I would describe myself as a plodder in regards to calcudoku (somehow plodded my way to #2 all time in Australia last week & also managed to finish my first 15x15 & managed to repeat again this week to prove it wasn't a fluke). I Will never trouble the top times in timed puzzles (probs happy with a top 50 for 4,5,6 & top 15-20 for 8 but have cracked the top 10 a few times for 8) but can usually work my way (all be it slowly) through the daily puzzles (helps with the extra time I have now being unemployed at the moment due to a redundancy a few Months back) with a bit of help from some previous posts on how to solve & djape.net combination helper but this weeks Tuesday 9x9 had me stumped.

My question is how to proceed with these type in future as I can normally work out combinations needed for the multiplication boxes but this week stopped me in my tracks as it had an extra multiplication box resulting in too many combinations (unless I stuffed up my maths in calculating column & row values). My normal plan of attack fills out the middle column & row & then calculate the values that the multiplication boxes add up to in the first & last 3 columns & the top & bottom 4 rows & working through possible combinations gets me there.

Thanks in advance for any information some of the superstars on this site can provide, A walkthrough would be amazing & much appreciated.

You follow the correct procedure.

I usually define the content of the central row and the three central columns.
It's easy to determine the operands in row 5 and columns e and f (of course 6+ = [2,4] and 14+ = [5,9]). And I am sure you quickly found that cage 20+ has a product of 192 [by dividing, in the three rightmost columns, (9!)^3 into the rest of products and into 45 (the product in cage 14+) and into the single digits 7, 2, 6, 6, 5]. The results is 192 and if you break this into its prime numbers … you find that, out of nine possible combinations, the only combination that adds to 20 is [1,1,4,6,8]. This information is useful for later.

The interesting thing is that knowing this product, 192, now you can find the product of d6*d7*d8*d9, by dividing, in the four bottom rows, (9!)^4 into the single numbers 1, 7, 6, 3, 5, and into the other 4 products including 192, and into 24, 20 and 72 which are the products of cages 9+, 9+ and 72x . The result of this operation is 756.

Now you have two possibilities (with all numbers different): [3,4,7,9] (with a sum of 23) and [2,6,7,9] (with a sum of 24). The sum of cages 11+ and 1- must be even (parity rule, for instance) so the solution is [2,6,7,9], so cage 1- add to 13 and must be [6,7] while 11+ = [2,9].

Obviously in column d the 8 is not inside cage 1- so cage 11+ = [3,8] and cage 1- = [4,5].

Next we can find the operands of cage 22+ following the same procedure as for cage 20+, using the three leftmost columns, arriving to a product of 480, what means that, out of six possible combinations, the only one that sums 22 is [1,2,5,6,8].

Of course, the analysis must continue until we find the final content of all cages and finally reorder the numbers in the proper places. Sometimes a digit can be placed directly, for instance, cage 6048x contains a 7, to be placed in a9-b8-b9 and since d8-d9 contains a 7, the only 7 in column a must be in a7, … .

Clm is very good at explaining procedures, soling strategies.
Clm describes the procedure that I use.

At one point, I went back thought old posts on solving strategies. I admit I didn’t always understand what was written, but what I did understand has been useful.

I thought that Tues (16 / 06 / 2020) was one of the harder 9 x 9s with that pattern. I didn’t get it solved until Wed.

Yes, eclipsegirl, and probably you have also continued like this:

Since 1344x contains (once) a 7, the only place for it is h2. And now, since 2268x contains also (once) a 7, the only place for it is i3.
Consequently, f3 = 1, f4 = 7.

Since cage 22+ = [1,2,5,6,8] contains a 1 and, looking to the ones in a6 and f3, you conclude that the 1 must be either in b4 or c4. At the same time 2268x is not a multiple of 8 so the 8 of row 4 must be in b4 or c4 too, then b4-c4 = [1,8]. And since 2268x is not a multiple of 5, the 5 of row 4 must be in d4, that is, d4 = 5 >>> d3 = 4. In row 4 we are left with g4-h4 = [3,4]. And, consequently: 2268 / 3 / 4 / 7 = 27 so h3-i2 contain the pair [3,9]. … .

When I started solving the puzzle, I got the 7s in the columns H and I very early, in what I would describe the basic set up. That forced the 7X cage in column F and forced the only possible solutions for the 2268X cage had to be.

But then I got stuck. I am jealous that you could solve it analytically. It may have been the first time with that pattern that I was forced to a methodical trial and error.

Thanks CLM, I'll work my way through it when I get a spare moment.

Totally agree eclipsegirl, Most of what I've worked out is from CLM's posts on solving strategies, But this one I just couldn't find the "in" that usually starts to reveal the rest of the puzzle. I gave up on it as had spent enough time searching, Could have just gone trial & error but I had too many options so didn't know where to start in the end.

DOH !!!! Had an absolute shocker on it looking at your workthrough. I just had it stuck in my head to do the uaual way when only the 2 x multiplication grids but had all the numbers there to work out what the addition grids summed up to. I feel like a right pillock now hahahahahaha Note to self take blinkers off & take a step back when stuck in future & look at what puzzle is showing & not what you think it should be showing.

For reference, here's the puzzle:

I used a new tactic for the decision. Maybe it’s an accidental success, but resolved pretty quickly, in about an hour.
I usually filled the center columns and row 5, and listed 7 and 5 in the possible places.

2268x cannot be two nines, but must be two three, i.e. 7x9x3x3x4.

1344 = 7x3x8x4x2, 720 = 5x9x8x2x1 so that the sum of the unused numbers in the GHI column is 20, i. 8 + 6 + 4 + 1 + 1.
I listed them in GHI columns, noting who can and cannot change their places.

1260 = 7x5x3x3x4 and 6048 = 7x9x6x8x2. At the time of writing, the locations and possible freedoms of the previously recorded figures had to be clarified. This has been done without leaving unresolved conflicts.

1728 = 9x6x8x4x1 so that the sum of the unused numbers in lines 1..4 is 22, i. 8 + 2 + 5 + 6 + 1.
For these numbers, the remaining places were in conflict with the previously recorded numbers. Only 3 change chains had to be made.

Thanks, Patrick, this is very useful for future reference.



The "methodical" TAE is practically an analytical process . TAE is part of the "scientific method" and it's difficult to stablish a clear dividing line between pure analysis and TAE (this has been discussed much in the past, here in the Forum). I am a defender of the "guess" or what I prefer to call the "theorem": "If a Calcudoku has a unique solution then it can be solved using only analytical means". That is, using known tools: addition rule, multiplication rule (specially in puzzles like this, where you have many products), parity rule, segmentation, … .

It's true that sometimes you can be lucky and solve a puzzle simply with TAE (and quickly) but this would generally occur with small puzzles, but thinking in the "big ones", like 25x25, 20x20, 19x19, 17x17, 15x15, … , it's better to be very systematic, otherwise the number of "branches" would be "infinite" . So it's better to get practice with analysis and forget TAE, generally speaking.

It's easy to determine that 1344x = [2,3,4,7,8] (analytical conclusion of mikas, for instance, considering the three rightmost columns), with a sum of 24. Now, considering only the four upmost rows, you find that the sum of cages 1728x and "1-" is 37 (addition rule to these four rows) and, applying the parity rule, the consequence is that sum in cage 1728x is even and, once you have reduced the possible combinations, you are left with [2,3,4,8,9] (sum 26) and [1,4,6,8,9] (sum 28). In the first case, "1-" has a sum of 11 so = [5,6] which is impossible since "1-" (d8-d9) would also have a sum of 11 >>> = [5,6] again. Then, "1728x" = [1,4,6,8,9] and "1-" (d3-d4) = [4,5].