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Saturday 20-Sep-2014

... by: newby

3J7, 8H5, 3H4, 9H2, 5H8, 5D7, 8E1
2A9>contradiction therefore 2C8
9F3, 2(E9, F9)>contradiction therefore 9D3
8G3, 2(E9, F9)>contradiction therefore 5G3........
3C3, 2(E9, F9)>contradiction therefore 3F3 solves

Saturday 20-Sep-2014

... by: JPF NAPERVILLE

#151 was readily solved by:
1. using just solver steps 1-6 to gain 7 fill-ins,
2, focusing on the sextet in row J and solving it via a 6 in J8 which
yielded 4 fill-ins and a pair.
3. then using only solver steps 1-6 then solved the puzzle.

Thus, the use of just basic 1-6 step logic plus test-solving a
combination allowed further use of only basic steps 1-6 to solve the puzzle. No need for solver steps 7-35!!

Saturday 20-Sep-2014

... by: ray

#1. Use basic strategies and bi-location/value tool to get working board.
#2. Choose cell J8{678} - bi-location cell for 7 and 8,tri-location for 6.
Forcing J8 = 6 gives the solution. Using this solution , other bi-location
1-cell solutions are found by forcing the following cell values: A7 = 1,
B1 = 1, E9 = 6 ; some 2-cell solutions found are: AB9 = (8,4),
DE8 = (1,9), F78 = (4,8), EF2 = (2,7).

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