by mattlondon » Thu Sep 29, 2005 2:35 pm
I am curious to know how you can be so sure about r6c7, without having got r8c7 first
My method was to add 10, 5 and 8 in bottom centre and deduct from total 45 leaving 22 which can only be formed in 2 ways: 9,7,6 and 9,8,5.
9,7,6 can be excluded because both the 9 and 7 cannot go in r6c7. Therefore, they have to go in r8c4 and r9c4 which is not possible as either a 9 or 7 must be r3c4.
9,8,5 is the other available group for the three spare boxes in the bottom centre group. Only the 5 can go in r7c6. This then gives you the value for r6c7 without knowing r8c7.