| From the introduction, each row and each column of the 5x5 Word Square
contains the characters BEAR and a blank cell in some order, so that no
character or space can be repeated in a row or a column. By clue 2, A is
the 2nd letter down column 2; while by clue 8, A is the 2nd letter across
row 2. One of two arrangements must then work: either the A in column 2 is
in row 3 and the A in row 2 is in column 3 or the A in column two is the A
in row 2. Trying the first possibility, B would be in column 3 of row 3
(clue 6), with the space in row 3 then in column 1. The space in row 2
would then be in column 2 (8). By clue 4, the R in column 1 would be in
row 5, and the R in column 4 in row 4, with the space in column 4 in row 5.
The B in column 4 would be in row 1 (10). However, there is now no way for
an A to fit into column 4--so the first arrangement for clues 2 and 8, that
the A in column 2 is in row 3 and the A in row 2 is in column 3, fails.
The A in clues 2 and 8 is the same one, in row 2, column 2. There are two
possible arrangements for the R's in clue 4: either the R in column 1 is in
row 5 and the R in column 4 in row 4, or the R in column 1 is in row 4 and
the R in column 4 in row 5. Trying the first arrangement, by clue 10, the
B in column 4 would be in row 1. By clue 9, then, the E in row 1 would be
in column 3 and the space in column 5. The A in row 1 would be in column 1,
with the R in column 2. The E in row 4 would be in column 5 (3). Then the
A in row 4 would be in column 3. However, the B would be the 1st letter in
row 4, a conflict with clue 1. So, in clue 4, the R in column 1 row 5 and
the R in column 4 in row 4 fails; the R in column 1 is in row 4 and the R
in column 4 in row 5. The space in column 1 must be in row 5. Since the A
in column 2 is the 2nd letter in that column (2), the space in row 1 isn't
in column 2. If the space in row 1 were in column 3, the E in row 1 would
be the 1st letter down column 4 (9)--no (10). So, the E in row 1 must be
in column 3 (9). By clue 10, B is 1st down column 4; so either B is in
row 1 or B is in row 2 with a space in row 1. Trying the B in column 4 in
row 1, by clue 9, the space in row 1 would be in column 5. The letter in
column 2 of row 1 would have to be R, with the A in column 1. By clue 3,
the E in row 4 would be in column 5. Then the E in row 5 would be in column
2. By clue 6, the B in column 1 isn't in row 3; it would be in row 2, with
the E in column 1 in row 3--a conflict with clue 7. So, in clue 10, the B
in column 4 isn't in row 1; it is in row 2, with a space in row 1 of column
4. The E in column 1 can't be in row 3 (7) and must be in row 2. The B
in row 3 also can't be in column 1 (6), so the A is; the B in column 1 is in
row 1. Then the A in row 1 is in column 5, with the R in column 2. In clue
3, if the E in row 4 were in column 4 and the space in column 5, the A in
column 4 would be in row 3, where we already have an A. Therefore, in clue
3, the E in row 4 is in column 5. Then in column 4, the E must be in row 3
with the A in row 4. By elimination, the E in column 2 must be in row 5;
and the A in column 3 must be in row 5. B in column 5 then completes row 5.
By clue 5, in row 4, the B is in the 2nd column and the space in the 3rd.
The space in column 2 is in row 3. In column 3, the B is in row 3 and the
R in row 2. Finally, in column 5, the space is in row 2 and the R in row 3.
In sum, Word Square 5 is as follows:
| | 1 | 2 | 3 | 4 | 5 |
| 1 | B | R | E | | A |
| 2 | E | A | R | B | |
| 3 | A | | B | E | R |
| 4 | R | B | | A | E |
| 5 | | E | A | R | B |
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