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For your daily dose of new KenKen® puzzles for all levels, visit: www.kenken.com/play_console.htm where 6 new puzzles are added daily. KenKen is a trademark of Nextoy, LLC.
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The Jan 2009 issue has detailed instructions about how to solve beginning KenKen®. Send me an e-mail by clicking the “contact Math Mojo” box (above and to the right in the right sidbar) and ask for a copy of the January 2009 Math Mojo Monthly. I’ll send you a copy as a PDF. There is a sample Kenken®, complete with strategy for solving it.
Here’s a hint for good Kenken® strategy – get great at factoring! You can learn how to factor at the “Pretty Good Guide” to Prime Factorization
How to solve a basic 4×4 KenKen
Here’s video (it’s about ten minutes long) that will explain not only the basic rules of KenKen (which are amazingly simple) and the basic tactics and strategy you will need to solve most basic 4×4 KenKen.
Please leave a comment after you have viewed it, and let me know if it helped, or how I can improve it.
How to solve a difficult 9×9 KenKen with no operation signs
“Abandon hope, all who enter here…” (“Not!”)
Below is a series of videos explaining how to do a truly sadistic 9 x 9 KenKen, which has no operations signs. In other words, there are no overt hints as to whether you should add, subtract, multiply or divide in order to find the digits that belong in the squares. Yowza! But don’t abandon hope! You can tackle it! We can try it together. You will understand much more than you may imagine.
I haven’t solved this one yet myself, so you can see the frustrating moments as well as the “thrill of victory” (I hope) when I get one right.
Each video will take us a step or two towards the solution. We’ll take it slow, because I’m not sure if each step is correct, until the puzzle is entirely solved, of course. Please feel free to add your comments after a vid (and remember to tell us which video you are referring to – they will be numbered) and your suggestions for how you think we might proceed. Don’t take us further than one step, though. We don’t want to get ahead of ourselves.
This lesson simply explains how KenKen works, and you might want to represent the numbers in the squares. There are many ways – this is just the one I prefer. It’s the same symbolism that many people use to solve Sudoku puzzles.
This is my first such video, so please be patient with my handwriting (I’m learning how to use a graphics tablet and stylus), and my technology in general. These lessons will become more streamlined soon, and you’ll also become more accustomed to learning this way.
Very soon, this page will be replaced by a blog, and each day we’ll have another step up, with a place for you to add your comments and suggestions on each step.
Let’s see how this thing goes.
Super-important:
The videos below are for very advanced KenKen®. If you are just beginning, take a look at them to show you you can learn to get to an advanced level with a some video lessons. But don’t get discouraged if you don’t understand what is going on! Remember, this is the advanced level. The beginning level is well within your reach with a few minutes of instruction. Leave a comment and nudge me to get the beginning level videos up soon,
By the way, this Kenken® was graciously provided by the people at KenKen.com, your source for the true, original KenKen®.
Something to keep in mind:
When I mention cells, that means the little individual squares of a puzzle. A 9×9 puzzle grid has 81 cells.
Cages are groups of cells that are outlined with darker borders. They have a number in the corner of one of their cells. That is the number that the cells must come to when the operation is performed on the digits in their cells. In the following puzzle, the first cage in the upper left hand corner is a four-cell cage which must amount to the number 630.
Something else:
Kenken is a trademarked term, and official rule from the Will Shortz books is:
“Cages with more than two squares will always involve addition or multiplication. Subtraction and division occur only in cases with two cells.”
Click on the puzzle below to download a printable .pdf of the puzzle.
Here’s the video for Step 1 of the KenKen Puzzle:
(If you have trouble seeing the videos at this size, click on the bottom right corner of the YouTube video, where the rectangle is, and that should open the video in full-screen view.)
It’s important to realize that we didn’t have to start where I started in this puzzle. There are lots of places to start. But we had to start somewhere. Even though some of the other places looked tempting, we can remember that they will still be there when we get to them. And maybe by that time we will have eliminated some obstacles that will lead to their solutions.
Patience is a big part of KenKen. If you don’t have it, you’re sure to learn it if you stick with KenKen .
Here’s Step 2 of the same puzzle:
In step 2, we start to tackle both 25 cages. I mention that we might go for the 17 or 13 cage next time.
I mention that “the larger the number you have, the less possibilities you have for factors, and the more you can narrow it down.”
That isn’t strictly true if you define “factors” in the typical mathematical sense. What I actually meant was that if you have a large number, say, 48 for a two-cell cage, you don’t have to take numbers that add up to 48 into consideration, because no two digits can add that high. In this case, you’d know that the only possibilities are 6 and 8, which multiply to 48.
On the other hand, if you had 15 for a two-cell cage, you’d have to take the factors 3 and 5, as well as the all the combinations of digits that add up to 15 into consideration.
This same principle applies to cages of more than one cell.
On to step 3:
Step 3 covers:
Getting the answer to the two-cell 17 cage very easily. Using that to eliminate some numbers in the 2 cage to the right of the 17 cage.
Approaching the 13 cage similarly to how we did the 17 cage. We can narrow it down a lot.
I suggested that we look at the 315 cage for next time, but I think we’ll try something much simpler first.
Step 4:
Warning: While editing the last video, somewhere I must have erased a 7 from the bottom, left cell of the puzzle (the left one of the 13 cage). I didn’t notice that until I got to step 14. It doesn’t effect what we are going to do, but I just wanted to let you know it was there, and I noticed it, in case it confuses you. So just imagine there are all the digits from 4 through 9 in that cell. Sorry, my bad.
Tackling the 48 cage helps us solve the 25 cage to the right of it. And that helps us eliminate some digits from the 2 cage in the top right.
Step 5:
In Step 5 we do a little factoring. We tackle the 630 cage. When we do that, we also notice that it gives us the factors for the 315 cage, because 315 is a factor of 630.
(Remember – there should be a 7 in the bottom, left cell of the puzzle (the left one of the 13 cage).
Note: YouTube viewer “TurkishGamer” noticed that you could eliminate all 9s from the top row outside the 17 cage. I didn’t catch that till lesson 7.
Step 6:
See that nasty looking 189 on the right side of the puzzle? Don’t jump to conclusions – it’s an easy one. Let’s do it now.
(Remember – there should be a 7 in the bottom, left cell of the puzzle (the left one of the 13 cage).
Step 7:
In this step I clear up some sloppiness I have overlooked in some of the past steps. (Well, I guess that clears up any doubt you might have had about if I was an “expert” or not!) Doing that, and doing an ecology check on the 630 cage gives us the final digits (but not their order) for that cage. And that helps us eliminate some digits in the 2 cage in that row.
An ecology check of the 315 cage gives us some good results which we can use to further narrow down some of the other cages in one of its rows.
(Remember – there should be a 7 in the bottom, left cell of the puzzle (the left one of the 13 cage).
Step 8:
We’ll learn that we can eliminate almost all of the digits for the remaining empty cell in the top row.
Then we attack the 5 cages along the right side. We’re not done with those yet. In the next vid we’ll find out something interesting about them.
And I’ll work on my speaking skills for next time (okay?)
(Remember – there should be a 7 in the bottom, left cell of the puzzle (the left one of the 13 cage).
Step 9:
In this step we work a bit on the 15 cage in the bottom right corner. We don’t make that much progress, although we are setting things up for some progress soon. It’s like a battle. Not every step of every battle must be an overwhelming victory – sometimes you’re just setting things up for the final push.
(Remember – there should be a 7 in the bottom, left cell of the puzzle (the left one of the 13 cage).
Step 10:
In this KenKen lesson, we continue narrowing down the 15 cage. Then, as we narrow down some more possibilities in the two 5 cages, we manage to solve the 2 cage in the right upper corner. So some of our preparations are finally paying off!
(Remember – there should be a 7 in the bottom, left cell of the puzzle (the left one of the 13 cage).
Whoops!
On the youtube version of this video, Mattachoo83 commented:
When you erase the 4 from Row 6 Col 9 and the 4 from R2C9, you say you are allowed to do this because there is no 1 in the cage to combine with the 4. However, you can’t erase the 4 because it doesn’t have to be addition, it can be subtraction, and if you use 9 – 4, you will satisfy having a 4 there. So you can’t eliminate the 4 just yet.
And he is absolutely correct. So I made a bunch of videos (Steps 10 a, b and c).
Step 10a:
There are two horizontal 5 cages along the right side of the puzzle. In this lesson, I mistakenly say that you can erase the 4s in the right hand cells of each of those cages, because there is nothing to combine it with that will make a 5 (once I’ve gotten rid of the 1s in those cages). That is wrong. An astute reader noticed that you could still have subtracted the 4s from the 9s in the other cells. So I shouldn’t have erased them.
After having erroneously erased the 4s, I prematurely concluded that the only place a 4 could appear in the right hand column was in the 2 cage at the top.
It turns out that it is correct anyway, and it doesn’t effect the outcome of the puzzle. I know this because since I’ve started making these videos I’ve solved the puzzle on paper.
But that is not important. What is important is that my reasoning was faulty. I don’t want to leave faulty reasoning in these lessons. It’s one thing to be sloppy, but it’s another thing not to correct it once you know better, and I don’t want to be guilty of that kind of sloppiness. I would be a disservice to you – and that’s not what Math Mojo is about. It’s about meaningfulness – so I’ve backtracked a bit to find a way to get the same results, but with steps that don’t rely on luck.
So I made a bunch of videos (Steps 10 a, b and c) to prove that the 4 must appear in the right cell of the 2 cage at the top right of this Kenken puzzle, in order for us then to be able to continue on with Step 11 to the end, without having to second-guess if we are on the right track.
In the video for Step 10 a:
Jump forward and watch the videos for steps 19 and 20. Neither of those steps assume anything that we haven’t already covered, so if you follow their logic, you can use their conclusions as if they were the first part of this step.
Lesson 19 eliminates the 1 + 2 + 1 possibility for the 4 cage. You can jump to it and watch it now. Nothing that happens between this step and that step will interfere with how it works.
Lesson 20 will explain why the 2 x 1 x 2 combination won’t work. Like lesson 19, it does not change anything that we’ve done in between.
That proves that there must be a 4 in the corner of the cage, because the only remaining possibility is the 1 x 4 x 1 combination. And that would preclude a 4 from being anywhere else in the second row of the puzzle.
So fill in that combo, and do an eco check for rows and columns.
Now we have to deal with the other 5 cage on the right ( row 6, columns 8,9). Not a huge problem, just long. The quick way to deal with it is to simple go down the rabbit hole. Let’s try putting using a 4 in the right cell of that cage. That puts a 9 in the other cell in that cage.
Step 10b:
n step 10a we put a 4 in the right cell of that cage. That put a 9 in the other cell in that cage.
That would make the top cage, right cell a 2, and the one next to it a 4.
Checking the far right column, you must cross out the 2s in the 15 cell, leaving 6 and 1 mandatory for it, so you can cross out the 1s in the rest of the column.
That leaves you with only an 8 in the upper 5 cage, and a 3 in that cage’s other cell.
When you check that row, notice that the 24 cage on the far left only had the possible combinations of 3 x 8 and 4 x 6, and we can now narrow it down to only 4 x 6. But as we know from having jumped ahead before to the rabbit holes of lessons 19 and 20, there is a 4 in the corner of the 4 cage, so the 4 is out. That would leave the only possibility for the top of the 24 cage as a 6.
That has us using 1,3,4,6,and 8 in the row. The only digits left are 2,5,7 and 9 for the empty cells.
7 doesn’t factor into the 36 cage, and 7 is out for the 1 cage, because it’s already being used in that column. That means the 7 would have to go in the top of the 29 cage.
2, 5 and 9 are left. Since 5 doesn’t factor into 36, the only place it can go is the top of the 5 cage. Leaving 1 and
Step 10c:
(note: in the video I misspoke and called it lesson 10d. It is 10c.)
In step 10b we filled in the entire second row.
See lesson 12 to see how we prove that there can be no 2 in the right wing of the 192 cage. Since there is already a 2 in the second row, no 2 can to there, either, which leaves the only remaining possibility for a 2 to be in the bottom cell in the vertical 5 cage in column 8.
A 7 is the only possibility to be in the top of that 5 cage, because 3 already appears in that column.
Now when you do your eco check of that row, you can get rid of the 7 in the top cell of the 189 cage, leaving 3 and 9 as the only two contenders for that cell.
Look at the 29 cage. If our rabbit hole is true, and there is a 7 on top, there is no way you can have a 3 in the cage, because a 7 + 3 would give you 10, requiring the remaining two cells in the cage to total 19, which is not possible.
The vertical 5 cage in the 7th column cannot possibly contain a 3.
Therefore, the only remaining possibility to place a 3 in that row is in the top cell of the 189 column. Put it in there and do a column eco check.
Also, as you do a column eco check for column 6, you’ll find that the only digit that can go into the bottom cell is an 8.
Doing a row check for the bottom row, you can cross out the 8s in the 13 cage.
Here we finally reach our conclusion of this rabbit hole, because that leaves 5 as the only digit that can go in the left cell of that 13 cage. That would make 8 the only possibility for the right cell in that cage. That is clearly impossible, because an 8 already appears in that row.
Whew! We can now go back and erase all the light pencil marks we made since we began in step 1a, because we have proven that it is not possible for a 4 to appear in the right cells of either of the horizontal 5 cages along the right side of the puzzle.
That means the only place a 4 can appear in column 9 is at the top, as I prematurely summized in lesson 10. Now we have proven it, though, and we can continue along to step 11.
Some of the moves in future videos will be easier now, as we have already seen them in lessons 10 a, b and c.
Step 11:
Ah, the eleventh step! I like number eleven, not simply because it is the first multiple-digit prime number, but for the fact that it is the first repunit, as well. Why not become the first kid on your block to know what a repunit is, by checking out http://www.squidoo.com/multiplication? There you can also learn how to multiply by repunits in your head, quicker than most people can even get their hands on a calculator. But I digress…”
In this KenKen lesson, we’ll clean up a loose end or two.
(Remember – there should be a 7 in the bottom, left cell of the puzzle (the left one of the 13 cage).
Step 12:
In the vertical 5 cage next to the 189 cage we narrow the possibilities down to two digits, but we can’t get which goes where yet.
In the 5 cage that is above the one we just worked on, we get a clue to narrow it down. That clue comes when we work on the 4 cage to the left of it, and in the same row.
So we do get something solid (the 5 cage) out of this lesson, and also narrow down something else, for future use (the 4 cage).
After this, we enter the “grunt-work” stage, in which we narrow down many cages, without immediate final results for any of them. When you do a large, difficult KenKen, this stage can last awhile.
Remember, though, we are setting up our position for major gains in the future. KenKen is a great lesson in patience and perseverance. It’s one of the reasons to learn KenKen. In this sense, KenKen is like math in general – when students ask, “Why do we have to learn math?” the answer is, “You don’t, of course.” But one of the reasons one should learn math is to learn the great life-lessons in a safe environment.
What I mean is, if you mess up a KenKen puzzle, or an equation while you’re learning math, big deal! (Of course schools do their best to ruin it and make it a big deal by making impersonal, high-stakes testing – but that is school, not math itself that is the problem.)
So this way, you can learn the value of patience and perseverance in the comfort of your own home, or wherever. Then, when you need those traits in school, at work, on the streets, in battle, or wherever, you have trained yourself to be the kind of person who knows how to keep on keeping on.
Hotcha!
Step 13:
Brian waxes philosophical about KenKen and strategy. (Big deal.)
We attack both of the 24 cages. At some point I mention that “this doesn’t help eliminate anything else…” , which is not necessarily the case, as we’ll see in one of the next steps. Can you see what I’m getting at?
We also start dealing with the 2 cage along the left side of the puzzle. I’ve noticed something that may be interesting in it. Can you notice it? (Or am I making it up?)
We end by taking a look at the 13 cage at the bottom left.
Step 14:
Brian finally gets around to filling in the 7 that’s been missing in the bottom cell since about lesson 4.
We find an interesting way to eliminate a digit from one of the cages in the furthest-left row.
Step 15:
We get a sneaky hint from a group of 4 adjacent cells, all of which are in the same row but don’t all share the same cage.
Step 16:
Discussion about how to avoid being overwhelmed or out of our leagues.
We talk about “grunt work” and “rabbit holes.”
We go down a shallow rabbit hole and get a bit of results. A lot of work for a small reward, but as is often the case, this small reward will bring us big dividends later.
“It ain’t what it don’t know that will hurt you, it’s what you know for sure, that ain’t so. ” – Mark Twain
Step 17:
Grunt work. We fill in the 36 cage and the 3 cage.
Step 18:
More grunt work. Heavy lifting. Can you hear us snort as we tackle the 29 cage?
Step 19:
More grunt work, but it starts paying off quicker than I imagined.
At this point, I’d like to mention that I have gone out of my way not to make this look artificially easy.
What I mean is, although I have already solved this Kenken, and I’ve done it over and over as I make this tutorial, and have found “streamlined” ways to solve it that I didn’t find originally, I’m refraining from making it look like I am more clever than I am by showing the streamlined ways.
I probably will do that in some other videos. If I do, I will tell you that I’m doing that, so as not have you think I’m better at this than I actually am.
Why am I doing it like this? Because I think that 99% of the internet is about “showing off” and making the authors sound like the ultimate experts. That is public relations, not teaching.
I don’t believe in “show-and-tell” teaching. It belittles the learner. It make you feel inadequate to the “great one” on the web. The truth is, the experts make mistakes, go down rabbit-holes and get frustrated themselves. I prefer to see how this is solved “warts and all.” It keeps it human, and keeps it from being intimidating. It keeps it “real.”
The main purpose of education should be encouragement. Who cares if you learn a bunch of facts and techniques that someone else shows you if it doesn’t lead you to exploring more on your own?
It’s much cooler to see someone do something then strike out on your own and pursue your own fun and techniques, with their experience as a “jumping-off” point.
What do you think? Leave a comment at the bottom of this post.
Let’s get to work:
Fill in the apparent possibilities for the 1 cage (the vertical one that appears in the second and third rows. Keep in mind that the top cell cannot have a 1 in it, because a 1 must a appear in that row, in the 4 cage. Also, no 7s can appear in the 1 cage, because there is already a 7 in that column.
Now the second row from the top has preliminary numbers in all of its cells. At this point I like to do a full row check. Just scan the row to see if you missed anything, or if you see any new possibilities.
Notice the lonely 4 in the 4 cage. Somehow I liked it. I imagined it might work. So I tried other rabbit hole first, in the hopes that I’d find out quickly that it was wrong, and then could pursue that 4 with impunity.
There are two possible combinations in the 4 cell the don’t involve the 4, but they are very similar. You could have:
- 1 + 2 + 1 with the 2 in the corner
or you could have
- 2 x 1 x 2 with the 1 in the corner.
The only effective difference is that one would leave you with a 1 in the lower row, and the other would leave you with a 2 in the lower row.
Let’s go for the first one first, with the 2 in the corner, and the 1s in the “wings.”
That immediately gives us a 3-8 combination in the 5 cell on the other end of the row.
Doing an ecology check for that row, (keeping in mind that the other combination which we could have used would give us the same results for an ecology check for this row) we could eliminate a 3 from the 24 cell, as well as the 2 and 3 from the 36 cell and the top of the 1 cell.
Notice that now in the 36 cell, the only combinations in that row that could give us 36 would be 4 x9 or 6 x 6. Clearly 6 x 6 is impossible (can’t have two sixes in the same row), so the 4-9 combination is the only one possible, leaving us a 1 in the bottom of the 36 cell.
But wait! Did you catch it? There is a 1 in the bottom of the 4 cell in the same row. Great! We were lucky. This was a very shallow rabbit hole – only about arm-deep.
Step 20:
We still can’t be sure that the 1 x 4 x 1 belongs in the 4 cage yet, because we have the 1 + 2 + 1 combination to check. That will be a bit of a deeper rabbit-hole, I’m afraid.
Let’s the only thing we have to change is the order of the digits in the 4 cage. Now the two is on the bottom, and we can keep going.
As I’m doing this I’ve just noticed something that could have proved that any combination that had a 2 in the 4 cage wouldn’t work. It would have saved the last step, but as I said, I want to keep this real, so I’ll leave that step in. Besides, it wasn’t a bad lesson.
OK, here’s what I noticed:
If you put a 2 in the 4 cage, you must have the 8-3 combination in the 5 cage at the other end of the row, as we had already noticed.That means the only place for a 7 to appear in that row would be at the top of the 29 cage.
We also know that we cant have a 1, 2, 3, 4, 6, 7, 8 or 9 in the top of the 1s column. so it has to be a 5. In the bottom half of that row the only thing left that would combine with the 5 to make 1 is 4 or 6, but the 4 is taken, so it must be a 6.
Let’s take time to do an ecology check for the row. Get rid of all the 4’s besides the one in the 24 cell. Get rid of all the 6s besides the one in the 1 cell.
Look at the 5 cell – we know that the 4 has been taken, so it must be a 9. Get rid of all the other 9s in the row.
At this point we can notice that the only place a 5 could appear is in the bottom of the 29 row. This is excellent intelligence, Captain! It means that we already know that a 7 nd a 5 appear in that cage, giving us 12. That means that 17 is the missing sum of the remaining cells.
Is there any combination of remaining digits in that cage that could give us 17? No, so we have reached the end of this fairly deep rabbit hole, and you win the prize for patience.
What’s the prize? You now know that the 1 x 4 x 1 combo is the only one that can work in the 4 cage, where we started.
Want to fill that in and work on your own from there until the next lesson? Print out the pdf file for this lesson and give it a try.
See you in lesson 21 to check how you did!
Note – As I was doing the video, using these notes as a guide, I found a slightly different way to prove the last part. So the end of the video doesn’t exactly match these notes. Both prove the same outcome, though.
This just illustrates that there are many ways to come to the same conclusion. And the having two ways is good – it is like checking your work.
Step 21:
We ended the last lesson proving the 1 x 4 x 1 combination in the 4 cell. We still need to do an ecology check for it’s columns and rows, though.
After we do that, we have to check all the cages that we changed for their own eco check.
Then notice the vertical 1 cage in the seventh column. Because 7 does not appear in it, the only way you can use an 8 or 9 in there is if you use both. But if you use both, you leave no possibilities at all in the cell of the 17 cage above it. So the 8 and 9 are out in the 1 cage.
That leaves the only possibility for a 9 in the third row to be in the top of the 29 cage.
When you do the column check for that 9, you’ll have to get rid of the 9 in the 25 cage, leaving the 5 x 1 x 5 combination as the only solution for that cage.
Now do the row and column check for the 5 and 1.
Step 22:
We ended the last lesson having done an eco check for the 25 cage.
In this lesson we are going to make a lot of progress, fast, as soon as I get some trivial digressions out of the way.
We notice that a 6 must appear in the center row of the 36 cage, which let’s us get rid of all the other 6s in that row. We also deduce that there can be no 6 in the lowest row of that cage.
Then we notice that either a 2 or a 3 must appear in the 5 cage in that row. That means that the combination for the center row of the 36 cage must contain the digit of those two that does not appear in teh 5 cage. In other words, the 2 and the 3 will be split between the 36 and the 5 cages, so neither of them can appear elsewhere in that row.
Therefore we know that the top cell of the 24 cage near the top left of the puzzle must be an 8, and the digit below it must be a 3.
When we do a row check, we see that the combination for the 5 cage must be the 7,2 combo.
Eco-checking that row gives us the 3,6 combo in the center row of the 36 cage, and a 2 in the lower cell of that cage.
Note: At this point I’m sort of “on a roll.” That is good and bad. I’ve noticed a tendency to get giddy, and therefore sloppy, when answers are coming too fast. I’ve got to force myself to sort of “sober up.” Anyone else notice that about themselves? Leave a comment, please.
After filling in the 2 in the 36 cage, a column check gives you a 4 in the right cell of the 2 cage in the seventh row, and a 2 in the cell next to it.
The follow-up is to get rid of some digits in the 24 cage near the middle of the puzzle.
Eco-checking column 9 gets rid of some 2s, which proves there must be a 1 in that column in the 15 cage, so we can dump the 1 in the 5 cage the sixth row. That makes that cell and 8 and the one to the left of it a 5.
The only place a 6 can appear in the eighth column is in the right wing of the 192 cage. Then get rid of the other 6s in that row, which makes a 3,8 combination in the the 24 cage of that row.
Now checking the bottom row, we get rid of the 8s in the 13 cage, which means we also must get rid of the 5s.
Time to get some sleep.
Step 23:
23 is a good number. It’s the first prime number with two consecutive digits in it. There are two other two-digit prime numbers with two consecutive digits in them. I name one more in the video. Can you name the other? Leave it in a comment.
Eco-check of the fifth row, we can get rid of a six in the vertical 3 cage.
Since 3 and 8 must appear in the 24 cage of that row, there must be a 4 in the corner of the 192 cage, leaving an 8 for the top of that cage.
After eliminating the 4s and 8s from that column, we know that the right cell of the 17 cage at the top must be a 9, and the cell on the left of it must be an 8.
Checking the sixth column, we see that there can only be a 3 in the right cell of the 24 cage near the middle of the puzzle, so there can only be an 8 in the left cell.
Eliminate extraneous 8s in the fifth column, and 3s in the sixth column. That puts a 6 in the right cell of the middle row of the 36 cage, and a 3 in the left cell.
An eco-check of that row (the second row) leaves only one empty cell, wich is the top cell of the 1 cage. It can only be the remaining unused digit, which is 5. Of course, that means that only a 6 could remain in the bottom cell.
Time for an eco-check of the third row. Get rid of extraneous 3s and 6s.
Since we got rid of the 3 in the top of the 189 cell, the only place for a 3 in that cage would be in the center cell, so fill it in.
Look at the third row again. The only place for a 5 is in the right cell of the 29 cage.
Picking a random cage to fill in, I chose the vertical 5 cage in the third column. It’s getting to be time to start filling in some of the empty cages with the information that we have so far.
Step 24:
… in which we begin by alienating the less discriminating members of our readership.
This lesson is like shooting fish in a barrel.
First we notice that the only place an 8 can appear in the third row is in one of the ambiguous cells in the 29 cage. By process of elimination, we can tell that 7 will be the digit in the remaining ambiguous cell. We just don’t know which of those two digits goes in which of those two cells yet.
We can now eliminate 7s from the rest of that row, giving us a 9 in the top of the 189 cage, and a 4 in the tip of the 5 cage to the left of it. So we can also put a 9 in the 5 cage below the 4.
There is only one possibility left fort the bottom of the 189 cage, so let’s put the 7 in there. Checking that row we eliminate the other seven from the top cell of the 5 cage.
Notice that there is a 2 and a 9 in the second and third cells of the five row. That means we can eliminate the 2 from the first cell of that row, leaving just a 1 in the bottom of the 2 cage in the middle of the left side of the puzzle. That also means that there must be a 2 in the cell above it.
You can eliminate the 2 in the cell that is to the immediate right of the 2 you just filled in.
Eco-checking the left colum, you can eliminate the 8, 1 and 3 in the left wing of the 18 cage.
Eco-checking the 18 cage, we can eliminate everything but an 8 or 9 in the corner cell.
Checking the fourth row, I noticed a 9 could be eliminated from the top cell of the vertical 3 cage.
Check the second column. Notice that you can eliminate the 6 in the top cell. Also notice that 7,5 combo now appears in two cells of that row, so they cannot appear elsewhere. So we can erase the 7 from the bottom cell.
(I just noticed that I could have filled in a 2 in the bottom of the vertical 3 cage in the second column, as the only possibility in the column. I don’t know how I overlooked it in the video. But never mind, the procedure in the video works just as well.)
At this point the 4 and 7 were easy to fill in to the horizontal 3 cage in the fourth row.
Filling in the 6 at the top of the vertical 3 cell in the second column makes us realize that there can’t be a 6 in the bottom cell of that row, which leaves us a 9 in that cell. Obviously the only digit that will combine with 9 for that 13 cell is 4, which we can now put in its rightful place in the bottom-left cell of the puzzle.
Step 25:
…. in which we once again show our age by our nostalgic musical references.
Something I should have mentioned in the previous video (I mentioned it in the previous blog post, but not the video) – In the second column, there’s only one place for a 2, and that’s in the bottom of the vertical 3 column.
That mean’s that the cell to the right of it (top of the 5 cage) can only be a 9, and the cell below it is a 4.
Checking the first column, we can eliminate the 3 from the furthest left cell of the 630 cage.
Checking the second column, we see that there is already a 9 in it, so we can eliminate the 9 in the corner of the 18 cage, which leaves an 8, which only can combine with a 7 in the left wing cell of that cage.
The 7 in the left wing of that cage let’s us erase the 7 in the corner cell of the 315 cage, giving us a 5 for that cell. Now the only digit left in the row is 6, which can only go in the furthest left cell of the 630 cage
The 5 that we just put in the corner of the 315 cage gives us a 7 in the right wing of the same cage. By putting in that 7 we leave a 5 as the only possibility for the cell at the top of that row (in the 630 cage).
Look in the top left corner of the puzzle. We can erase the 7 in that cell because 7 already appears in that column. Same goes for the 5, so a 6 must go in that cell.
Check that row and erase the extraneous 6s and 5s, leaving only the 7,3 combination as a possibility remaining in the 630 cage.
At this point I did a random check of all the rows and columns to make sure I didn’t overlook any duplicates. So far, so good.
So I filled in the possible remaining digits in the empty cells in the sixth row, and then eliminated what I could by checking what was already used in rows and columns. Then I checked the possible combinations that would work.
That left me in the sixth row wit with a 2 in the left of the 3 cage, and a 6 on the right. And a 2 on the left of the 2 cage and a 4 on the right.
Step 26:
…. in which we wrap this puppy up. This is the final step of this KenKen puzzle. You must be quite patient to have come this far. It’s been a pleasure for me, and I hope the same for you. If you’d like to see more of these, and have even more learning materials, leave a comment. I’ll be starting a membership blog for KenKen soon, where we can help and support each others’ progress. Stay tuned…
The final step!
In the seventh column only a 2 and 3 are missing, but we don’t know which goes where yet.
In the sixth column, the 5 and 1 are missing, and there’s already 5 in the eighth row, so the 1 must go there, and the 5 goes in the bottom row.
We just put a 1 in the 8 cage, so there must be a 9 (to subtract from) or a 7 (to add to) in the other cell. There’s already a 7 in that row, though, so the only remaining contender is the 9.
Filling in the 15 cell at the bottom of the puzzle is a cinch. 7 is the only contender for the middle cell (because it’s the only digit missing in that column). The 7 and 5 that are now in this cage add up to 12, so the missing digit must be 3.
When we fill in the 3 in the 15 cage, we can eliminate the other 3 in that row (over in the 1 cage to the right). That makes the bottom cell of that cage a 2, and the top cell a 3.
Now, the only digits missing in the bottom row are 6 and 1. There is already a 1 in the third column, so the digit that must go in the cell in the bottom row of the column must be a 6, which makes the cell in the furthest right cell in that row (the corner 15 cage) take a 1.
Now the only digit missing from the ninth column is a 6, so let’s fill it in.
Eco-checking the fourth column, we see that we can eliminate the 3 as a contender for the top cell, making that cell take a 7. That makes the cell to the immediate right of it a 3.
Checking that fourth row again (eco-checking again because solved the top cell), we see that we can eliminate the 7 of that row in the 29 cage, making it an 8. That makes the cell to it’s immediate left an 8.
Eco-checking that row once again, we notice that the seventh row can’t take an 8, so it must take the remaining 6. Of course that means that the digit to its immediate left must be a 6.
Savor the moment! It’s the final cage!
In the fourth row, only a 4 is missing, so fill it in.
That shows us that the 12 cage is an addition problem, and the missing digit must be 2. Checking the row and column, we see that this, indeed is the case.
Time to check every row and column to see that our efforts were not wasted by having double digits in any of them.
It checks!
You’ve been so patient. I hope you had fun while playing along.
In the future, I’ll have downloadable pdfs of each step, with the progress of the puzzle so far, so you can fill in the steps for yourself before you watch the video, and see how you’d attack it. If you can’t find a way, then watch the video. And if your way is better (or different) than mine, help us all out by leaving a comment.
Hotcha!
Brian
Tags:
Please send me a copy of the January Math Mojo Monthly. Thank you.
Professor Homunculus sez:
Done.
Please send me the KeKen strategy newsletter
Thanks
Professor Homunculus sez:
Done. My pleasure.
Great video on the simple KenKen! I’ve been frustrated by the 6×6 puzzles my local paper started this week, but this has given me new hope. I think I’ll work on the 4×4’s until I build up some confidence. Thanks!
This is pretty tough. I’ll try too solve a kenken puzzle one of these days.
Thanks for dropppin by.
Just completed this video series- Excellent. My game has certainly improved, thanks. Looking forward to your membership blog and future videos.
I’m fairly new to KenKen, so please forgive if I miss something obvious. I just viewed Step 5 of your 9×9. Why did you not list 1 as a possible factor for you 630 cage? Also, when you completed your 315 cage, you said that you could not erase any factors at this time. I thought you could already erase the 5 and the 7 as possibilities for the 7th row of the 315 cage, because they are already definites elsewhere in that row. That would mean that cell must contain the 9.
By the way, thanks for the videos. I am enjoying your logic.
Professor Homunculus sez:
Jan,
Those are pretty good questions. The reason why you can’t have a 1 in the 630 cage is because you have 4 cells, and you have 4 prime factors. Each of those prime factors must be be used in at least one of the cages. The 2 must be used – be it in the form of a 2, or a two as a factor of another number, like 6. Same for the 3, it must be used as a 3, 6, or a 9, but it must be used.
Just try to make 630 by multiplying four digits and see if you can finagel a 1 amongst them. If you can, let me know how you did it!
As far as your second point – the one about getting rid of the 7 and 5 in the bottom row of the 315 cage – you are 100% right.
The problem with making these videos is that there are many possibilities at any time, and different people see different ones at different times. There is no “official order” as to how to solve a kenken puzzle. (Well, maybe there is for a computer program, but I imagine that there is not even an optimal universal algorithm for computers, either ).
Considering that I try to keep the vids to under 10 mins, I basically only show what I focus on during that time.
It is true though, that I should have seen that immediately. You’ll find that no matter how good you get, there is always chance for oversight. That’s why I occasionally do a random ecology check of a row, column, or the whole puzzle.
When you get to video # 7, you’ll see that I mention that I’d overlooked what you caught much earlier. Good thinking! And thanks for the comment.
Hotcha!
Brian (a.k.a. Professor Homunculus)
Me again. Still working through your videos. I’m checking to see how much of my Sudoku logic applies to KenKen as well. Here’s the question. I just watched Step 13, where you filled out the cells for possible combos to make 24, and got the possible entries of 3,4,6,8. When I look further across the row, I see another cell that only contained some of those numbers in it. In other words, 3,4,6,8 HAVE to be in one of those four cells. Doesn’t that mean one can eliminate them at this time as possibilities for any other cells in that row?
Or would I be getting myself in trouble by doing that?
(By the way, I do realize that there are multiple ways of solving any kenken, so this isn’t meant as a criticism.) :)
Professor Homunculus sez:
Jan,
No worries – your input is very constructive. The real reason I started these vids is to get input exactly like what you are giving me, so I can learn as well.
Yes, the rules for sudoku hold as well for Kenken, except for the “quadrants” (which in the case of a 9×9 cell would be the “ninths” or whatever a square of 9 cells would be called).
You’re way ahead of me, though. Be patient. You are right, of course, and we get to it in lesson 15.
Can you do me a favor? Keep track of your questions, and when you get to the end of the series, let me know if any of them never get addressed. Some may not, because sometimes I’ll attack a problem from a different angle, but if you see an angle that I missed, let me know in a comment, so other people can see that there is more than one way to skin a KenKen.
You may find that some of your questions are answered in a video further down the line, though, so wait till you’ve been through the bunch of them, OK?
Your input is very, very welcome.
All the best,
Brian
This is great, I had a lot of fun, thank you. But you shouldn’t write anything down. I do everything in my head. It takes more time but is much better for your brain.
One more thing. Another great exercise is to memorize the grid first in your head and do everything in your head then, without ever looking back at the screen, everything memorized. Can’t do it with this one though, but I’ve doen up to 7×7 like that. Takes a reaaally long time, but again, real good for your brain in lots of different ways. Writing everything down is kind of wasting the whole point of the puzzle in my opinion since you’re ultimately looking for brain stimulation.
Professor Homunculus sez:
Jeremy,
Thanks for your thoughts. I also do most of these without writing. I am have been doing mnemonics for quite a few years, though. I feel, as you do, that doing it like that is a great brain-building tool.
Wow, just started and this really helped. Thanx soo much, now i understand.
Brian -
First, thank you for all your excellent work on describing KenKen puzzles. This is one of the first places I came to when I got “hooked”.
As much as I enjoyed playing KenKen, I soon grew tired of the awkward players available, so I developed my own. You can view this at the URL above.
The reason I am writing you is that I’ve used both of the puzzles that you have on this page within my list of puzzles, and I want to get your permission for that.
So please check out what I’ve done, and let me know what you think.
Thank you,
John Gilbrough
Professor Homunculus sez:
Thanks for your comment at the Chronicles. I checked out your player, and it is very cool.
The puzzles on my site (all of them) were given to me to use by the people at Nextoy, LLC (they own the KenKen trademark), so I can’t give you my permission to use them.
If you contact them, you can ask them. They seem very reasonable, and are really trying hard to get the “message” of math with KenKen out there.
hi, I didn’t know where to contact you but your layout looked rearranged on firefox and internet explorer. Anyways, i just suscribd to your rss.