, 17, 20, 21, 22. B5 is 14 or 15. D1 and D3 are both 14 or 15 or 17. A3 is over 20 and B3 is over 19. D3 can be anything but 20, since it is diagonally adjacent to 19. The only remaining cell that can be 20 is B3.

12.) Column A must sum to 73 or 74. Column B must sum to 69 or 70. Column C sums to 67. Column E sums to 44. Column D is the question… Realize that Column D cannot contain both 14 and 15, because one of them must go in B5. So the smallest numbers available for Column D are 14, 17, and 21. That would bring the total to 70. The greatest numbers available for Column D are 15, 17, and 22. That would bring the total to 72, which is still not enough to equal Column A. The only way is if Column B and Column D are equal, and that only works at 70 Therefore B5 is 15 and the rest of Column D is 14, 17, 21, in some order. The 21 cannot go between 5 and 18, nor can it go between 19 and 9, so D2 is 21.

13.) Now A3 must be 22.

14.) A3 is referring to D2-E2. The difference there is 19. This means D1 cannot be 17 or the positive difference between C1 and D1 would be 12. Therefore D1 must be 14 and D3 must be 17.

(Back to puzzle)


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