**(1/25)**

The letters that rhyme with “day” are A, J, and K. Since columns are in alphabetical order, you know that A must go in the top row somewhere. Since cell one does not rhyme with “day,” the three remaining corners are A, J, and K. Of those then, A must go in Cell 21.

**(2/25)**

Since J and K are in Cells 5 and 25, and A and F are already used, you know that B-I (minus F) must be the letters in the first and last columns. This is because you have 9 cells, one column ending in J, the other in K, and you only have 9 letters before J/K in the alphabet with A and F out. So now you know, according to A, that B is in the first column. And obviously it must go in Cell 1.

**(3/25)**

If Cell 22 is C, Cell 23 could be G at earliest (according to B). That would mean Cell 24 would be I and Cell 25 would be K, since Cell 25 can go no farther than K. So if Cell 23 could be G at the earliest, and that would still make Cell 25 be its latest possibility, then that combination must be right. Cells 23-25 are G, I, K.

**(6/25)**

F now tells you that Cell 5 is J. H can’t go in the fifth column, so it has to go in Cell 4. C and D need to be in Cells 2 and 22, or vice versa. So E must go in Cell 3.

**(9/25)**

Given the remaining letters, If P is in Row 2 or 4, that row would have to go L M N O P, since nothing before L is left. However, according to K, O cannot be in Row 4. This means that P must go in Cell 15.

**(10/25)**

The only letters remaining between F and P (to finish the middle column) are L, M, N, and O. If O were to be used, it would have to go in Cell 14, since O is immediately before P. Again, K forbids this. So the middle column must go F L M N P.

**(13/25)**

You know that C and D are in Cells 2 and 22 or vice versa. This means that O is the earliest remaining unused letter. That means it would have to come at the top of a column, but again, K forbids this. So now you know that O is not in the puzzle.

The Z, obviously, must be last in whatever column it is in. So it is either Cell 10 or 20. Z also must be adjacent to at least one vowel, according to M. If Z were in Cell 10, Cell 9 would have to be a vowel, but again, K forbids this. Therefore Cell 20 is Z.

**(14/25)**

G tells you that D does not share a diagonal with Z. This means D cannot be Cell 2. Therefore Cell 2 is C and Cell 22 is D.

At the moment there are no letters that are adjacent to more than one vowel. But you haven’t placed the U yet. The U, because of K, can either go in Cell 7, 10, or 17. But if the U goes in Cell 17, then there are four letters that are adjacent to more than one vowel: Cells 16, 22, 18, and 23. Z says there are only two letters adjacent to more than one vowel. So U is not Cell 17. If U is in Cell 10, the second column would have to go Q R S T U, since Q is the earliest remaining letter. That would dictate the fourth column being V W X Y Z, which would work, except V rhymes with D; according to G, Cell 16 cannot be V. So U is not Cell 10 either. Therefore U is Cell 7.

**(17/25)**

With three cells under U and only four letters after U remaining (V, W, X, Y), choices are limited. Cell 8 can only be V or W. Any later and you’d run out of alphabet by Cell 10. Z says that Cells 2 and 8 are one-syllable letters. The only thing that does is rule out W, but that’s all you need. Cell 8 is V.

**(18/25)**

Cell 6 must be Q, R, S, or T. If it is Q or R or S, then Column 4 will have to start with the other two of those, and then T. in other words, if Cell 6 isn’t T, Cell 18 is T. G says that Cell 18 cannot be T. So Cell 6 is T.

**(19/25)**

Q, R, and S can’t go after anything on the board. Therefore Column 4 starts Q R S.

**(22/25)**

S tells you that Cell 9 is W. With Z already used, Column 2 must finish with X Y. This leaves W for Cell 19.

TRY HENRY820’S OTHER LOGIC PUZZLES!