Answer to the Fiendish Second Conway Puzzle, The Ten Divisibilities!

John_H_Conway_2005_(cropped)

Last month, in honor of mathematician and puzzly spirit John Horton Conway, we shared two of his favorite brain teasers and challenged our fellow PuzzleNationers to crack them.

Two weeks ago, we shared the solution to puzzle #1The Miracle Builders, and offered a few hints for puzzle #2, The Ten Divisibilities.

Now that we’ve heard from a few solvers who either conquered or got very close to conquering the second puzzle, we happily share both the solution and how we got there.


The Ten Divisibilities

I have a ten digit number, abcdefghij. Each of the digits is different, and:

  • a is divisible by 1
  • ab is divisible by 2
  • abc is divisible by 3
  • abcd is divisible by 4
  • abcde is divisible by 5
  • abcdef is divisible by 6
  • abcdefg is divisible by 7
  • abcdefgh is divisible by 8
  • abcdefghi is divisible by 9
  • abcdefghij is divisible by 10

What’s my number?

[To clarify: a, b, c, d, e, f, g, h, i, and j are all single digits. Each digit from 0 to 9 is represented by exactly one letter. The number abcdefghij is a ten-digit number whose first digit is a, second digit is b, and so on. It does not mean that you multiply a x b x c x…]

And here are the hints we offered to help:

-If you add all the digits in a number, and the total is divisible by 3, then that number is also divisible by 3.
-If the last two digits of a number are divisible by 4, then that number is divisible by 4.
-If the last three digits of a number are divisible by 8, then that number is divisible by 8.


The solution is 3816547290.

So, how do we get there?

First, we use process of elimination.

Any number divisible by 10 must end in a zero, so j = 0.

Any number divisible by 5 must end in a zero or a five, so e = 5 (because each digit only appears once).

That gives us abcd5fghi0.

But that’s not all we know.

If a number is divisible by an even number, that number must itself be even. So that means b, d, f, and h must all be even numbers (i.e. some combination of 2, 4, 6, and 8). That also means that a, c, g, and i must all be some combination of the remaining odd numbers (1, 3, 7, and 9).

That’s a lot of information that will come in handy as we solve.

So, where to next? Let’s look at one of those even-numbered spots.

We’ve been told that abcd is divisible by 4. But any number is divisible by 4 if the last two digits are divisible by 4. So that means cd is divisible by 4.

So, if c is odd, d is even, and cd is divisible by 4, that limits the possibilities somewhat. cd must be 12, 16, 32, 36, 72, 76, 92, or 96.

So d is either 2 or 6.

That will be helpful in figuring out def. And knowing def is the key to this entire puzzle.


One of the clues we offered in our last post was that if the sum of a number’s digits is divisible by 3, then that number is also divisible by three. We know abc is divisible by 3, so that means a + b + c is also divisible by 3.

And if something is divisible by 6, then it’s also divisible by 3, so a + b + c + d + e + f is divisible by 3.

Here’s where things get a little tricky. Since a + b + c + d + e + f is divisible by 3, and a + b + c is divisible by 3, then when you subtract a + b + c from a + b + c + d + e + f, the result, d + e + f would also be divisible by 3.

Why is that helpful? Because it means we can look at def instead of abcdef, and we know a lot about def right now.

d is either 2 or 6. e is 5. f is either 2, 4, 6, or 8. And the sum of d + e + f is divisible by 3.

So that gives us two possibilities to deal with, either 2 + 5 + f, where the sum is divisible by 3, or 6 + 5 + f, where the sum is divisible by 3.

Since each number is only used once, that’s six possible equations:

  • 2 + 5 + 4 = 11
  • 2 + 5 + 6 = 13
  • 2 + 5 + 8 = 15
  • 6 + 5 + 2 = 13
  • 6 + 5 + 4 = 15
  • 6 + 5 + 8 = 19

Only 258 and 654 have sums divisible by 3, so they’re our two possibilities for def.

We’ll have to try both of them to see which is the correct choice. How do we do that?

Let’s start with the assumption that def is 258.


That would mean our answer is abc258ghi0. We know b and h have to be even numbers, and only 4 and 6 are left as options. Since fewer numbers are divisible by 8 than by 2, let’s look at abc258gh.

One of the other hints we offered was that if the last three digits of a number are divisible by 8, then the whole number is divisible by 8.

So that means if abc258gh is divisible by 8, then 8gh is divisible by 8. That’s much more manageable.

So, f is 8, h is 4 or 6, and g is either 1, 3, 7, or 9. That gives us eight possibilities for 8gh: 814, 834, 874, 894, 816, 836, 876, and 896.

Dividing each of these by 8 reveals only two possible choices: 816 and 896. That means, in this scenario, h is 6, b is 4, and our number is a4c258g6i0.

What’s next? Well, remember that trick we did with abcdef before? We’re going to do it again with abcdefghi.

Any number divisible by 9 is divisible by 3. Our rule of sums tells us that a + b + c + d + e + f + g + h + i is also divisible by 3. And since a + b + c + d + e + f is divisible by 3, subtracting it means that g + h + i is also divisible by 3.

With 816 and 896 as our possibilities for fgh, that means our possibilities for ghi are 16i and 96i. That gives us the following possibilities: 163, 167, 169, 961, 963, 967, where the sum of our answer must be divisible by 3.

  • 1 + 6 + 3 = 10
  • 1 + 6 + 7 = 14
  • 1 + 6 + 9 = 16
  • 9 + 6 + 1 = 16
  • 9 + 6 + 3 = 18
  • 9 + 6 + 7 = 22

963 is the only one that works, which gives us a4c2589630. With only 1 and 7 remaining as options, our possible solution is either 1472589630 or 7412589630.

But, if you divide either 1472589 or 7412589 by 7 — which is faster than running every one of the 10 conditions through a calculator — neither divides cleanly. That means 258 is incorrect.


I know that was a lot of work just to eliminate one possibility, but it was worth it. It means 654 is correct, so our solution so far reads abc654ghi0.

And we can use the same techniques we just employed with 258 to find the actual answer.

We know b and h have to be even numbers, and only 2 and 8 are left as options. Again, since fewer numbers are divisible by 8 than by 2, let’s look at abc654gh.

4gh is divisible is 8. So, f is 4, h is 2 or 8, and g is either 1, 3, 7, or 9. That gives us eight possibilities for 4gh: 412, 432, 472, 492, 418, 438, 478, and 498.

Dividing each of these by 8 reveals only two possible choices: 432 and 472. That means b is 8, and our number is a8c654g2i0.

Now, let’s look at ghi.

With 432 and 472 as our possibilities for fgh, that means our possibilities for ghi are 32i and 72i. That gives us the following possibilities: 321, 327, 329, 721, 723, 729, where the sum of our answer must be divisible by 3.

  • 3 + 2 + 1 = 6
  • 3 + 2 + 7 = 12
  • 3 + 2 + 9 = 14
  • 7 + 2 + 1 = 10
  • 7 + 2 + 3 = 12
  • 7 + 2 + 9 = 18

Okay, that leaves us four possibilities for ghi: 321, 327, 723, and 729.

Stay with me, folks, we’re so close to the end!

Let’s look at our four possibilities:

  • a8c6543210 (79)
  • a8c6543270 (19)
  • a8c6547230 (19)
  • a8c6547290 (13)

Next to each number, I’ve placed the only digits missing in each scenario, two for each.

That means there are only 8 possible ways to arrange the remaining numbers:

  • 7896543210
  • 9876543210
  • 1896543270
  • 9816543270
  • 1896547230
  • 9816547230
  • 1836547290
  • 3816547290

So let’s do what we did last time, and divide each chain at the seventh number by 7.

  • 7896543 / 7
  • 9876543 / 7
  • 1896543 / 7
  • 9816543 / 7
  • 1896547 / 7
  • 9816547 / 7
  • 1836547 / 7
  • 3816547 / 7

Only one of the chains can be cleanly divided by 7, and it’s 3816547.

Which means the solution for abcdefghij is 3816547290.


I know this was a monster of a solve — it rivals our Brooklyn Nine-Nine seesaw puzzle solution in complexity — but it’s one that every one of our fellow PuzzleNationers are capable of puzzling out.

How did you do on this diabolical brain teaser, folks? Let us know in the comments section below. We’d love to hear from you!


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The Strange Mystery of Florida’s Coral Castle

coral castle entrance

[Image courtesy of Bitter Southerner.]

At the center of every great mystery, there is a puzzle. When people look at the pyramids of Egypt or the Moai statues of Easter Island, the puzzle at heart is obvious: how? How were these incredible objects created?

A similar, and no less puzzling mystery, can be found much closer to home for most Americans: the Coral Castle of Florida.

The-Coral-Castle-14

[Image courtesy of The Bohemian Blog.]

Less a castle and more a varied arrangement of stones into walls, shapes, doorways, and more, the Coral Castle is composed not of coral, but of oolite limestone blocks weighing literal tons. More than 1,000 tons of rock are part of the Coral Castle’s elaborate layout, which was assembled and expanded from 1923 to 1951.

Some of those monstrous stones are seamlessly joined into different structures. Others are so perfectly balanced that they can open like a revolving door with the gentlest push.

There is a sundial, a telescope, and even stone rocking chairs carved from single pieces of rock.

coral castle moon

[Image courtesy of Bitter Southerner,]

It’s an engineering marvel, to be sure, but what separates the Coral Castle from some of those other creations we mentioned above is the fact that we know who built the Coral Castle.

One man. Ed Leedskalnin.

coral castle tools

[Image courtesy of LiveScience.]

Using basic tools like picks, winches, ropes, and pulleys, Leedskalnin created the Coral Castle in secret, allowing visitors to ponder just how he was accomplishing this remarkable feat.

It’s particularly remarkable when you consider that Leedskalnin only had a fourth-grade education, having gone to work at a young age.

The-Coral-Castle-3

[Image courtesy of The Bohemian Blog.]

Of course, it’s also worth noting that Leedskalnin was a bit of a kook, claiming he had learned the secrets of the architects of King Solomon’s temples by studying books about the pyramids at the local library.

And yet, he created something amazing. So amazing, in fact, many people attribute the Coral Castle to supernatural efforts, not merely the engineering prowess, cleverness, and determination of a hardworking man.

coral castle stairs

[Image courtesy of LiveScience.]

Over the years, many peculiar theories have circulated surrounding the Coral Castle and Ed Leedskalnin. Unreliable eyewitnesses claimed to see coral blocks floating in the air like balloons while Leedskalnin worked at night.

Some believe Leedskalnin levitated the blocks with telekinesis or psychic powers, or by singing the stones into place. Others attribute the Castle to some sort of strange manipulation of gravity, antigravity, magnetism, ley lines, or earth energies. And, of course, alien technology has been floated as a possibility as well.

(Some people even believe there’s a hidden cipher lurking in several tracts written by Leedskalnin, just waiting to be found to reveal his secrets.)

coral castle chairs

[Image courtesy of Bitter Southerner.]

A friend of Leedskalnin’s wrote a book about the physics and engineering of the Coral Castle, entitled Mr. Can’t Is Dead. It’s one of many books that claims to explain how the Coral Castle came to be.

To me, the Coral Castle seems like one giant mechanical brain teaser, a math problem more about leverage and patience than the paranormal.

And yet, I can’t help but stare at some of these creations with awe. Maybe this one of those puzzles that’s better left unsolved.

The-Coral-Castle-1

[Image courtesy of The Bohemian Blog.]


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A Conway Puzzle Solution (And Some Hints for the Other Puzzle)

John_H_Conway_2005_(cropped)

Two weeks ago, in honor of mathematician and puzzly spirit John Horton Conway, we shared two of his favorite brain teasers and challenged our fellow PuzzleNationers.

So today, we happily share the solution for puzzle #1, The Miracle Builders.

I had a window in the north wall of my house. It was a perfect square, 1 meter wide and 1 meter high. But this window never let in enough light. So I hired this firm, the Miracle Builders, who performed the impossible. They remodeled the window so it let in more light. When when they’d finished the window was a perfect square, 1 meter high and 1 meter wide.

How did they do it?

Both windows are perfect squares, 1 meter wide and 1 meter high. So how can there be a difference in the amount of light?

The trick of this puzzle is in the description. Although the original window was a perfect square, the dimensions of the square aren’t 1 meter by 1 meter. No, it was a square placed like a diamond, with one corner directly above its opposite. So the 1 meter dimensions were the diagonals, not the sides.

All the Miracle Builders had to do was build a square window in the usual arrangement (two sides horizontal, two sides vertical) with dimensions of 1 meter by 1 meter. That creates a larger window (with a diagonal of √2m) and allows more light.

Very tricky indeed.


We had several solvers who successfully cracked the Miracle Builders puzzle, but there was less success with puzzle #2, The Ten Divisibilities.

So, in addition to the original puzzle, we’re going to post some solving hints for those intrepid solvers who want another crack at the puzzle.

The Ten Divisibilities

I have a ten digit number, abcdefghij. Each of the digits is different, and:

  • a is divisible by 1
  • ab is divisible by 2
  • abc is divisible by 3
  • abcd is divisible by 4
  • abcde is divisible by 5
  • abcdef is divisible by 6
  • abcdefg is divisible by 7
  • abcdefgh is divisible by 8
  • abcdefghi is divisible by 9
  • abcdefghij is divisible by 10

What’s my number?

[To clarify: a, b, c, d, e, f, g, h, i, and j are all single digits. Each digit from 0 to 9 is represented by exactly one letter. The number abcdefghij is a ten-digit number whose first digit is a, second digit is b, and so on. It does not mean that you multiply a x b x c x…]

Here’s a few hints that should help whittle down the possibilities for any frustrated solvers:

-If you add all the digits in a number, and the total is divisible by 3, then that number is also divisible by 3.
-If the last two digits of a number are divisible by 4, then that number is divisible by 4.
-If the last three digits of a number are divisible by 8, then that number is divisible by 8.

Good luck, and happy puzzling!


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Two Brain Teasers, Courtesy of Conway

John_H_Conway_2005_(cropped)

Last week, we penned a post celebrating the life and puzzly legacy of mathematician John Horton Conway, and several of our fellow PuzzleNationers reached out with their own thoughts or questions about Conway.

One recurring subject was about his love of puzzles and what kind of puzzles he enjoyed solving. So, naturally, I went hunting for some of Conway’s favorite puzzles.

As it turns out, Alex Bellos of The Guardian had me covered. Alex has a recurring puzzle feature on The Guardian‘s website where brain teasers and other mental trickery awaits intrepid solvers.

Years ago, Alex had asked Conway for suggestions for his column, and Conway offered up two tricky puzzles.

And now, I happily share them with you.


#1: The Miracle Builders

I had a window in the north wall of my house. It was a perfect square, 1 meter wide and 1 meter high. But this window never let in enough light. So I hired this firm, the Miracle Builders, who performed the impossible. They remodeled the window so it let in more light. When when they’d finished the window was a perfect square, 1 meter high and 1 meter wide.

How did they do it?


#2: The Ten Divisibilities

I have a ten digit number, abcdefghij. Each of the digits is different.

The following is also true:

  • a is divisible by 1
  • ab is divisible by 2
  • abc is divisible by 3
  • abcd is divisible by 4
  • abcde is divisible by 5
  • abcdef is divisible by 6
  • abcdefg is divisible by 7
  • abcdefgh is divisible by 8
  • abcdefghi is divisible by 9
  • abcdefghij is divisible by 10

What’s my number?

[To clarify: a, b, c, d, e, f, g, h, i, and j are all single digits. Each digit from 0 to 9 is represented by exactly one letter. The number abcdefghij is a ten-digit number whose first digit is a, second digit is b, and so on. It does not mean that you multiply a x b x c x…]


Did you solve one or both of these fiendish mind ticklers? Let us know in the comments section below! We’d love to hear from you.

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Forget 3-D Chess and Try 4-Directional Tetris!

[Image courtesy of Eurogamer.]

Tetris is the ultimate puzzle game. If you somehow haven’t played it, you’ve at least heard of it. The brainchild of Alexey Pajitnov conquered the world, not only making Nintendo’s Game Boy a bestselling video game platform, but turning millions of puzzlers into gamers and gamers into puzzlers as well.

Those iconic little Tetromino shapes are instantly recognizable, and the music can still induce panic and nervousness in players decades after the first time they heard those infamous notes.

And it serves as a brilliant template for ambitious game designers and puzzlers to add their own twists to the Tetris formula.

The basic concept is simple: try to arrange the constantly falling Tetromino blocks so that they make complete lines in the play area. If they do, that line disappears.

Of course, just because the concept is simple, that doesn’t mean the game is. As your play area fills up with Tetrominos, the music speeds up, amping up the tension. And as you progress through different levels, the pieces fall faster and faster. You need quick reflexes and ice in your veins to handle the higher difficulty levels.

Thankfully, you only need to worry about the blocks falling from one direction.

But a new variation on Tetris quadruples the gameplay area in a very devious way.

Say hello to Schwerkraftprojektionsgerät, Stephen Lavelle’s take on Tetris. In Schwerkraftprojektionsgerät, Tetromino blocks appear in the center of the play area, and you control where the piece is placed in each of the four directions.

No, the blocks do not fall automatically, nor is there a time limit that forces you to place the blocks quickly. Yes, you can spin each piece before placing it.

But each block goes into all four play areas simultaneously, and in the same position on the opposing sides.

schwerk1

As you can see, the straight piece lands flat in the east and west grid spaces, but standing upright in the north and south grid spaces. You have to work along each axis to find the best case scenario for all four of your play areas.

Yes, you’re trying to complete lines in four different directions at the same time. The T-shaped Tetromino will land in four different arrangements (flat, on one side of the T-bar, on the other side of the T-bar, and on the long end of the T-bar) with a single keystroke.

schwerk2

It’s mind-blowing how challenging this makes the game, but it’s challenging in a good way.

I mean, in a regular game of Tetris, you need several dozen completed lines to conclude a level and feel like a champion. In Schwerkraftprojektionsgerät, you feel like a genius if you can get two completed lines in each of the four play areas before the game is over!

Just as addictive as the original, yet offering a totally new twist on the familiar style of puzzling, I foresee a lot of office hours being lost to this engaging four-directional experiment in space efficiency.

You can try Schwerkraftprojektionsgerät for yourself here, and check out all of Stephen’s games here. (He also has a YouTube channel featuring some of his creations.)

After 35 years, it’s cool to see there are still new ways to make Tetris feel fresh again.


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Have You Been Playing Uno Wrong All Along?

unocards

The holiday season is one of the few times in a given calendar year I play games with most of my family members.

The necessity of gathering for multiple events — it takes two or three days to see everyone around Christmastime, based on geography, family obligations, and such — creates opportunities for group gameplay that simply don’t exist other times of the year.

This got me thinking about house rules.

Every family has house rules for games and activities. Maybe it’s where you stand and throw in a round of cornhole, or what’s fair in a game of Horse, or how many do-overs younger kids get during a trivia game. It could be whether you call all shots during a pool game or only the 8-ball shot. That sort of thing.

I virtually guarantee that every household has some house rules for Monopoly, whether it’s doubling your $200 if you land directly on Go or collecting previously-paid fees when you land on Free Parking.

As it turns out, a lot of us have been playing Uno with house rules as well.

Get this:

unocards3

That’s from the official Uno Twitter account, which I didn’t know was a thing.

This was also a total surprise to me. Growing up, I learned that you can stack Draw 2 cards or Draw 4 cards. Apparently, in some households, you can add to Draw 2 with a Draw 4 or a Draw 4 with a Draw 2, making a Draw 6 for an opponent.

In any case, that sort of stacking has never been allowed in the official rules.

Gasp! That means many heartbreaking Uno moments from my childhood could have been avoided!

So, I decided to dig a little further. Were there other rules I didn’t know about?

As it turns out… there were.

unocards2

In this Facebook post from January of 2018, an astonished Uno player discovered this little gem in the Uno rule book:

Did y’all know that you can only play the Draw 4 Wild card IF you have NO other cards of the same color that can be played??! AND if you suspect that someone has illegally played this card, they have to show you their hand. AND if they in fact played the card illegally they must draw 4, but if not, the person who challenged the play must DRAW 6?

How am I only learning about these rules now?! I, for one, never knew that you could force someone to show you their hand if they broke the honor system Go Fish-style.

Have these revelations changed the way you play Uno, fellow puzzlers? Or am I in the minority as part of a group that thought we knew the rules, but were very much mistaken?

Let us know in the comments section below! We’d love to hear from you.


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