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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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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Celebrating the Puzzly Legacy of John Horton Conway

The worlds of puzzles and mathematics overlap more than you might think. I’m not just talking about word problems or mathy brain teasers like the Birthday Puzzle or the jugs of water trap from Die Hard with a Vengeance.

For twenty-five years, Martin Gardner penned a column in Scientific American called Mathematical Games, adding a marvelous sense of puzzly spirit and whimsy to the field of mathematics, exploring everything from the works of M.C. Escher to visual puzzles like the mobius strip and tangrams. He was also a champion of recreational math, the concept that there are inherently fun and entertaining ways to do math, not just homework, analysis, and number crunching.

And on more than one occasion, Gardner turned to the genius and innovative thinking of John Horton Conway for inspiration.

John_H_Conway_2005_(cropped)

[Image courtesy of Wikipedia.]

Conway was best known as a mathematician, but that one word fails to encapsulate either his creativity or the depth of his devotion to the field. Conway was a pioneer, contributing to some mathematical fields (geometry and number theory among them), vastly expanding what could be accomplished in other fields (particularly game theory), and even creating new fields (like cellular automata).

Professor of Mathematics, Emeritus, Simon Kochen said, “He was like a butterfly going from one thing to another, always with magical qualities to the results.” The Guardian described him in equally glowing terms as “a cross between Archimedes, Mick Jagger and Salvador Dalí.”

lifep

[Image courtesy of Cornell.edu.]

His most famous creation is The Game of Life, a model that not only visually details how algorithms work, but explores how cells and biological forms evolve and interact.

Essentially, imagine a sheet of graph paper. In The Game of Life, you choose a starting scenario, then watch the game proceed according to certain rules:

  • Any live cell with fewer than two live neighbors dies, as if by underpopulation.
  • Any live cell with two or three live neighbors lives on to the next generation.
  • Any live cell with more than three live neighbors dies, as if by overpopulation.
  • Any dead cell with exactly three live neighbors becomes a live cell, as if by reproduction.

The process plays out from your starting point completely without your intervention, spiraling and expanding outward.

It’s the ultimate if-then sequence that can proceed unhindered for generations. It is a literal launchpad for various potential futures based on a single choice. It’s mind-bending and simple all at once. (And you can try it yourself here!)

JHC-GOL-600x170px

[Image courtesy of Sign-Up.To.]

But that’s far from Conway’s only contribution to the world of puzzles.

Not only did he analyze and explore puzzles like the Soma cube and Peg Solitaire, but he created or had a hand in creating numerous other puzzles that expanded upon mathematical concepts.

I could delve into creations like Hackenbush, the Angel Problem, Phutball/Philosopher’s Football, Conway’s Soldiers, and more — and perhaps I will in the future — but I’d like to focus on one of his most charming contributions: Sprouts.

Sprouts is a pencil-and-paper strategy game where players try to keep the game going by drawing a line between two dots on the paper and adding a new dot somewhere along that line.

The rules are simple, but the gameplay can quickly become tricky:

  • The line may be straight or curved, but must not touch or cross itself or any other line.
  • The new spot cannot be placed on top of one of the endpoints of the new line. Thus the new spot splits the line into two shorter lines.
  • No spot may have more than three lines attached to it.

Check out this sample game:

sprouts

[Image courtesy of Fun Mines.]

It’s a perfect example of the playfulness Conway brought to the mathematical field and teaching. The game is strategic, easy to learn, difficult to master, and encourages repeated engagement.

In a piece about Conway, Princeton professor Manjul Bhargava said, “I learned very quickly that playing games and working on mathematics were closely intertwined activities for him, if not actually the same activity.”

He would carry all sorts of bits and bobs that would assist him in explaining different concepts. Dice, ropes, decks of cards, a Slinky… any number of random objects were mentioned as potential teaching tools.

Professor Joseph Kohn shared a story about Conway’s enthusiasm for teaching and impressive span of knowledge. Apparently, Conway was on his way to a large public lecture. En route, he asked his companions what topic he should cover. Imagine promising to do a lecture with no preparation at all, and deciding on the way what it would be about.

Naturally, after choosing a topic in the car, the lecture went off without a hitch. He improvised the entire thing.

Of course, you would expect nothing less from a man who could recite pi from memory to more than 1100 digits? Or who, at a moment’s notice, could calculate the day of the week for any given date (employing a technique he called his Doomsday algorithm).


Conway unfortunately passed away earlier this month, due to complications from COVID-19, at the age of 82.

His contributions to the worlds of mathematics and puzzles, not to mention his tireless support of recreational math, cannot be overstated. His work and his play will not soon be forgotten.

MAC31_BOOKS_COVERS_POST02

[Image courtesy of Macleans.]

If you’d like to learn more about Conway, be sure to check out Genius at Play: The Curious Mind of John Horton Conway by Siobhan Roberts.

[My many thanks to friend of the blog Andrew Haynes for suggesting today’s subject and contributing notes and sources.]


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A Coin Puzzle: My Two Cents (Plus 97 More)

Our friends at Penny Dell Puzzles recently shared the following brain teaser on their social media:

Naturally, we accepted the challenge.

Now, before we get started with this one, we have to add one detail: which coins we’re allowed to use. It’s safe to assume that pennies, nickels, dimes, and quarters are available, but the question doesn’t say anything about half-dollar coins.

So we’re going to figure out the correct answer without half-dollar coins available, and then with half-dollar coins available.

Let’s begin.


[Image courtesy of How Stuff Works.]

The easiest way to get started is to figure out the smallest number of coins we need to make 99 cents, since that’s the highest number we need to be able to form. Once we have that info, we can work backwards and make sure all the other numbers are covered.

For 99 cents, you need 3 quarters, 2 dimes, and 4 pennies. That’s 25 + 25 + 25 + 10 + 10 + 1 + 1 + 1 + 1 = 99.

Right away, we know we’re close with these 9 coins.

You don’t need more than 3 quarters, for instance, because your possible totals are all below $1.

Now, let’s make sure we can form the numbers 1 through 24 with our chosen coins. (If we can, we’re done, because once we’ve covered 1 through 24, we can simply add one quarter or two quarters to cover 25 through 99.)

Our four pennies cover us for 1 through 4. But wait, there’s 5. And we can’t make 5 cents change with 4 pennies or 2 dimes. In fact, we can’t make 5, 6, 7, 8, or 9 cents change without a nickel.

So let’s add a nickel to our current coin count. That makes 3 quarters, 2 dimes, 1 nickel, and 4 pennies. (Why just 1 nickel? Well, we don’t need two, because that’s covered by a single dime.)

Our four pennies cover 1 through 4. Our nickel and four pennies cover 5 through 9. Our dime, nickel, and four pennies cover 1 through 19. And our two dimes, one nickel, and four pennies cover 1 through 29. (But, again, we only need them to cover 1 through 24, because at that point, our quarters become useful.)

That’s all 99 possibilities — 1 through 99 — covered by just ten coins.

[Image courtesy of Wikipedia.]

But what about that half-dollar?

Well, we can apply the same thinking to a coin count with a half-dollar. For 99 cents, you need 1 half-dollar, 1 quarter, 2 dimes, and 4 pennies. That’s 50 + 25 + 10 + 10 + 1 + 1 + 1 + 1 = 99.

Now, we make sure we can form the numbers 1 through 49 with our chosen coins. (Once we can, we can simply add the half-dollar to cover 50 through 99.)

Once again, we quickly discover we need that single nickel to fill in the gaps.

Our four pennies cover 1 through 4. Our nickel and four pennies cover 5 through 9. Our dime, nickel, and four pennies cover 1 through 19. Our two dimes, one nickel, and four pennies cover 1 through 29. And our one quarter, two dimes, one nickel, and four pennies cover 1 through 54. (But, again, we only need them to cover 1 through 49, because at that point, our half-dollar becomes useful.)

That’s all 99 possibilities — 1 through 99 — covered by just nine coins.


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Brain Teaser: A Curious Way to Tell Time

We love tackling brain teasers, riddles, math puzzles, and mind ticklers here at PuzzleNation Blog.

Our friends at Penny Dell Puzzles also enjoy putting their puzzly skills to work by posting various brain teasers..

And one of our mutual readers was hoping we could explain how to solve a puzzle that recently made the rounds on Penny’s social media accounts.

We’d be happy to!

Today, let’s take a look at a brain teaser all about time.

So, where do we begin?

Let’s start by breaking that sentence down to make it easier to parse: “The number of hours left today is half of the number of hours already passed.”

Well, a simpler way of saying that is “the number of hours already passed is twice the number of hours left.”

So if you have the number of hours left — let’s call that X — then the number of hours already passed is twice X, or 2X. Between X and 2X, that’s your entire day covered.

Sorting that out gives us the simple formula of X + 2X = 24, since there are 24 hours in a day.

That easily becomes 3X = 24, and simple division tells us that X = 8.

So X, the number of hours left today, is 8. Which means that twice X, or 16, is the number of hours already passed.

And if there are 8 hours left in the day (or 16 hours already passed), that means it’s 4 PM.


Most of the time, brain teasers are all about efficiently organizing the information we have.

That allows us to figure out how best to use that information to move forward and solve the puzzle. This is just as true with a relatively straightforward brain teaser like this as it is with a complicated logic puzzle with all sorts of pieces to put together.

It’s all about figuring out what we know, what we don’t, and how what we DO know can lead us to what we don’t.

That’s just part of the fun.


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