#### Otis

##### Senior Member

- Joined
- Apr 22, 2015

- Messages
- 2,098

- Thread starter Otis
- Start date

- Joined
- Apr 22, 2015

- Messages
- 2,098

just the one username all along. lookagain, I see you tried to give him/her "an out" by deleting that post # 16 in that

other thread. And this username, "greg1313," would not exist in response to the creation of "Otis."

But live and learn must mmm4444bot. (With apologies to Yoda.)

For convenience, I let A = 1, ... , J = 10.

We have the equivalent of:

\(\displaystyle * \)

\(\displaystyle 6 \ \ \ * \)

\(\displaystyle * \ \ * \ \ * \)

\(\displaystyle \ \ \ \ \ 2 \ \ \ 8 \)

\(\displaystyle * \ \ \ *\)

The only place for the 7 is the far left bottom:

\(\displaystyle * \)

\(\displaystyle 6 \ \ \ * \)

\(\displaystyle * \ \ * \ \ * \)

\(\displaystyle \ \ \ \ \ 2 \ \ \ 8 \)

\(\displaystyle 7 \ \ \ *\)

The only potential places for the 1 and the 3 are above the 6 and to the right of the 6.

This forces the 9 to go directly below the 6:

\(\displaystyle * \)

\(\displaystyle 6 \ \ \ * \)

\(\displaystyle 9 \ \ * \ \ * \)

\(\displaystyle \ \ \ \ \ 2 \ \ \ 8 \)

\(\displaystyle 7 \ \ \ *\)

Numbers 1 and 3 are committed between two squares. We still need to place 4, 5, and 10. Look at the

square immediately to the right of the square that has 9 in it. It cannot be 10 or 5, because that would

violate the squares having 9 and 6 in them. So, it must be 4:

\(\displaystyle * \)

\(\displaystyle 6 \ \ \ * \)

\(\displaystyle 9 \ \ \ 4 \ \ \ * \)

\(\displaystyle \ \ \ \ \ 2 \ \ \ 8 \)

\(\displaystyle 7 \ \ \ *\)

The rest of the numbers fall in place. The number above 4 must be 1 and the number above 6 must

be 3. The 5 must be directly below 2, and finally, 10 is to the right of 4:

\(\displaystyle 3 \)

\(\displaystyle 6 \ \ \ 1 \)

\(\displaystyle 9 \ \ \ 4 \ \ \ 10 \)

\(\displaystyle \ \ \ \ \ 2 \ \ \ \ 8 \)

\(\displaystyle 7 \ \ \ 5\)

Then, convert the numbers back to the letters of the alphabet.