Maths Logic Puzzle

Welcome to the Maths Logic Puzzle walkthrough. This website takes you through the puzzle step by step. Got your math skills packed? Then let’s go. First of all, behold the grid, with A1 given.

20 needs to be on a corner, as it's divisible by 10. It can’t be in A5 or E5, as it isn’t a prime, so it has to go in E1.

9 needs to go in a cell from which eight knight moves are possible (as to reach eight cells containing 1-8). The only cell on the grid that provides this possibility is C3.

Important to note: because there are exactly eight cells you can reach this way, every one of those eight cells (clockwise: B1, D1, E2, E4, D5, B5, A4, and A2) must contain a single-digit number.

A2 and A4 must contain single-digit numbers (clue E1). The highest ones still available are 7 and 8. That means that the maximum sum of A1+A2+A4 is 10+7+8=25. Therefore, A3+A5 must be at least 73-25=48. That leaves two possibilities: 23 and 25, resulting in 48, or 24 and 25, resulting in 49. Because A5 has to be a prime (clue A1), neither 24 nor 25 can be there. So A5 has to be 23, leaving A3 with 25. – This means that A2 and A4 have to be 7 and 8, in any order.

Consider clue A3. Basic math is to add the same number to both sides of an equation. Here, you can make the equation easier by adding E3 to both sides. Then you'll have E1*E2-E3+E3 and E4+E5+E3. The -E3 and +E3 on the left side 'diminish' each other, so you'll have E1*E2 = E3+E4+E5.

The highest possible numbers for E3, E4 and E5 are 24 (highest number), 6 (highest single-digit number, 7 and 8 being in column A) and 19 (highest prime), respectively. This means a maximum value for E3+E4+E5 of 49. The lowest possible numbers for E3, E4 and E5 are 12 (11 must be in column D per clue A5), 1, and 13, respectively. So the minimum value for E3+E4+E5 is 26. The only number between 26 and 49 that is divisible by 20 (because E3+E4+E5 must equal 20*E2) is 40. This means that E2 has to be 2.

Clue A5 tells us that 22 has to be in column D. It can’t be in D1, D2 or D3, because if it were, it would be adjacent to a number that ends in the same digit (2 being in E2). Neither can it be in D5 (both the prime rule and the single-digit rule). Thus D4 has to be 22.

Consider E2’s clue. The square numbers in the grid are 1, 4, 9, 16, and 25. A3 and C3 already contain one. The only square number that can be in column C (along with 9) is 16, because 1 and 4 have to be somewhere in column B, D or E (remember the cells that contain single-digit numbers?).

The equation that D4’s clue gives us must be 10+7=17 or 10+8=18 (because A2 must be either 7 or 8). Column C holds one even number at most (clue C3). We’ve just seen that 16 has to be in column C, so 16 is the one even number there. This means that D4’s equation has to result in C2 being 17, with A2 being 7. Because A2 is 7, A4 has to be 8 (clue C3).

The lowest number that could go in C1 is 13 (11 has to be in D, and apart from 16, only odd numbers in C – so not 12). The highest number that could go there is 21. This means that the value of C1+C2 must lie somewhere between 30 and 38.

The lowest and highest possible numbers for C4 are 13 and 21 as well. For C5, the lowest and highest available primes are 13 and 19. This means the value of C3+C4+C5 is at least 9+15+13=37, and at most 9+21+19=49.

The sides of the equation must be equal, and the only overlapping values between C1+C2 versus C3+C4+C5 are 37 and 38. This means C1 has to be either 20 or 21. Because 16 is the only even number in column C, C1 has to be 21.

We’ve seen (clue E2) that 16 has to be in column C. It can’t be in C5 (the prime rule), so it has to be in C4. We now know that the sides of the D4 equation result in 38, so C5 has to be 38-9-16=13.

We’ve seen that C1 contains 21. This helps us in determining where 11 goes. As a number with repeated digits, it has to be in column D but cannot be adjacent to 21. This leaves D3 as only option.

Let’s think prime numbers. The only ones left to place in the grid are 3, 5, and 19. Because of the single-digit rule, 19 cannot be in B5, nor can it be in D5. Therefore it has to be in E5.

Earlier, we’ve rewritten the A3 equation to E1*E2=E3+E4+E5. This means that E3+E4+E5 is 40. With 19 being in E5, E3+E4 has to be 21. E4 is either 1, 4 or 6 (the only single-digit numbers left, 3 and 5 being in row 5). This means that E3 has to be either 20, 17 or 15. Both 20 and 17 are already in the grid, so E3 has to be 15, which leaves E4 being 6.

Wait, what, now there’s perfect numbers too? We are looking for a perfect number with divisors 2, 7, the-number-in-B3, and 1 and 4 (1 and 4 are the only possible numbers for B1 and D1). The perfect number that fits the bill is 28. This number’s only divisor not aforementioned is 14, so B3 is 14.

Apart from the top and bottom row, three cells (B2, D2, B4) are still to be filled. The only numbers left for these three cells are 12, 18 and 24. Here E5’s clue comes in. 24 can’t be in B4, because it would be adjacent to both 23 and 25, which makes over 1 adjacency of successive numbers.

With 24 in B2 (thus adjacent to A3’s 25), 12 would have to be adjacent to either 11 (when placed in D2) or 13 (when placed in B4). This makes two adjacencies, so 24 can’t be in B2 either. That leaves 24 being in D2. Putting 18 in B2 and 12 in B4 would make both numbers have successive adjacencies (18 to 17, 12 to 13), so that can’t be the case. This means that B2 has to be 12 and B4 has to be 18.

Still to be determined are 1 and 4 (in the top row, as per clue E4) and 3 and 5 (in the bottom row, as per clue A1). Although we used the adjacency rule quite a lot in the step before this one, there aren’t two adjacent successive numbers up to now. If 1 and 5 are in column B, there are no adjacencies whatsoever in the grid. If 1 and 5 are in column D, there are two. Both situations are not allowed, so 1 and 5 go in different columns. Thus one column holds 1 and 3, and the other holds 4 and 5. B2’s clue tells you that B is the column with 4 and 5, and D is the column with 1 and 3. Done!