Frank Puzzles About Σ | Challenging Variant Sudoku

Added: 2026-05-18

[00:00:02]Hello everyone, I'm Frank and welcome to Frank Puzzles. And this is today's challenging level puzzle. And the puzzle is called Sigma by Kakuslav. And Sigma is usually a well, it's a symbol that usually represents a sum and I think that's what it's representing here.

[00:00:23]Anyways, let's get to the rules. We have normal sudoku rules apply which means in every cell in the grid place a single digit from 1 through nine such that there are no repeats in any row column or designated nine cell region.

[00:00:43]Each pink line is a renban that is it contains a set of consecutive digits in any order. So for example, if this were say a six, this would have to be five or seven because these have to form a consecutive set of digits in any order.

[00:01:03]But we also have no two pink lines have the same digit sum. I don't know what that means by digit sum. I think it just means basically the sum of its digits, right? It should be just the you should be able to just say like this. No, two pink lines have the same sum, right?

[00:01:24]Oh, I guess it does make sense because you're not usually adding digits on a renb band. So, saying it has the same digit sum, I guess, is a is a quick way of of saying just add up all the digits and then sum them together and then none none of the renban lines have the same sum in that instance. So, I guess that makes sense because usually you're not summing the sum the digits on a reband line. So because of that then it makes sense to basically say the the digit sum right.

[00:01:57]Um so no two no two pink lines if you sum up their digits on the line they will not no two lines will have the same sum.

[00:02:11]So keep that. So for example this would be the digit sum of 13 here for this line if this were six and seven.

[00:02:20]And those are all the rules. If you like to play along, there's going to be a link in the description. And now I'm going to get started.

[00:02:30]Let's go.

[00:02:33]Okay.

[00:02:35]So for each um so for every three cells and since these are all in the same row this must be the sum of you have to have 1 2 3. So one has to be on the renb band and it must be with two and three nine has to be on a run band. It must be with eight and seven. So the remaining renb band is four five and six. So that means this combination is B one of them's going to be the sum is going to add up to six.

[00:03:08]So one of them's adding up to six here.

[00:03:10]The other the other ones are adding up to 15 and 24 or these sets. How many lines do we have? We have 1 5 6 7 8 9 10 11 12 13 14 15 16 17 18.

[00:03:34]So we have 18 lines.

[00:03:40]Um for the twodigit ones, you can start with one and two. You have 1 2 2 3 3 4. They're all going to be odd numbers, right? So 1 2 adds up to three.

[00:04:03]2 three adds up to five.

[00:04:06]3 4 adds up to seven.

[00:04:11]And then we have all the way up to 17.

[00:04:13]But we definitely have 15 in one of these three renb bands because it has to be four, five, and six adding up together.

[00:04:25]Um, for rands on the same row, it has to be either 1 2 3 4 and 5 6 7 8 1 2 3 4 and 6 7 8 9 or 2 3 4 5 6 7 8 9.

[00:04:48]Oh, so actually that's all the combinations we have. 1 2 3 4 with 5 6 7 8 and 2 3 4 5 with 6 7 8 9. Right? So then these have to be 1 2 3 4. So one of them is 10 with 5 6 7 8 because the thing is you can't have one two through. So 2 through 5 guarantees one has to be from 6 through 9.

[00:05:21]um 6 through the 5 through 8 guarantees that one has to be 1 through 4. And since we can't have a repeat sum, you can't have 6 7 8 9 twice. You have to have one has to be 1 2 3 4, one has to be 2 3 4 5. One has to be 5 6 7 8 and the other one has to be 6 7 8 9 because they're in the same row. So 1 2 3 4 and 2 3 4 5 is four. So it would not be with 2 3 4 5. It would have to be with 5 6 7 8 specifically which is 26.

[00:06:01]And then you have 2 3 4 5 which is 14 and 30 for 67 8 and 9.

[00:06:12]So those are the definite sums you have.

[00:06:17]Um twodigit sums always are going to add.

[00:06:21]Okay. So fourdigit sums.

[00:06:29]Interesting.

[00:06:31]Two-digit sums always will sum up to an odd number and four-digit sums always add up to an even number. Threedit sums can be either odd or even. So we're not sure about that.

[00:06:56]So the combination of digits here is interesting.

[00:07:02]For the two digits we can have 3, five, 7, 9, 11, 13, no 15 and 17.

[00:07:15]So there's seven possibilities and we can we only have six of them accounted for for four digits also.

[00:07:32]Um okay so these combinations are either 1 through 4 and 5 through 8 which means 9 is left over or 2 through 5. So this has to be a one nine pair here. So sorry not a one nine pair but one of these is one or nine and this could has to be one or nine guaranteed.

[00:07:54]Um we have two more sets of four digits.

[00:08:16]So we have one accounted for two accounted for. So starting at one as the lowest digit two starting at two at those lowest digit five and six. So the other ones must be starting at three and starting at four.

[00:08:36]Okay.

[00:08:38]Um starting at three is 3 4 5 6 7 3 4 5 6 which is 18.

[00:08:46]So one of these is 18 and then the other one must be 4 5 6 7 which is 22.

[00:08:56]So for the fourdigit numbers we have all of them accounted for.

[00:09:02]Um, so we know these are guaranteed to be what?

[00:09:27]3 4 5 6 or 4567.

[00:09:31]Which means we have a 3 4 5 6 7 quadruplet. Actually, this has to be whatever digit is not here. So, this is guaranteed to have four, five and six, right? Because these are combinations of it either is 1 through 4 with 5 through 8 or 2 through six two sorry one through four with 5 through 8 or 2 through 5 with 6 through 9.

[00:10:00]Um so this is guaranteed to have four, five and six. So it cannot be four, five and six here. Which means this has to be a four, five, six triplet here because it has to be either 3 4 5 6 or 4 5 6 7 here and the three extra 3 or 7 is here. That means the remaining digits in this box are 1 2 8 9. So we are guaranteed this cannot be one or this is one or nine. Here this is 109. Here we have a 1 2 8 9 quadruplet in box 9. So this can't be 1 2 8 9. This is 3 4 5 6 7.

[00:10:38]This can't be 2 or 8 because the two or eight in the row has to be in one or in this box has to be in these cells here which means that can't be in the row. So this digit is the same as that.

[00:10:52]Um and it cannot be the same as that digit there. So these digits are the same and it has to then be one of these two digits here. Interestingly enough.

[00:11:25]So that means this one obviously cannot be here. So it has to be one of these three here.

[00:11:36]Okay.

[00:11:38]So it must be one of these three.

[00:11:43]Obviously we cannot have four five six here. So the remaining digits in this column are 1 2 3 7 8 9.

[00:11:54]So this could be this is six total.

[00:12:00]Um so what do we have for certain for the fours? We have 10 10 14 18 22 26 and 30.

[00:12:27]We're We have for the threes we definitely have 65 and 21 24 sorry 65 and 24.

[00:12:41]So this can never be four five and six.

[00:12:59]For the remaining three cell digits, what can they be?

[00:13:05]So you have 1 2 3 adding up to six. 2 3 4. Okay. So let's just let's list down all of them right now. For the twodigit numbers, we have what? We have 3, 5, 7, 9, 11, 13, and 17.

[00:13:27]For the remaining threedigit numbers, we have so we have 1 2 3 already. 2 3 4 is 9, which repeats there.

[00:13:38]2 3 4 is 9. 345 is 12.

[00:13:44]four five six we already have 567 is 18.

[00:13:50]Okay. So we don't have 18 available because we are guaranteed all of these in the for the four-digit numbers. So then it's 21 and then yeah so the remaining digits.

[00:14:04]Okay. So then that gives us all the information we need then.

[00:14:10]So for the remaining three-digit numbers we have 9 12 and 21 which is 2 3 4 3 4 5 and 678 because so again we have to have all six four fourlength rim bands which is 10 14 18 22 26 and 30 30 not uh three.

[00:14:55]So and the three that we definitely have for three digits is 6 15 and 24. The two digits can be 3, five, 7, 9, 11, 13, and 17 because 15's already definitely taken. And for the remaining three-digit sums, we have not it's 6 9 12. So we have 92 available. 15's already guaranteed.

[00:15:21]18 is already taken by the four-digit numbers. And then the other the other one left is 21.

[00:15:30]So again we have 2 3 4 and 3 4 5 available and then 6 7 8.

[00:15:38]Well these two cannot be both 2 3 4 and 345 because they would have five digits that would be different when 2 3 4 and 345 are definitely not different. Same with these. These would have this. So this has to be some combination of either 2 3 4 here and the problem is so okay if we try to have 2 3 4 here that would mean one of these would have to be 3 4 5 but this having three digits here and two digits different here does not work because 2 3 4 5 is only four digits.

[00:16:14]So this can't be 2 3 4. Same with this being 345 because then it shares the same row with this. So this is a combination of 2 3 4 or 3 4 5 with 678 and therefore this has to be 678 here.

[00:16:27]So this since this is 678 this must be one of these is 2 3 4 or 345.

[00:16:34]So we have all the so we know all the three digits are 9 12 6 15 and 24.

[00:16:43]That means we cannot have a nine for the two-digit number. So, all the twodigit numbers are 3 5 7 11 131 17.

[00:16:53]Um, okay.

[00:16:57]So, then let's remove these so we can put all of these in the same place. So we have 6 9 12 15 21 and 24 for the three-digit numbers for sorry for the yeah threedigit ren bands for the twodigit remands we have 3 five 7 no 9 11 13 and 17.

[00:17:28]So this is guaranteed to be 678. We have to have this is 2 through five here and this must be 2 through five here.

[00:17:38]Um either way this can't be three or four.

[00:17:50]And the 678 triplet eliminates 678 in the rest of the box.

[00:17:56]Therefore 9 can't be here.

[00:17:59]Where is nine in box one? It has to be here. So this is nine which makes this one.

[00:18:07]This cannot be nine. This must be 1 2 3 as a triplet here which eliminates 1 2 3 and the rest of the row. So no 2 three here. This is four five as a pair which forces this to be three. So then this can't be three here. This has to be 2 3 4 as a triplet here because we have two and four required to be there and we already have a 345 renb band here. So then this is three here. This is two or four. So this cannot be 4 56 here. This is 7 8 9 here.

[00:18:42]Okay.

[00:18:46]So that means this is 7 8 9 and this must be 456 here.

[00:18:53]Uh this has to be a different digit. So these are both nine. So we have nine here, nine here.

[00:19:00]Um this must contain five 6 through 9 or 2 through 5. And therefore this must be 2 through five here because we can't have an eight or a nine there. That 345 triplet eliminates 3 4 5 in the rest of the box. This is then a two. So this is two. This has to be 6 7 8 9 and this is 67 which which must be with five six four five here sorry that eliminates four five in the rest of the box.

[00:19:38]Um, this has to be 1 through.

[00:19:42]No, this is not 4 five. It's it's 58 here.

[00:19:50]This is 58 to be adding up with 67. That eliminates nine here. This cannot be five. This is 1 through four here. So here we have 1 through four. No four. No one here. So, this has to be one here, which means no reds there.

[00:20:16]So, we can remove the coloring.

[00:20:20]No four here.

[00:20:24]Um, this can't have four here. It must be six through nine.

[00:20:29]So, no nine here. No eight here because of a five a pair in the box. So this is 67 as a pair which eliminates 67 in the rest of the box. It also eliminates 67 here. So this is 3 four as a red band which makes this 56 which forces this to be seven.

[00:20:48]Seven eliminates seven here. So that's eight which eliminates eight in the rest of the box.

[00:20:54]Uh this can't be one. So this must be a two three pair here.

[00:21:02]Okay, remaining digits in row two, we have 1, five and nine.

[00:21:09]So this and this can't be 7 8 9 because of a 789 triplet. And this can't be one because of one in the column. So this is five. That leaves one and nine remaining. This can't be nine because of nine in the column. So that's one, which makes that nine. That eliminates one there.

[00:21:27]um the remaining digits in two eliminates two here. So this is four and two.

[00:21:38]This cannot contain a one because one is only consecutive with two and we already have a two in the column. So one has to be here which leaves this as a six. Oh, so that does become six with five or seven.

[00:21:51]That's 12 or 11 or 13. So that's fine.

[00:21:57]Remaining digits of box four. We have 1, two, three, and seven, eight, nine.

[00:22:06]Two cannot be here because that would guarant we need to have a one or three to be consecutive with that. So there's no two here, which means this has to have a two here.

[00:22:20]One could have one could have a two here.

[00:22:24]Three could have a three eliminates three here. So this is four which makes that three.

[00:22:30]So we have only one out of one or three here. So this can't be two because of two in the column. This can't be two because of two in the column.

[00:22:44]So here the remaining digits are one of them's two which can be with one of one out of one or three.

[00:22:53]But the only remaining digits can be here is 6 through 9. So two out of 7 8 or nine are here. This is eight or nine.

[00:23:01]So 2 out of 7 8 or 9 are here. And then one of eight of nine is there. So this cannot be 7 8 or 9. It's six which makes that seven. Six eliminates six here. So it's five and six. Five eliminates five here. So it's four and five. That eliminates four here.

[00:23:19]Um so what are the twodigit numbers remaining? We have three which is 1 and two.

[00:23:27]Five which is 2 and three. Seven which is three and four.

[00:23:34]But three and four can't be here. And three and four can't be here. So where's the three and four? They have to be here. which eliminates three and four in the rest of the column. This can't be three or four. It's five. That eliminates five there. So it's six, which eliminates six here. So it's seven, which makes that six. So these can't be five anymore. This three eliminates three here. So that's four, which makes that three. This can't be four or six. So it's five, which means this can't be six or five. It's four.

[00:24:04]We need a 56. 54 is not valid as the two-digit red band because we already have 2 3 4 adding up to 9. So this must be this can't be five here. This can't be five here. This must be five there.

[00:24:20]That eliminates five in the rest of the column. So this is 8 and five. 8 and 9 eliminate eight and nine in the rest of the column. So that's seven. There's no seven here. No seven here.

[00:24:32]So we have 11 counted for and seven. We still need 8 9.

[00:24:39]Okay. Remaining digits in row three, we have 278.

[00:24:43]Okay. So, no seven here because of a seven in the column.

[00:24:49]Uh 3 4 as a pair means this can't be three four. So, it's one two or eight.

[00:24:58]Um renb band here could be one two.

[00:25:06]7 8 or 8 9 H.

[00:25:14]Okay. Remaining digits in column six are 1 and three.

[00:25:22][snorts] Five here eliminates five in the rest of the box. We have a 1 34 pair in row. So this is in row five and six. So we have guaranteed a three in one of these cells. We are guaranteed a three in one of these cells which eliminates three in the rest of rows five and six.

[00:25:48]So where are where's three in box four?

[00:25:50]It must be here.

[00:25:52]And in addition uh this this is two three as a pair. So this can't be three.

[00:25:58]This is four. In addition, we still need 32 and one two as a pair for the ren band. Well, this must then be 32 because that's the only other place for it. So that's three, that's two, which means this can't be two or three here.

[00:26:16]And this can't be two, but we need a one two pair. These can't be two anymore. So this can't be one and two. So that means this must be a one two pair here. So this is one and this is two, which means this can't be one or two. It's eight.

[00:26:28]This can't be two or eight. It's seven.

[00:26:30]This can't be 1 2 7 or 8. It's nine. So seven eliminates seven here. Eight eliminates eight here. So that's two, which makes this one. Two eliminates two here. So it's eight, which makes that two. This can't be one or two. It's eight. One elimin one and eight are eliminated here. Seven eliminates seven here. So this can't be one or eight.

[00:26:51]It's nine. Therefore, there's no more nine in the rest of the box. Three here eliminates three here. So that's two, which makes this three. Three eliminates three here. So that's two, which eliminates two here. And this three eliminates three here. So that's one, which makes that three. And this can't be two because of two in the row. So it can't be eight because of the seven eight pair in the box. So this must be one, and this must be two.

[00:27:18]Nine eliminates nine here. So it's eight and nine.

[00:27:22]And again, this cannot have a nine here.

[00:27:26]We cannot add up to 15. So 7 and 8 are removed. So it must be 6 7 and 7 9 or 8 9 remaining. Well, this can only be 8 9 here. So then this must be seven and six.

[00:27:40]One eliminates one here. So it's three and one. Three eliminates three here. So it's four and three.

[00:27:46]And we have nine here eliminating nine here. So it's eight and nine. Six eliminates six here. So it's seven and six. And the rest we do just from normal. Sodoku.

[00:27:57]Um, no four here. No six here. We should have all the ones. We do. We have all the twos. We have all the threes.

[00:28:09]Fours are not quite restrictive enough.

[00:28:13]Neither are fives. What about sixes?

[00:28:16]Neither are sixes.

[00:28:19]Um okay remaining digits in H.

[00:28:30]We do have enough eights. So we have eight eights. So the remaining eight has to be here. We have all the nines though. So this must be four five as a pair in row six. But this can't be four because a four in the column. So this is five which makes this four which makes that six which makes that five.

[00:28:48]And so remaining digits we have eight fours.

[00:28:53]This must be four here. So we did five.

[00:28:56]We have all the fives. Now do we have all the sixes? No. But this is the last six which leaves these as seven.

[00:29:03]And that's the solution. Um not too bad.

[00:29:08]The logic was fairly straightforward.

[00:29:16]I don't know. May maybe it's a little harder for those who don't have these sort of like I've been I've done so many like killer cage sodokus or um arrows or you know basically mainly actually probably killer cage sodoku because you have to add up all these very quickly and be able to sort of spot these what can be where right and so the ren bands were fairly easy to determine what the subs were.

[00:29:49]I don't know. Seems pretty straightforward.

[00:29:53]Anyways, I hope you enjoyed this solve and I'll see you next time on Frank Puzzles. Take care.