Note: Please keep scrolling down to see every step.
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.