Hallo, I would like to introduce my ”step by step“ solution of this hard nut, and compare to yours if…

(The order of steps is not necessarily chronologic.)

First (basic) eliminations/thoughts:

0 a) CD4 ≠ [8,9] (due to CD6 and “rule” of unique solution; it would already be eliminated after step 1) (later) though);

b) interaction between FG4 and FG6 –

**F4 ≠ 4** and

**G4 ≠ 1**;

c)

**H4 ≠ 6**;

d) the right 1512× cage has combinations [3,7,8,9] and [4,6,7,9];

e) rule of parity says that three rows or columns in a 9×9 calcudoku must get an odd result (3 × 45). So if the right 1512× cage is [3,7,8,9], G4 = [3,7]; if is [4,6,7,9], G4 = [4,6,8];

f) the central 324× cage can by composed of digits 1, 2, 3, 4, 6 and 9.

1) (Im)possible combinations for the central 324× cage: [1,1,4,9,9], [1,1,6,6,9], [1,2,2,9,9], [1,2,3,6,9], [1,3,3,4,9], [1,3,3,6,6], [2,2,3,3,9], [2,3,3,3,6] and [3,3,3,3,4].

Of these, only two are valid: [1,2,3,6,9] and [1,3,3,4,9] (number 4 can only be in E5!). It means that the column E contains 1 in E4 or E6, the row 5 contains 9 in D5 or F5. Thus 9 in the right 1512× cage must be placed in the row 4 –

**I4 = 9**.

2) Needed a long chain on 7 in H6, but it was very useful.

If H6 = 7 → H8 = 1, B7 = 7 (due to no other place for the third 7 in lower three rows), B6 = 3, G6 = 1, E6 = 2, A6 = 5, I6 = 6, H5 = 4, DEF5 = [3,6,9], AB5 = [2,8], A4 = 6; H5678 = [1,4,6,7], thus H34 = [2,3].

Now the sum of known numbers in the columns G, H and I is 117, so G2 + H2 + G4 = 18. G4 ≠ 6; if G4 = 8, GH2 is blocked; if G4 = 4, GH2 = [6,8].

This with three 7's in the columns A, B and C means that BC2 = [2,9]. The “rule” of unique solution is again on the scene: BC8 ≠ [2,9] = [5,6].

Now factorial counting the three left columns: 9!^3 ÷ (cages by rows: 2160 × 18 × 7 × 480 × 21 × 1512 × 3 × 30) = 128.

So the cells B3, B4, C4 and C6 must get product of 128 = 2^7; C6 = 8, C4 = 2, B34 = product of 8, which is impossible →

**H6 ≠ 7**It gives

**sure 7 for the 21+ cage**. Also

**the 10+ and the upper 11+ cage ≠ 7**.

3) If B6 = 2 → H6 = 6, FG6 = [1,4], A6 = 5, E6 = 3, I6 = 7; if HI5 = [4.6], the 324× cage is blocked; if [3,8], DEF5 = [2,6,9], AB5 = [4,7], A4 = 5. A4 = A6 – contradiction, therefore

**B67 ≠ [2,8] = [4,6]**.

**B34 = 1–3** and a little bit eliminations…

4) If H6 = 6 → G8 = 7, G6 = 3; so G4 is even and the right 1512× cage must be [4,6,7,9]; H5 = 4, H8 = 1, H34 = again [2,3]. And again the sum of the columns G, H and I: this time 119, so G2 + H2 + G4 = 16.

All possibilities of numbers 5, 8, and 9 in H2 lead to contradiction, thus

**H6 ≠ 6 = 5**.

5) Another long chain:

If the right 1512× cage = [3,7,8,9] → in cooperation with B5 = [3,7,8], DEF5 must be [2,6,9], A5 = 4; E46 = [1,3], therefore HI5 = [3,8], B5 = 7, I6 = 7, A6 = [2,4,6]; A4 + A6 = 10 and A5 = 4, thus A6 = 2, A4 = 8; G6 = 1, G8 = 7, G4 = 3 (rule of parity), F4 = 2, E4 = 1. But now exists no number for B4, so

**the right 1512× cage = [4,6,7,9]**. Which created

**naked triple of 4, 6 and 7 in I567**…

6) a)

**G4 = [4,6,8]**;

b)

**the 21+ cage contains 8**; therefore

**remaining digits are 1, 2, 3, 4 and 5 and A4 = 2–5**;

c)

**E8 ≠ [2,5] = [4,6]**;

d) I789 = min. 15 → A)

**I9 ≠ [1,2,5] = [3,8]**; B) if I7 = 4 → A7 = 7, G9 or H9 = 7 – no combination for the sum of 21 here, thus

**I7 = [6,7]**;

e) H5 = [6,7] →

**H34 = 1–4**7) If E5 = 4 → DF5 = [3,9], HI5 = [6,7], AB5 = [2,8], A4 + A6 = 11, which makes combination [4,7] here; E8 = 6, A8 = 4. A4 = A8 – contradiction, therefore

**the central 324× cage = [1,2,3,6,9]**.

8) If FG6 = [1,4] → B6 = 6, HI5 = [4,6], but it forces numbers 1 and 6 to one cell (E4).

**FG6 = [2,3]**.

9) Due to the 324× cage,

**the 21+ cage ≠ 3** (otherwise three 3's in two rows). It is applicable the rule of parity for the three central rows: sum of known cells/cages is odd, so BH4 must be even… [1,3] or [2,4]. If BH4 = [2,4] → A4 = 5, the two subtraction cages are blocked. Thus

**BH4 = [1,3]**,

**B3 = 2**,

**H3 = 4**.

10) At the end the rule of parity helps again:

**A4 = even ≠ 5** →

**the 21+ cage = [2,4,7,8]**,

**A78 ≠ 4**.

**Etc…**