There are only two rules of KenKen and they are very simple. All of the numbers from 1 up to the grid size must appear in every row and every column. So in a 6x6 KenKen puzzle, for example, each of the numbers 1-6 must appear once and only once in every row and every column. The other rule is that in each block that has a thick line around it, the target number in the top left-hand square of that block is calculated from the numbers in all the squares in the block, using the addition, subtraction, multiplication or division operation indicated by the symbol.

Solving KenKen puzzles requires a mixture of logic and simple arithmetic. The logical process makes use of the information you have as to what numbers are, or could be, in which squares of the grid. Arithmetic enables you to narrow down the possibilities. You might find it helpful to pencil possible numbers into a square as the possibilities are reduced.

We start by explaining the most important simple methods employed in solving KenKen puzzles.

“Unused” numbers

Take a look at this row from a 6x6 KenKen puzzle.

5

4

2

6

1

We can see that five of the squares have already been solved, with the numbers 1, 2, 4, 5 and 6. One of the numbers from 1-6 is currently “unused” in this row, and that is 3. There is only one empty square in the row and so we can deduce that it must contain a 3.

Now look at this row.

A

B

4

2

6

1

Here we can see that only four of the squares have been filled in, with the numbers 1, 2, 4 and 6. The other two squares, A and B, have no numbers in them yet. The numbers 3 and 5 are currently unused. We know that 3 and 5 must both be in this row, because every number from 1 to 6 must be in this row, so we can be sure that one of A and B is 3 and the other is 5. We do not yet know which is which, so it would help us to use a pencil to write both 3 and 5 in square A, and both 3 and 5 in square B, until we have enough information to decide which is which.

Only one number in a block is missing

Since we know the total for every block, if we find a block that has only one missing number we can often calculate that number by knowing the total and the other numbers in the block.

19+    3

5

6

4

Here the total in the block is 19, and is created by addition. We already have four of the five numbers in the block: 3, 5, 6 and 4. These four numbers add up to 18, so the empty square must contain 1.

A similar approach will always work for blocks with multiplication – there will be only one number that can possibly go in the empty square. But where the symbol is subtraction or division we might not be certain which of two numbers it could be, as in the next example.

2-    A

4

Here we wish to calculate A. It could be 6 (because 6─4 = 2) or it could be 2 (because 4─2 = 2). Remember that the subtraction symbol means that the total is the difference between the two numbers in the block – it does not indicate which square contains the larger of the two numbers.

Similarly, the division symbol means that the total is created by dividing one of the numbers in the block by the other, but it does not indicate which of the squares in that block contains the larger of the two numbers.

Only one way to make the correct total with a set of numbers

Sometimes we will find a block for which there is only one set of numbers that can possibly make the correct total.

4+    A

B

Here we have a block of two numbers that add up to 4. The only way to make a total of 4 using two numbers is 2+2 or 1+3. But A and B cannot both be 2 because we cannot have the same number more than once in any row or any column. So the two numbers must be 1 and 3. We do not yet know which is which, so it would help us to use a pencil to write both 1 and 3 in square A, and both 1 and 3 in square B, until we have enough information to decide which is which.

Here is another example of this method.

10x A

B

In this case we have a block of two numbers that we know multiply to make 10. The only way we can do this in any KenKen puzzle, using two numbers, is with 2 and 5. So we know that one of A and B is 2 and the other is 5. We do not yet know which is which, so it would help us to use a pencil to write both 2 and 5 in square A, and both 2 and 5 in square B, until we can decide which is which.

The only possible solution to a block that covers more than one row or column

Sometimes we will come across a block that has only one possible way to make the given total with however many numbers we are allowed to use, and it will also be possible for us to decide exactly which numbers go in which squares of that block.

The most commonly seen example of this is an “L” shaped block with three squares.

4+    A B
C

Here we have a block of three squares in which the total, created by addition, is 4. It is easy to see that the only three numbers that add up to 4 are 1, 1 and 2. But we cannot have two number 1s in the same row (the upper row here), and we cannot have two number 1s in the same column (the left-hand column here). So the only way to arrange 1, 1 and 2 that complies with the rules is for A to be 2 and for B and C both to be 1. In this example, in one fell swoop, we have not only worked out what the three numbers must be, we have also worked out exactly which boxes have which numbers. You will not always be this fortunate, but you will often be able to decide exactly what set of numbers must go in a block.

Solving a KenKen Puzzle Step-by-Step

Now we are going to work through two examples of solving whole KenKen puzzles step-by-step. This first example is extremely simple because the only arithmetic operation involved is addition.  Remember, all the numbers 1, 2, 3 and 4 must appear in every row and every column. And to help us start the puzzle we are told that the number in square C is 1.

Step 1

Q and R add up to 3, so one of them must be 1 and the other must be 2. But Q cannot be 1 because C, in the same column, is 1. So Q must be 2 and R is 1.

Step 2

G and L must be 3 and 4 (but we do not yet know which is which), because these are the only two unused numbers in their column. If L were 4 then M would have to be 1, because L+M totals 5, but M cannot be 1 because R, in the same column, is 1. So L cannot be 4 and therefore L must be 3 and G is 4.

Step 3

M must be 2 because L+M totals 5.

Step 4

D and H must be 3 and 4 (but we do not yet know which is which), because they are the only two unused numbers in their column. But H cannot be 4 because G, in the same row, is 4. Therefore H must be 3 and D is 4. We can verify the three numbers in this block by noting that the total of 3, 4 and 4 is 11, which is the correct total for this block.

Step 5

E and J must be 1 and 2 (but we do not yet know which is which), because they are the only two numbers that add up to 3. But J cannot be 2 because M, in the same row, is 2. So J must be 1 and E is 2.

Step 6

F is 1 because that is the only unused number in this row. K must be 4 because that is the only unused number in this row. We can verify these two numbers because 1+4 totals 5, which is the correct total for the FK block.

Step 7

A and B must be 2 and 3 (though we do not yet know which is which), because they are the only two unused numbers in their row. But A cannot be 2 because E, in the same column, is 2. So A must be 3 and B is 2.

Final Step   

N is 4 because it is the only unused number in that column. P is 3 because it is the only unused number in that column.

More Hints:

Here is another fairly simple example. You can see that this puzzle uses all four arithmetic operations. Remember, all the digits 1, 2, 3 and 4 must appear in every row and every column, and in this puzzle we are told that the number in square M is 4.

Often a good way to start is to look for a shape in which you can already deduce exactly which numbers go in one or more of the squares, as in the first step below.

First Step

The total of squares C, D and H is 4. The only way to make this total with three numbers, using addition, is 1+1+2, but the two number 1s must be in different rows and different columns. So C and H must be 1, and D is 2.

Second Step

R is 3 because that is now the only “unused” number in the rightmost column.

Third Step

Q must be 2 because QxR is 6 and we now know that R is 3.

Fourth Step

G and L must be 3 and 4 (but we don’t yet know which is which), because they are the only two unused numbers in their column. L cannot be 4 because M (in the same row as L) is 4, so L must be 3 and therefore G is 4. (Another reason is that G cannot be 3, because we could not then use division to create a total of 2 for the FG block. Yet another reason is that if L were 4 the total in the KL block, created by addition, could not be 4.)

Fifth Step

Since G is 4, F must be 2, because the total of the FG block is 2 and is created using division.

Sixth Step

K is 1 because L is 3 and K+L totals 4.

Seventh Step

E is 3 (the only unused number in its row) and J is 2 (the only unused number in its row). We can verify these because E+J is 5, which is the correct total for their block.

Eighth Step

A and B must be 3 and 4 (but we do not yet know which is which), because they are the only two unused numbers in this row – a reason based on logic. Another reason is that 3x4 is the only way in this size of KenKen puzzle that we can make a total of 12 with two numbers – a reason based on arithmetic. But E (in the same column as A) is 3 so A cannot be 3. Therefore A must be 4 and B is 3.

Final Step

N is 1 (the only unused number in its column) and P is 4 (the only unused number in its column). We can verify these from the total of 3, which is created by subtraction.

Using only the methods of reasoning you have learned so far should enable you to completely solve easy KenKen puzzles and to make progress in solving the more difficult ones.

A Very Useful Trick

As you progress to KenKen puzzles with larger grids, you will find that there are other tricks that help you make certain types of deduction. Here is the most common of the slightly more advanced tricks. If you learn this, you will find that a lot of puzzles larger than 4x4 grids will become easier for you to solve.

1 or 2

B

C

2 or 3

E

F

3 or 1

H

J

Take a look at this row from a 9x9 puzzle on which the solving process has already started and some possible numbers have been pencilled in. We can see that there are three squares that contain, in various combinations, the possibilities for the three numbers 1, 2 and 3.

Since we have three squares and three numbers for those squares, we can be certain that one of these numbers must be in one of the three squares, another of the three numbers must be in another of the three squares, and the third number must be in the remaining one of these three squares. It is therefore impossible for any of the other squares in this row (B, C, E, F, H or J) to contain any of the numbers 1, 2 or 3. If any of the numbers 1, 2 or 3 are already pencilled in to any of those squares (B, C, E, F, H or J) then we can delete them quite safely. And during the remainder of the solution process we can take care to avoid pencilling in 1, 2 or 3  as possibilities for any of the currently empty squares in this row.

This same trick works just as well for combinations of two numbers spread across exactly two squares in a row or column, and for combinations of four numbers spread across exactly four squares, etc. When you try solving larger puzzles you will soon realise how helpful this trick can be in eliminating certain possibilities.