Full Analytical Solution (FAS) for the 9x9 on 8/18/2015:

**The empty puzzle:****1.** Pre-filled cells.

**2.** The 9: at 7d must be [9,1].

**3.** The 45x at 6b must be [9,5].

**4.** The 4: at 8b must be [2,8].

**Through step 4:****5.** Subtracting the known sums from rows 1-3 leaves 17 for the 2- at 2d, the 12x at 2f, and the two leftover cells at 3b and 3h. The leftovers must sum as least 3 and the 12x sums either 7 or 8, leaving a max value of 7 for the 2-. With the 1 blocked in column d, that means the 2- is [2,4].

**6.** The 12x cannot be [2,6] because this would block the 2s in rows 2-3, while forcing the leftovers at 3b and 3h to be [1,2], which is impossible. So the 12x at 2f must be [3,4] and the leftovers must be [1,3].

**7.** Since both leftovers are part of 3- cages, the corresponding 4b and 4h cells must be [4,6].

**8.** The 14+ at 4c must therefore be [9,5] due to the blocked 6s in row 4.

**9.** The 15+ at 2b must be [7,8] due to blocked 9s in columns b-c.

**10.** The 15+ at 2g must therefore be [6,9]

**Through step 10:****11.** Subtracting known sums from columns a-c leaves 15 for the 3- at 3b and the leftover cell at 6c. The only values that work are a [3,6] for the 3- and a 6 for 6c.

**12.** The 3- at 3h is therefore [1,4].

**13.** The 3- at 6c is also therefore [6,3] due to the 9 at 7d.

**14.** The 10+ at 6h must be [2,8] due to blocked 1s and 4s in column h as well as blocked 3s in rows 6-7.

**15.** Subtracting known sums from rows 4-6 leaves 18 for the 6- in 4f and the leftover cell at 6h. The only values that work are a [2,8] for the 6- and an 8 for 6h.

**Through step 15:****16.** Subtracting known sums from columns d-f leaves 22 for the 1- at 7f and the leftover cells at 3f and 6f. The only values that work are an [8,7] for the 1-, a 2 for 4f, and a 5 for 6f.

**17.** The 7 in row 7 must be in 7a, which also makes 7i a 4.

**18.** Subtracting known sums from rows 7-9 leaves 36 for the 3240x at 7i and the 1- at 8g. The only values for the 3240x that include a 4 sum at 27 and 29, which mean that the 1- must be 7 or 9, which means that 8g must be a 4.

**Through step 18:****19.** 3a and 3i must both be [6,9].

**20.** With the 9s, 8s, 6s, 5s, and 3s blocked in columns b-c, the only choices for 9b-c are [7,4,2,1].

**21.** 9b-c must contain a 7 because no other combination allows for usable values in 8-9a. ([1,2] leads to sum 18 which is impossible, [1,4] leads to sum 16, which is blocked by 7a, and [2,4] leads to sum 15, which is blocked by 7a and 3a.)

**22.** 8a cannot be a 6 because it allows for no usable values for 9a. (With two 7s and a 6 in the 28+ cage, 9a and the non-7 in 9b-c would sum 8: [1,7] is blocked by both 7s, [2,6] is blocked by the 6, and [4,4] is impossible.)

**23.** The only place for a 6 in row 8 is 8i.

**24.** The only option for the 3240x cage is now [5,4,9,6,3] which makes 8h 5 to balance of the sum of rows 7-9.

**25.** This forces a resolution of the cells in the 3240x at 7i, the 15+ at 2g, as well as 3a and 3i.

**Through step 25:****26.** 8a cannot be a 3 because it allows for no usable values for 9a. (As in step 23, the remainders in the 28+ cage would sum 11: [1,10] is impossible, [2,9] is blocked by the 9 in row 9, and [4,7] is blocked by both 7s.)

**27.** 8a must therefore be 9 and 8e must be 3.

**28.** 9a-c must be [1,4,7] due to the blocked 3 in row 9.

**29.** The remainder of [2,6,8] are easily resolved in 9d-f.

**Through step 29:****30.** 1g must be a 2.

**31.** The only place for a 2 in column c is 8c, resolving the 4: at 8b, the 15+ at 2b, and 9c.

**32.** The only place for a 2 in column b is now 5b.

**33.** 1c must be a 4, which means 1b must be 1, further resolving 9a-b.

**34.** 4-6a must be [3,4,2] and 1-2a must be [8,5].

**Through step 34:****35.** The rest of the puzzle is fairly simple.

**Final solution:**