Calcudoku is a popular Sudoku Variation. It’s a bit more challenging than a traditional Sudoku puzzle because it requires using some math.
How to Play Calcudoku
Calcudoku is played on a square grid. The most common Calcudoku grid sizes are between 4×4 and 9×9.
The object of the puzzle is to fill all the empty squares with the numbers 1 through X (where X is the grid size). Each number must appear once (and only once) in each column and each row.
For example, in a 4×4 Calcudoku puzzle, X is 4. This means the numbers 1, 2, 3, and 4 will appear once in each column and once in each row.
But wait! There’s more (sorry, couldn’t resist).
Every Calcudoku grid will have blocks in it that are surrounded by bolder lines. Each block will contain a result and math operator in its top-left corner. The numbers in the block must total the result using only the given operator. A number may be used more than once in the same block, as loing as it doesn’t appear more than once per row/column.
What does “result” and “math operator” mean?
Result means total. In Calcudoku, a math operator could be addition (+), subtraction (-), multiplication (x), or division (÷).
For example 2×3=6; ‘6’ is the result and ‘x’ is the operator.
Which operators can be used in a Calcudoku puzzle?
In Calcudoku, there are 5 different types of puzzles:
- addition only
- multiplication only
- addition and subtraction
- multiplication and division
- any operators (add, subtract, multiply, divide)
Example Calcudoku Puzzle (6×6, any)
There are a variety of puzzle sizes and types, so before you get started make sure you know what puzzle you have.
This example is a 6×6 puzzle that could have any operators. This means, our puzzle won’t use any numbers larger than 6 and it could use add, subtract, multiply, and divide operators.
The very first thing you should do is check for any single square blocks. Our example puzzle has four (6, 4, 5, 4).
The next thing you should do is look for clues that have only one or two possible solutions.
Let’s start with 4÷ (under the 6) and 6x (to the left of the bottom 4).
For 4÷, what are our options? There’s 2 squares, we must divide, and the result needs to be 4. We only have one option here: 4÷1=4 (remember, we can’t do 8÷2=4 because our largest number will be 6). So one of these squares will be 4 and the other will be 1, we just don’t know which is which… yet.
For 6x, what are our options? There’s 2 squares, we must multiply, and the result needs to be 6. We have two options: 1×6=6 and 2×3=6. Can we narrow it down? Yes, we can.
Look at the first square. There are 4 possibilities: 1,2,3,6. One of the rules of Calcudoku is that each number (1 through 6) can only appear in a column once. If we look up this column, we already know this square cannot be 6. But we also know there will be a 1 and a 4 in this column, which means this square cannot be 1.
So our 6x has to be 2×3=6 even though we don’t (yet) know which square will be 2 and which square will be 3.
Let’s take a look at one more ‘easy’ one: 15x (3 squares in the second column).
For 15x, there are 3 squares so there’s only one option: 1x3x5=15. We can’t fill out any of these squares yet (although we do know that the middle one cannot be a 5). However, since they’re all in one column, we do know that nothing else in that column can be a 1, 3, or 5.
This means our bottom square of the column (part of 6x), cannot be 3 so it must be 2. And the corner square of 6x must be 3.
The more numbers you fill out, the easier the remaining numbers get because you can cross out more options.
Let’s look at 9+ (top-left corner) with 2 squares. There are two options: 4+5=9 or 3+6=9. If you look at the far-left column, you can see 3 and 6 are already used so we know this block must be 4+5.
And it gets better, we know that the left square cannot be 4 because there’s already one accounted for in the column, so it must be a 5. And the second square must be 4.
Let’s check out 7+ which is a 3-square block in the top-right corner.
There are many options: 1+2+4, 1+3+3, 1+1+5, 2+2+3. Because this block has squares in multiple rows/columns, you can use the same number twice as long as they aren’t going in the same row (or column).
Let’s start with the easy eliminations. We know this cannot be 1+2+4 because there are already 4s in (or planned to be in) both rows. We know this cannot be 1+1+5 because the 5 would have to be the top-right corner but there’s already a 5 in that row.
So we’re left with 2 options: 1+3+3 or 2+2+3.
Let’s go to the 3÷ in the middle of the top row. It’s two squares so it could be 6÷2 or 3÷1.
Here’s the thing though, we know from the 7+ that there will already be a 3 in the top row. So our 3÷ must be 6÷2.
This leads us to filling out the 7+ entirely. The first square cannot be 2, so it must be 3, 1, 3.
We can also know that the 3÷ under the 7+ must be 6÷2. Which tells us that the last two squares in the column are 4 and 5.
A 4 is already used in the bottom row, so the bottom-right corner must be 5 and above it must be 4.
Now we can easily see that the remaining 2 blank squares in the bottom row have to be 1 and 6.
If we look at the 2-, there are 2 options that contain a 1 or 6: 6-4 and 3-1. If we use 6 in the bottom row, 4 would have to go in the top square but it cannot because there’s already a 4 in that row. This must be 3-1.
So we can also fill in the 6. Row complete!
And there’s one more easy box given to us:
120 = 4x5x6x? well, 4x5x6=120 so to fill in the fourth box we can only use a 1.
The puzzle is about half complete. Can you finish it up?
There are a couple of really easy squares left to get you started:
- The 5th column and 4th row square can only be one thing (because it’s the only empty square in the column).
- The 1st column and 5th row square can only be one thing (because it’s the only option remaining for the column).
(If you want to check your solution, the completed puzzle is at the very end).