If anybody would like a more detailed description of the steps taken to solve the puzzle I am happy to provide more information.

d7=5, d8=3

cage g8(7/) = 1,7

>> f8=8

cage c6(72x) = 8,9

cage e4(324x) : row 5 in cage must have 9

>> i4=9

i9 ≠ 7 (row 6 has no 7)

i9 ≠ 1 (row 9 has no 4)

i9 ≠ 2 (cage a7 has no possible solution)

g9 ≠ 3 (row 8 has no 4)

a8 ≠ 2 (row 8 has no 9)

a9 ≠ 9 (col b has no 8)

b9 ≠ 9 (col b has no 8)

c9 ≠ 9 (row 2 has no 8)

>> h9=9

a8 ≠ 6 (row 9 has no 7)

>> a8=9

>> cage b8 = 5,6

cage e8(120x) must have 5 in row 9

>> i9 ≠ 5

i9 ≠ 3 (cage h6(13+) has no solution )

i9 ≠ 6 (row 2 has no 8)

>> i9=8

g9 ≠ 7 (cage c6 (72x) has no solution)

From here the puzzle can be completed easily with basic level steps.

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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 –

c)

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 –

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 →

It gives

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

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

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

6) a)

b)

c)

d) I789 = min. 15 → A)

e) H5 = [6,7] →

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

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

9) Due to the 324× cage,

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

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However, I always solve the hardest puzzles by guessing, so I don't know how hard it was to solve it analitically, although I only used one guess.

This guess was a really bad guess, too, I was still able to fill in the other option and change a few numbers as the guess turned out to be wrong, having already filled in almost the complete puzzle.

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it is know as the hardest 9 x 9 pattern to solve, at least I think that is what Patrick has stated.

What I was surprised at , is that we had this pattern about a month ago (sometime in late Feb or March) and I thought we would be done with it for the year.

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paulv66 wrote:

Yesterday's difficult 9x9 was not the usual type of pattern. I found it extremely difficult to,solve and have only just managed to complete it a few minutes ago. I note that it's only been solved 70 times, so it looks like I'm not the only one who has struggled with it..

I had a look at the number of times the Tuesday difficult 9x9 was solved over the past month. The numbers are as follows:

26/03/19 235

02/04/19 217

09/04/19 169

16/04/19 259

23/04/19 73 (so far)

I know there's nearly six hours left within which yesterday's puzzle will earn points, but it seems likely that the final number will be a lot lower than average. I'd be curious to know what's the lowest number of solvers for the Tuesday 9x9 in recent years.

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Thanks for explaining. I was indeed making a logic error. Mentally I had put parentheses around e3 and e4 - doing their subtraction first and then subtracting that result from 0. I used to know that subtraction is not commutative. Funny how my old brain can get stuck on such a simple thing.

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nashvillebudd wrote:

This puzzle seems unsolvable to me.

I start by filling in the givens...d1, c5, e5.

Now looking at column d, d2 cannot be zero since that would force c2 to also be zero, d4 cannot be zero since either b4/c4 must be a zero, and d5 cannot be zero since either a5/b5 must be zero.

That leaves only d3 to be a zero.

But in order for that clusters subtraction to equal zero that would force e3 and e4 to be the same number.

Stalemate.

What am I missing? Can anyone help?

You are reasoning wrongly because when you have negative numbers in a puzzle like this not necessarily e3 and e4 must be the same number, What about a -1 and a 1, for instance?: 0 - (-1) - 1 = 0

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I start by filling in the givens...d1, c5, e5.

Now looking at column d, d2 cannot be zero since that would force c2 to also be zero, d4 cannot be zero since either b4/c4 must be a zero, and d5 cannot be zero since either a5/b5 must be zero.

That leaves only d3 to be a zero.

But in order for that clusters subtraction to equal zero that would force e3 and e4 to be the same number.

Stalemate.

What am I missing? Can anyone help?

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pnm wrote:

I'm thinking probably a volume V before the end of the year..

Sounds like a plan!

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marblevolcano wrote:

FINALLY! I solved A2V1 by figuring out the proper combination in the top right 1- cage.

What a relief that is... but now I have no unsolved book puzzles... I guess I need a new book. What a

I'm thinking probably a volume V before the end of the year..

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