by **carlosr** » Sun Mar 12, 2006 10:21 pm

How did I get the 7s?

The sum at r6c4 is 23 so it must be 689. There's a 6 at r6c3 (which I forgot to show) so r6c4 must be 8 or 9.

Adding up the bottom-middle and bottom-right nonets you get r6c4 + r6c6 - r7c5 = 7.

Now consider where the 8 and 9 go in the bottom-middle nonet. One of them goes in r7c4/r8c4 and the other goes in r7c5/r7c6. The 9 can't go in r7c6 (because that only leaves 2 in the 11 sum) so suppose r7c6 is 8. This leaves either 1 or 2 in r6c6. It can't be 1 (you'd have 2 in both r7c5 and r7c7) so suppose r6c6 is 2. This implies r7c5 is 3, implying r8c5/c7c6 is 25, implying r9c5/c9c6 is 14, implying r9c7/r9c8 is 25, implying r7c8/r7c9 is 63. Which clashes with the three in r7c5; so r6c6 isn't 2 and r7c6 isn't 8. Hence r7c5 is 8 or 9, and equal to r6c4, so r6c6=7.

r2c5 + r2c6 + r3c6 = 24 (by summing the top-right and middle-middle nonets), so they must be 789. r2c5/r2c6 can't be 89 (that would leave zero in r2c7) so there must be a 7 in r2c5/r2c6.