KenKen Strategies

Using Kenken to Build Mathematical Thinking Skills

HAROLD B. REITERUNIVERSITY OF NORTH CAROLINA

CHARLOTTE

Circles on the Road, Houston, March 19, 2011

KenKen is a puzzle whose solution requires a combination of logic and simplearithmetic and combinatorial skills. The puzzles range in difficulty from very simpleto incredibly difficult. Students who get hooked on the puzzle will find themselvespracticing addition, subtraction, multiplication and division facts. This paper isdivided into three main sections. The first section is a sequence of one-step problems.We attempt here to include most of the common ideas used in solving KenKenpuzzles. The second section solves a 6× 6 KenKen step by step, and includes somemathematical notation that is needed only for problems that are much harder thanthese, but nevertheless, may lay a foundation for a formal approach to puzzles ofthis type. The final section consists of Prize Problems.

Websites of interest:

1. http://www.math.uncc.edu/~hbreiter/KenKen

2. http://www.geometer.org/mathcircles/kenken.pdf

3. http://www.kenken.com/

4. http://www.nytimes.com/ref/crosswords/kenken.html

5. http://www.stanford.edu/~tsnyder/kenken.htm

6. http://en.wikipedia.org/wiki/KenKen

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KenKen Strategies

1 Introduction

1.1 Puzzle 1

Consider the 6× 6 multiplicative Kenken r© puzzle below. Find the digits that go inthe three cells. Find all integers k for which the L-cage with clue k× has a uniquesolution.

150×

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KenKen Strategies

1.2 Puzzle 2

Consider the 6× 6 additive Kenken r© puzzle below. Find the digits that go in thethree cells. Find all integers k for which the L-cage with clue k+ has a uniquesolution.

17+

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KenKen Strategies

1.3 Puzzle 3

KenKentorics. Consider the 6× 6 additive KenKen r© cage below. What are thepossible values of k. For each such value, find the number of ways to fill the cells ofthe cage so that the clue is satisfied.

k+

4

KenKen Strategies

1.4 Puzzle 4

Consider the 6× 6 additive Kenken r© puzzle below. Find the digit that goes in thecell with the ?. Find all integers k for which the rectangular 5-cage with clue k+uniquely determines the other cell in that row.

20+

?

5

KenKen Strategies

1.5 Puzzle 5

Consider the 6× 6 Kenken r© puzzle below. Find the digit that goes in the cell withthe ?. Find all integers k for which the rectangular 5-cage with clue k+ uniquelydetermines the other cell in that row.

37+

?

6

KenKen Strategies

1.6 Puzzle 6

Using Faultlines and Parity. Consider the 6 × 6 KenKen r© puzzle below. Afaultline is a horizontal or vertical cage line that extends across the whole KenKengrid. Numbering the seven lines 0, 1, 2, 3, 4, 5, 6, horizontal line 2 is a faultline. Theparity of a cage is either even or odd, depending on the parity of the sum of theentries in the cage. Initially each cage can be categorized into one of three categories:odd, even, or undetermined (that is, it could be odd or even). In the problem below,the odd cages are 1−, 5−, 5+, 120× while the even cages are 15×, 18+, 2−, 3÷, 8+.The other three, 2÷, 2÷ and 12× are undetermined. Its worth spending a little timeto determine the parity of cages of each type {+,−,×,÷}. Of course the + and −cages are easy: k+ cages are even if k is even and odd if k is odd. Is the same truefor k− cages? Now for k×, we have to consider case by case. What about the 3-cellcage 12×. Could be either even or odd, right: {1, 2, 6} would be odd and {1, 3, 4}would be even. What about a three cell cage k×, like 5×, 9×, or 15×. They mustbe odd cages. Why?

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KenKen Strategies

1−

15×

18+

5+

12×

2÷

2−

2÷

2−

120×

5−

2−

3÷

10+

8+

8

KenKen Strategies

1.7 Puzzle 7

The X-wing strategy. Consider the 6× 6 additive Kenken r© puzzle below. Findthe digit that goes in the cell marked ?. The reasoning we use here is called theX − wing strategy: suppose a candidate can only be in two squares in a row (orcolumn), and the same candidate can only be in the same two squares of a differentrow (or column). Then that candidate is eliminated from any other positions in thecolumn (or row) containing either of the two squares.

3

5−

3+

40×

11+

16+

3−

2÷

1

15×

8×

6

1−

5

11+

7+

2÷

?

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KenKen Strategies

1.8 Puzzle 8

Consider the 4 × 4 KenKen r© puzzle below. Instead of using the digits 1, 2, 3 and4, use the letters A,B,C and D. Can you arrange it so that the cage k× is themultiset {A,A,A, B,B,C}? If so, can you find k such that the two-cage KenKenpuzzle with the single clue k× has a unique solution.

k×

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KenKen Strategies

A

B

C A B

A

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KenKen Strategies

1.9 Puzzle 9

Consider the 4× 4 additive Kenken r© puzzle below. Find the clue that goes in thecell k+ and solve the puzzle.

144× 96×

12×

k+

12

KenKen Strategies

1.10 Puzzle 10

Consider the 6×6 KenKen r© puzzle. Notice that there are two fault lines, and thatthe distinguished column has two [2÷] cages. Since the sum of each column is 21and each [2÷] cage contributes either 3, 6, or 9 to the column sum, the [1−] cagemust also have a sum that is a multiple of 3. Only 1, 2 and 4, 5 work. If the [1−]cage is 1, 2, then the sum of the two other cages is 18, which we can see is impossible.Solve this puzzle taking into account that the rightmost two columns have togethera sum of 42, and these two columns include the cages [10+], [11+], and two [2÷]cages. What does this say about the [1−] cage at the bottom of the region?

2− 72×

30×

5+3÷ 2÷2÷

2÷

3−

3−

5− 1−

1−

11+

10+

2÷

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KenKen Strategies

2 A Formal Approach

We’ll begin with an example

9+ 2÷4− 1− 18×

1−

4−

2÷

3−

12+

12× 3−

2−

3÷

15+

3−

3−

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KenKen Strategies

In order to identify the cages, we need to establish some notation. We’ll usealgebraic notation, with digits for columns and letters for rows. A fault line is aheavy line that goes entirely across a puzzle. Notice that the puzzle above has twofault lines, 45 and ef .

9+ 2÷4− 1− 18×

1−

4−

2÷

3−

12+

12× 3−

2−

3÷

15+

3−

3−

a

b

c

d

e

f

1 2 3 4 5 6

Let’s denote the sum of the entries of a cage k∗ by∑

k∗ . Here the ∗ refers to oneof the four binary operations on real numbers. Similarly, the product of the entriesof the cage will be denoted by Π k∗ . Thus

∑15+ = 15 and Π 18× = 18. We

can also reason that, for this puzzle Π 15+ is either 108 or 120 while∑

18× = 10,because the three entries must be 1, 3, and 6. We also need the notation for theset of entries of a cage, and for this we use the set builder notation {}. Thus{ 18× } = {1, 3, 6}. Now look at row e: Since 5 /∈ { 12× } and 5 /∈ { 2÷ }, it

follows that { 3− e5} = {2, 5}.

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KenKen Strategies

At interactive websites like the two KenKen sites mentioned above, we wouldsimply establish that 2 and 5 are candidates for the entries e5 and e6. This wouldlook like

9+ 2÷4− 1− 18×

1−

4−

2÷

3−

12+

12× 3−

2−

3÷

15+

3−

3−2, 5 2, 5

a

b

c

d

e

f

1 2 3 4 5 6

The next step in solving is to notice that each row and column sums to 1 + 2 +3 + 4 + 5 + 6 = 21, so the two columns 5 and 6 sum to 42. We capture this by

∑1− +

∑18× +

∑15+ +

∑3− +

∑3− = 42.

Replacing each entry that we know yields

∑1− + 10 + 15 + 7 +

∑3− = 42.

Now the cages 15+ and 18× both have 6’s, so 15+ can have only one 6, in which

case it must have a 5 as well. This implies that the cage 3− in row f cannot have a

5 or a 6. Thus { 3− } = {1, 4}. Now the 1 in the 18× cage forces the 1 in the 3−cage to be in cell f5. From this it follows that the 1− cage in column 5 satisfies∑

1− = 42− 32− 5 = 5. Taking all this into account, we have

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KenKen Strategies

9+ 2÷4− 1− 18×

1−

4−

2÷

3−

12+

12× 3−

2−

3÷

15+

3−

3−

4, 6

4, 6

2, 3

2, 3

1, 3, 6

1, 3, 6

1, 3, 6

a

b

c

d

e

f

1 2 3 4 5 6

41

5 2

5

For the next step we concentrate on the cage 2÷ in row e. There are just three

possible sets for a 2÷ cage in a 6× 6 Kenken: {1, 2}, {2, 4}, and {3, 6}. Since boththe first two require a 2, we conclude that our cage must be {3, 6}. This means thatthe set of 12× cage must be {1, 3, 4} with the 3 in the position d2. One possible

next step is to consider the cage 2− f3. Cages of the types 2− and 4− must

contain digits of the same parity. In a 6×6 KenKen, every cage 2− contains eithera 3 or a 4. Since the 4 can be ruled out and the 1 already appears in the row, itfollows that { 2− } = {3, 5}. Then in row f we have { 4− } = {2, 6}.

At this stage, let us summarize what we know:

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KenKen Strategies

9+ 2÷4− 1− 18×

1−

4−

2÷

3−

12+

12× 3−

2−

3÷

15+

3−

3−

2, 3

2, 3

1, 3, 6

1, 3, 6

1, 3, 6

1, 4 1, 4

3, 5 3, 5

3, 6 3, 6

2, 6 2, 6

1, 4 1, 4

a

b

c

d

e

f

1 2 3 4 5 6

6

4

41

5 2

53

Our final observation deals with the cage 3÷ . The two possible sets for it are{2, 6} and {1, 3}. We can rule out the later using what is called the X-wing strategy,which requires that once two instances of a digit have been assigned to a pair ofcolumns (or a pair of rows), that digit cannot occur in any other cells of that pairof columns. Since the bottom two cages of columns 3 and 4 both have 3’s, the setof the 3÷ cage must be {2, 6}.

At this stage there are several ways to complete the puzzle. The complete solu-tion is given below.

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KenKen Strategies

9+ 2÷4− 1− 18×

1−

4−

2÷

3−

12+

12× 3−

2−

3÷

15+

3−

3−

6

4

5 1 2 4 3 6

4 5 1 6 2 3

3 6 5 2 4 1

2 3 4 1 6 5

1 4 6 3 5 2

415326

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KenKen Strategies

3 Prize Problems

3.1 Backwards KenKen

In this problem we start with the ‘solution’, and we try to find the cages that willproduce a unique result. Here’s the distribution of digits.

3

2

1

4

1

4

3

2

2

1

4

3

4

3

2

1

Now here’s the catch. The only clues you can use are 12× and 12+ with theexception that there can be a single one-cell cage. The puzzle must have the distri-bution above as the only solution.

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KenKen Strategies

3.2 Extremal KenKen

Here’s a very mathematical puzzle, one that requires some ideas from elementarynumber theory. The puzzle is 5× 5.

86400× 38+

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KenKen Strategies

3.3 Turbo KenKen

In standard KenKen, you’re given an n × n grid of squares with cages shaded byheavy lines. Each cage has a mathematical clue, and the objective is to put thedigits from 1 to n in each row and column so that each row and each column getseach of the n digits, and so that the clues are all satisfied. In TurboKenKen, thesolver must also select the n digits that can fill the grid. Only one n-digit subsetof S = {0, 1, 2, 3, 4, 5, 6, 7, 8, 9} can be made to satisfy all the clues. We’ll agree atthe outset that 0 can appear only in × and + cages. In the 5 × 5 TurboKenKenbelow, we’re looking for five digits selected from S that can be used to satisfy therequirements.

12× 13+ 8+

1 1−

4− 12×

24×

12×

1−

7+

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KenKen Strategies

Our second Turbo KenKen is a harder challenge.

12× 2− 6×

16+ 3−

3−

24×

2−

1− 2÷

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KenKen Strategies

3.4 Maximal KenKen

Once again we given the distribution of digits in a 5 × 5 array. This time we wantto build the cages so the product of the sizes of the cages is as large as possible. Ofcourse we still insist on the cages and clues to be consistent with just one distributionof the digits. As an example of what we want here, define the value of a puzzle asthe product of the sizes of the cages. Then the puzzle

55+ 22+

has the value v = 15 · 10 = 150 because it has one cage with 15 cells and anotherwith 10 cells. Your prize problem is to partition the puzzle below into cages thatare just the right size so that when we multiply these cage sizes, we get the biggestvalue possible.

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KenKen Strategies

4

5

3

1

2

5

2

4

3

1

1

4

5

2

3

2

3

1

5

4

3

1

2

4

5

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KenKen Strategies

3.5 Using another arithmetic

Consider the 6 × 6 KenKen r© puzzle. Thanks to John Thornton for this beautifulproblem. We call this one Cyclic KenKen.

3−

2÷ 2÷14+

2÷

2÷

2÷ 2÷19+

2−

1− 1−

13+

12×

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KenKen Strategies

3.6 Even Harder

3−

2÷ 2÷14+

2÷

2÷

2÷ 2÷12+

60×

1−

18+

13+

6

27