Sums Plus is a Sudoku variant where, in addition to standard Sudoku rules (placing digits 1-9 without repeats in rows, columns, and 3x3 regions), the gray cross-shaped region must also contain all digits 1-9 once, and the digits along each arrow must sum to the digit in the attached circle. The solving strategy involves using the sum constraints to eliminate possibilities and place digits systematically, starting with cells that have the most restrictive constraints (like arrows requiring minimum sums of 3+ unique digits) and working through the puzzle by identifying where digits must go based on the combined Sudoku and sum rules.
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Frank Puzzles About Sums Plus | Easy Variant SudokuAdded:
Hello everyone. I'm Frank and welcome to Frank Puzzles. And this is today's easy level puzzle. And as always, take that easiness with a grain of salt because it's considered easy by the people on the Logic Masters Germany forum. And there's a lot of really smart people there.
Today's puzzle is called Sums Plus by Cane Puzzles. And here are the rules. We have classic 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.
The gray region must contain each digit 1 through nine once. So this also has to have the digits 1 through nine without repeats in this gray plus.
And the digits on each arrow must sum to the digit in the attached circle. So for example, if this were say three and three, this would have to be six because the digits along the arrow must add up to the digit in the attached circle.
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. Let's go.
Okay, this is a sum of at least three unique digits which is at minimum 1 2 and three plus another digit which has to be at least one. That means this has to be seven, eight or nine here because it's 1 2 3 6 + 1 at minimum. But it cannot be seven or nine because of seven or nine in the row. So this can't be seven or nine. It's eight.
And this must add up to either six or seven. So it's either 1 2 3 or 1 2 4 here.
And this must be one or two here to add up to eight, right? It's 1 2 3 and two to add up to eight here or 1 2 4 and one here to add up to eight there.
Either way, this cannot be 1 or two. We have 3 through 7 as possibilities here.
Um, and this could be 1 through six.
So, this adds up to either four, five, 6, 7, or 9 because this could be one here and three here. Actually, no, it cannot. If we try to have one here and three here, that would force this to be two and one two four here, which adds up to nine. So this cannot be four here.
Um where is eight in column two then? Can't be here or here because of eight in the box?
Actually where is eight in row three?
Can't be here because of eight in the boxes. Can't be here because that must add up to eight. So eight has to be here.
Um, where is eight in column two? Can't be here or here because of eight in the boxes. Can't be here because 8 plus another digit can only add up to nine as another digit and we already have a nine in the column.
So, eight has to be one of these two digits here.
Um, none of these can be eight.
Okay, seven's here. Seven's here. So, the seven in column nine has to be one of those three.
Um, this digit cannot be one, two again.
So, this is 3 4 5 6 7 9 as possibilities.
Um, eight cannot be any of these digits here. So, where is eight in this cross?
It has to be one of these four digits, which eliminates eight in the rest of that row. Then, so this cannot be eight, which means this must be eight here.
Um, so these four digits must be the same as these four digits here and these four digits must be the same as these four digits here.
That means um these must be the digits a certain amount.
So these have to be the digits 1 through nine. Again, these four are exactly the same as these four here. So let's make these yellow.
These are the same as these four here.
These four are the same as these four here.
That means these have yellow and orange have to have the digits 2 through 9 because one is never going to be the sum of two unique digits. And the only place where we can have two now is here. So this is two which is the sum of one and one.
um that means this can't be one or two. It can't add up to seven. So this it cannot be eight. Therefore, this must be nine here.
And this has to be either 3 6 or 4 5.
One eliminates one here.
Um so nine cannot be in the yellow digits.
Um, we can't have Well, nine has to be So, two has to be in one of the orange digits.
Um, we have to have one of these digits add up to three. And three requires one and two to be on the arrow. Well, this can't be these can't be threes because of the threes on the of the ones in those columns.
So, three has to be one of these two then, which means this is three. It could be one and two here or one and two here. We are guaranteed a three though in a yellow cell or also cannot be in a orange cell or it can't be in one of these orange cells because it must be one and three added together. So this must be three and four as a pair here.
That means we have guaranteed one and two or one and three here. And this can't be one because of one in the box.
So this is guaranteed to be one. And this is two and three. One eliminates one here. So that's two, which makes this one, two, three as a triplet.
And this can't be one or two. It's three. This can't be one. Three eliminates three here. So that's two, which makes that one. This is no longer three. This can't be two or three.
Um, so this can't be one. Where is the one in for to add up to two, three or four?
Here it must be here. And this is two or three cannot be three because of a three four pair in the box. So this is two which makes that three to add up. So this is four which makes this one and three to add up. This can no longer be three. So this can no longer be six.
We have three four eight and one out of five six seven here. And this digit has to be one because one cannot be the sum of two unique digits and therefore can't be part of yellow and orange.
One, where is one in row four? Cannot be here because one's in the box. Can't be here because of one in the column. So it has to be there.
Where is one in box one? Can't be here because of one in the row. Can't be here or here because one's in the column. So it must be there.
Where is one? We have eight ones left.
We know where the last one has to be. It has to be there. This is 5, six or seven, which means this cannot be seven there to add up to 5, six or seven there.
Um, this has to be a five, six, seven here.
Um, that means these add up to these cannot have a one in them. So, they're two through five there.
Okay.
And they do they they don't both have to have a two. Where is two in row one?
Can't be here because of twos in the boxes and the the rest are filled out.
So this is two there.
Um remaining digits in remaining digits in row eight. We have excuse me two five six and nine. And two cannot be here because of two in the column. This can't be two because of two in the box.
So we're guaranteed a two in row eight in one of these two cells which means this cannot be two here.
So we only have one two available here.
So that means one of these is going to be have have to be at minimum 3 4 which means we are guaranteed a seven here which means this cannot be seven which means this can't be six. So this is five or six here.
Um and this must be two with three or two with four adding up to five or six.
So five can't be any of these. This can't be five because it only adds up to seven at most. We have a 3 4 3 4 pair here.
Interesting.
Remaining digits in box seven then are 2 5 6 7 9. Obviously this can't be nine because of nine in the column. This can't be two because of two in the box.
This can't be two because of two in the box.
Remaining digits of box eight are 5 6 7 8 9. Eight cannot be here.
That's fine. We're guarant where is two in row seven then cannot be here because one of these has to be two. Can't be here because of two in the box. Can't be here because of two in the column. So two has to be here.
Therefore this is five through nine. But it cannot be seven or nine because or five because five seven in the box and nine in the column.
um remaining digits of box. This is guaranteed to be a two here.
Where is two in column five? This can't be two because one of these is guaranteed to have a two here.
And these can't be two and two because of twos in the boxes. So two has to be here.
Okay. Remaining digits. Uh this can't be eight because of eight in the box.
Remaining digits of box four. No seven no two. So we are left with one out of three four.
We have this can't be nine here.
One out of three four. We have one out of five six. And we have a nine here which and we have the nine guaranteed to be one of these three which means this cannot be nine here.
One of these is four or five. No, we have one out of three, four here and one out of five, six here.
Okay. Remaining digits of row two, we have what? 3 4 5 6. Okay.
Where is seven then in column three?
Seven cannot be any of these digits because one of these has to be seven.
can't be either of these because this adds up to five or six here. And we have a seven in the row here. So, this must be seven here, which means this can't be seven here. There's no seven here. Where is seven in box eight? It can only be here now. So, that's seven. Five eliminates five here. So, that's six.
This can't be five or six. It's nine.
That eliminates nine here and six being here. This is a two five pair, which eliminates two five in the rest of the row. So no five in the rest of the row.
This cannot be five here because that would force this to be 2 three and that would eliminate all possibilities for that cell. So neither of these can be five. So it must be six, seven here.
Therefore we have two and four here and then three and four here. So this cannot be three. It's two or four which eliminates two four in the rest of the box and 67 in the rest of the box. This is 359 which eliminates five here. So it's four. No 3 four five here it's six.
That eliminates six in the rest of the row.
That means 1 four add up to five. So this is five which eliminates five in the rest of the column.
This has to be six seven or nine.
No five here because of five in the column. Remaining digits of row nine are four and eight. That eliminates four and eight in the rest of the box. This can't be eight. It's six. This can't be six. It's nine. Six eliminates six here, so it's nine. Which eliminates nine in the rest of the column and box. This can't be nine, so it's six. No 6ix9 here. No 69 here.
Okay.
Um, we have a 2 3 4 five quint quadruplet in column one which eliminates 2 3 4 5 in the rest of the column. So this has to be six or seven.
We have a 67 pair now in row three which eliminates 67 in the rest of the row.
This can't be six or actually this six eliminates six here. So that's seven, which makes that six, which makes that seven. So this is six adds up with two and four, which makes this three. That's four.
Two eliminates two here, so it's five and two. This can't be four or five here, so it's three, which makes this four, which makes that five. There's no more five here, so this can't be four there to add up to nine. Nine eliminates nine here, so it's three or five.
Uh remaining digit in column two. This seven eliminates seven here. So that's six, which makes this seven.
Remaining digit in column two is a five.
Remaining digit in row three is a nine.
Uh we have four and seven remaining in box two. This can't be seven because of seven in the column. So that's four and seven.
We have 1 2 3 4 5 6 7 8 9 1's. 1 2 3 4 5 6 7 8 nine twos.
We don't have enough information on threes.
Four, we don't quite have enough information.
Five, we don't really have enough information. Okay, let's just do this normal Sudoku solving pattern. We have 369 remaining in row one. This can't be six or nine because of six or nine in the column. So that leaves that as three. This can't be three or nine because of nine in the column. Three in the box. So that's six, which makes that nine. Three eliminates three here. So it's four and five to add up to nine here. Four eliminates four in the rest of the column. So that's eight and four.
Five eliminates five here.
Remaining digits of this is seven here.
Remaining digits of column 9 are three and seven. This can't be seven because of seven in the row. So that's three.
That's seven.
And we have six and eight remaining here in column seven. This can't be six because of six in the row. So this is eight, which makes that six.
Um remaining digits of row four we have three 9 and six. This can't be six or nine because of six or nine in the column. So that's three here which makes this nine which makes that six. And nine eliminates nine here. This can't be three or nine. It's five which makes that three there.
That leaves the remaining digit as four and five in row five. And this can't be four because of four in the column. So it's five and four. Five eliminates five here. So it's eight and five. Which leaves this as a nine and this as an eight.
And that's a solution.
Um, yeah, pretty cool interaction between the the digits on the circles on the arrows and particularly this cross shape, right?
It definitely fulfilled the sums plus uh moniker.
Yeah. And it just flowed very well throughout. Um maybe it's because I'm used to puzzles like this, but um felt rather easy for me.
Anyways, I hope you enjoyed this solve and I'll see you next time on Frank Puzzles. Take care.
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