### Crazy Number Logic Puzzle – WALKTHROUGH

The beginning board:

A1 tells you that E1, E5, and A5 have to be 3, 5, and 7 respectively.

You know know that B1, B2, and A2 must add up to 10. B1, B2, and A2 have to be 2, 3, and 5, respectively. It is the only combination of numbers that work. (Try it!)

Let’s try to figure out the column that contains the numbers 1-5 (A2 clue). It cannot be A because there is already a 7. It cannot be B because the numbers have to get larger as you move down the column which means a 6 or higher must be there. It cannot be D because D4 has to be greater than 7 (B2 clue). It cannot be E because E3 has to be larger than B5 and you already know that B5 has to be a 6 or higher. So Column C contains 1-5. B2 clue says that C5 must be a 4 or a 6, so it must be a 4. Let’s go to C1. It has to be 1, 2, 3, or 5. It cannot be 1, 2, or 3 because those numbers are already in that row, so it must be a 5.

B3 must be a 5 because A5 clue says that C1 and B3 must be the same number

C5 clue says that C4 and A4 must add up to 4. They cannot be 2 and 2 because they are in the same row. So it must be 3 and 1. A4 cannot be a 1 because A1 is a 1, so A4 is a 3 and C4 is a 1. C2 and C3 have to be a 2 or a 3, but B2 is already a 3. So C3 is a 3 and C2 is a 2.

B1 clue says that C3 x A4 must equal E3. C3 and A4 are both 3′s so E3 is a 9. E5 clue says that E3 is one number larger than B5. So B5 is an 8. (excuse the absence of B3)

C4 tells us the column E adds up to 24. That means E2 and E4 have to add up to 7. The options are 6&1, 5&2, or 4&3. There is already a 5 and a 4 in the column, so it must be a 6 and a 1. C4 is a 1, so E4 is a 6 and E2 is a 1. (excuse once again)

This is the puzzle, with B3 included. (The previous 2 were a picture error)

D4 has to be an 8 because B2 clue says that it has to be greater than 7. The only 9 is already E3. Now B4 must be a 7 because B3 says that every 8 needs a 7 in the same row.

We’re almost done! We need one more 8 on the board, and it cannot be D1, D2, D3, or D5 because D4 is an 8. So A3 is an 8, and D3 then must be a 7.

The last of the 4 1′s (C1 clue must go in D5 because D1 and D2 already have 1′s in their row. To find D2, we reference to E3′s clue. The only number that would fit is a 4.

The final box, D1, has to be a 6 because the B column does not contain a 6 (A3 clue). This is the completed puzzle!