How to Solve Calcudoku Puzzles

Calcudoku was invented in 2004 as Kendoku by Tetsuya Miyamoto to assist in teaching math to his students. Another name, KenKen, was popular as well, but both of the names beginning with “ken” are trademarked by their publishers, so Calcudoku became an accepted common name as well.

Similar to Sudoku, you will be filling out the puzzle with numbers from 1 to X (where X is the length of one side of the grid) in such a way that no number repeats itself in any row or column. Duplicates are allowed in a region if it encloses more than a single row or column.

The grid is divided up into regions, each with a total. If an arithmetic operator is given, you must be able to reach that total by using only that operator on the numbers entered into that region’s cells. If no operator is given, you may use any operator, but only one type (you can’t subtract one number from another, and then multiply by a third to reach the result, for example).

Most of the time, classic mathematical symbols are used (× instead of * for multiplication, and ÷ instead of / for division). To make things a little easier to see, I color coded the clues in the example puzzle – green and red for addition and subtraction, and blue and purple for multiplication and division.

Choosing where to start can be tricky, but it helps to look for some traits to limit possible digits.

Small regions in a single row or column.
Multiplication and division operations with a very large or very small total.

Here, we have a 2-cell division problem with a result of 6. Because of the size of the grid, we’re only using the numbers 1-7, so the result can’t be from 12/2. However 6/1=6, so that must be the two numbers used here. We can’t know the exact placement yet, so we’ll just use some pencil marks for now.

Now we have a multiplication problem resulting in 56 from three cells. If you remember your times tables, you know that 7×8=56. But we can’t simply use 7x8x1, because the 1 in this column is already used in the region we discussed last step.

7 is a prime number, so we can’t break it down further, but 8 can be expressed as 2×4. There isn’t another combination of three numbers that results in 56 with only multiplication, so it must be 7x2x4. Again, we don’t know the order, so we’ll use more pencil marks for now.

This region is for a multiplication resulting in 5 from 2 numbers. Conveniently, 5 is a prime number, so our two digits must be 5×1=5. Again, we don’t know the order, so we’ll have to use pencil marks, but notice the green cell.

The only possible numbers that can be placed are 1 and 6. We now know that 1 and 5 are already used in that row, so the green cell must contain a 6, and the cell below it will hold a 1.

Let’s look at a couple of regions this time. On the right is a multiplication that results in 3, which is a prime number, so that’s an easy one – pencil in 1 and 3 in those cells.

Over on the left is a sum that results in 11. It can’t be 5+6, because both those numbers are already being used in a single row, so neither can be used again. It also can’t be 9+2 or 8+3, because our grid is only 7×7, which means all the numbers must be between 1 and 7.

This leaves 7+4 that will be pencilled in here.

Here are two regions, each a product totaling 30, although one of them uses three factors instead of two.

The yellow cells must be 5×6, because no other digits within our range can multiply together to get 30.

Using 5x6x1 is out for the green cells, because the 1 is already in use in that row. 5 is a prime number, so we have to break down the 6 giving us 2 and 3. So the green cells will have 5x2x3=30 pencilled in, because as usual, we don’t yet know the order.

This region is two cells to produce a sum of 4. We can’t use 2+2, because duplicate numbers aren’t allowed, so the only option must be 1+3=4. This will eliminate 3 as an option in the green cell, because it’s in the same column where both 1 and 3 must be used.

Here, we have an interesting case. This region contains a sum of 6, and we already know 3+3 is out because of duplicates. That means it is either 4+2 or 1+5.

Because of other regions in the first column, neither 1 or 4 can be placed in the yellow cell. We don’t yet know about 2 or 5, so they can be placed there.

This means that 4 and 1 are the two possibilities for the green cell to provide the other half of the numbers for the correct addition options.

And now, we have two cells in the same column that only have two possibilities left, a 2 and a 5.

This means that the green cell cannot contain a 5, because it will definitely be used in one of the yellow cells. Now we know the green cell contains a 6, and the other cell in its region must hold the 5.

We’re about to have a fun domino effect. First, 5 can now be eliminated from the pencil options in these two cells.

In the yellow cell, that will leave only 1 as an option, meaning that 5 must be the number in the other cell in that region.

5 can now be removed as a contender from the green cell’s pencil marks.

Meanwhile, 1 is in the same column as the yellow cell, so it can no longer hold a 1, meaning it must be a 4. That means that 2 is the matching number to reach a sum of 6 in that region.

Finally, the yellow cell can no longer contain a 2, so it must hold a 5.

And the green cell can no longer hold 2 or 4, because both are in the same row. That means it must be a 7, which can now be eliminated from the other two cells in the region that multiplies to 56.

In the third row, 1 and 5 are known, and while the exact placement of 2 and 3 aren’t certain, they are definitely in those two cells.

This means the only remaining numbers for this row are 4, 6, and 7.

The yellow region is a subtraction problem resulting in 1 so that must be where the 6 and 7 will be used, although again, we don’t know the order.

Therefore, the green cell must hold a 4.

This region is for a sum totaling 15. We already have a 4, which means we need an additional total of 11 in the remaining two cells.

Both 5 and 6 are already used in that row, so it can’t be 5+6. As we already discussed earlier, it also can’t be 8+3 or 9+2, because 8 and 9 are out of our allowed range of numbers.

Therefore, the two numbers here must be 7 and 4, which is allowed, because the duplication rule is only about being in the same row and column, not the same region. We simply are not allowed to place the 4 in the same column as the one in the third row. So, we place a 7 on the left, and a 4 on the right, and we end up with 4+7+4=15.

After making that placement, we can eliminate the 4 in the yellow cell, leaving a 2 as the only option, and a 4 will go in the bottom cell of the region to complete it.

Then, in the green region, we’re looking for a sum totaling 6. We can’t use 3+3, because of duplicates, and 2 and 4 are already both used in the same row. Therefore, we have to pencil in 1+5=6.

Since those two cells only hold two possible numbers, a 1 and 5, that means the yellow cell in the same row can no longer be a 1. Eliminating it makes that space a 3, and the cell above it must be a 1.

Now the only remaining number for the green cell is 3, because all other digits from 1-7 are used in that row.

In order to reach a sum of 9, the cell below it must be a 6, because 3+6=9.

At this point, the yellow cells can only contain either 3 or 5 because the rest of the middle column is known.

Looking at the top, we have a subtraction that results in 3. Using the 3 and 5 in the yellow cell, our options are: 6-3, 8-5, or 5-2. So the red cell will contain a 2, 6, or 8. We can pencil those in.

At the bottom, we have a sum of 7. Again, using 3 or 5, our options would be either 3+4, or 5+2. But wait! There is already a 4 in the column with the green cell, so we can’t use 3+4.

That means the region at the bottom must contain a 5 in the yellow cell, and a 2 in the green cell. Once we’ve established that, then the yellow space at the top has to hold a 3, and the only option for the red space is a 6.

Remembering the limit in your range of numbers can be helpful. In the yellow region, we have a subtraction totaling 5. We can’t use any digit higher than 7.While 7-2=5, a 2 is already used in the same row. That means we have to use 6-1=5. There is already a 1 in the last column, so it must be placed on the left, with a 6 on the right.

Then at the top, we have a sum totaling 9. We can’t use 3+6, because both of them are already used in the top row. This leave us with either 5+4, or 7+2 as options to pencil in, and the 5 can’t be on the bottom because there’s already one in that row.

We’ve eliminated the 1 in the two yellow cells, which lets us easily solve those regions.

Then, in the green, we have a multiplication resulting in 30. We’ve eliminated 1 as an option in all three cells by having it in those columns, so it can’t be 5x6x1=30. As we discussed earlier, that means we factor the 6, meaning the three cells contain 5x2x3. When pencilling them in, we can eliminate the 5, 2, and 3 where they are duplicated in those columns.

There are a couple more cells where one of the numbers is now eliminated because we’ve found the correct position in that column (3 for green, 6 for yellow), so we can complete those regions.

Highlighted in red, we have a multiplication totaling 42, which we know is 6×7. There is already a 1 in both columns, so that means we again factor the 6 down to 2 and 3. There is a 7 and a 2 in column 5, so that space on the second row must hold a 3. We’ll just pencil in the 7 and 2 in the sixth column for now.

The yellow cell is the only space left in its column, and eliminating the 3 we just placed means it must be a 5.

Within that region, it forces the red cell to contain a 2, and there must be a 3 in the middle.

In the green cells, placing a 2 below eliminates 7+2 as one of the options, leaving 5+4, and we already know the correct placement.

Next, the 4 eliminates that option in the yellow cell, making it a 7, and that region is now solvable.

After placing the 7 in the yellow cell, it will be removed as an option in the green cell, leaving a 2, and that region becomes solvable as well.

The last remaining options for the row with the yellow cells are 1 and 2, which of course are correct for 2-1=1. Because each is already represented in their columns, we also know that 2 goes on the left, and 1 goes on the right.

Together, the red and green cells add up to a sum of 24. The red cells must contain 6 and 7, while the green must hold 4 and 7. We can easily confirm that 4+7+6+7=24, so now we only have to worry about correct placement.

Looking at the red spaces first, there is already a 6 in the right column, so that means 6 must go on the left, and 7 on the right.

Now down to the green. Once we know that 7 is in the right column in the red cells, that must mean that 7 is on the left here, and 4 on the right.