I’m overdue to post a puzzle, but I’m momentarily tapped out. Here’s a curiosity in the meantime: You can provide a very good estimate of a conversion from miles to kilometers by choosing sequential Fibonacci numbers.  The conversion rate is 1.609344 kilometers to a mile. So this gives us:

This leaves you in pretty good shape if you need to get from Cincinnati, OH to Destin, FL at 610 Miles, but what if you need to convert some distance that doesn’t happen to be a Fibonacci number?  Just build it up from parts!

100 miles is 89+8+3.  So in kilometers, that’s 144 + 13 + 5 or 162 kilometers.  (160.9344 by conversion…)

OK.  Here’s a puzzle, sort of.  I found this interesting set of numbers recently:

{ 1, 2, 3, 4, 5, 6, 7, 8, 9, 53, 371, 5141, 99481 }

The series doesn’t continue.  That’s all of them.  What’s special about those numbers?

Man, what is it with puzzlers and prisoners? Jeff Yoak lines ’em up and the stakes are high in this week’s puzzle. 

Also, we are now twittering at MathFactor; each of the authors has an account of his own; mine is CGoodmanStrauss. You can tag solutions and comments with #mathfactor. See you there!

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How can three people, each required to guess the color of hat on their head, strategize and maximize the chances they’ll all be right?

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Jeff Yoak discusses the mathematical – and non-mathematical – nature of poker. Sitting at the table led him to wonder: Which numbers, precisely, are the sum of consecutive integers, and in how many ways?

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The Math Factor podcast catches up with Jeff Yoak, an author on the Math Factor website, to discuss his fantastic Find-the-Gold-Coin puzzle.

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What can you say about a sequence if you know that each term is a weighted average of the terms to either side? 

For example, in the sequence 1, 2, 4, 8, 16, … each term is exactly 2/3rds of the previous term, plus 1/3rd of the following term. What other sequences have exactly that property?

For a given value p, what sequences s₁, s₂, s₃, … s_n have the property that each s_k = p s_(k-1) + (1-p) s_(k+1) ? Does knowing s₁ also fix the remaining terms? 

Amusingly, this actually will help with last weeks puzzle!

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Rob Fathauer discusses the ins and outs of the mathematical toy business, and we ask: For which numbers is the sum of digits the same as the sum of digits of twice the number. For example:

The sum of the digits of 351 is 9 and the sum of the digits of 2 x 351 = 702 is also 9.

1) If a number has this property, can we always rearrange its digits and obtain another number with this property (513, 135, etc all have it)

2) Which powers of 2 have this property?

3) And most of all, can you give a simple characterization of the numbers with this property, in terms of just the digits themselves?

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There has been a theme in some of the recent posts and problems. It’s a little buried but almost enough to say its another of those Mathfactor agendas when we try to sneak some knowledge to you buried in the fun.  Never one to miss such an opportunity I will jump in with a post, and a problem.  This is a slight change to a classic problem that comes out of the work of one of my mathematical heroes:  Leonardo of Pisa, also known as Fibonacci.  He is responsible for changing how we count! Not many people can claim that. He introduced the system base value, also known as Arabic numerals that we still use today into Europe.  He is more famous however for talking about rabbits:

Imagine that you have immortal rabbits, Bugs Bunny’s version of Olympus perhaps.  Even if they are immortal however rabbits are famous for one thing.  They breed like, well rabbits.  Some of the rabbits are children and some adults and are divided into pairs.  Each month any child pairs become adults and any adult pairs breed to produce a new child pair.  They are immortal so no pair ever dies.  These rabbits are also a little odd.  They live on a line (don’t complain, this is no more ludicrous than that they are immortal!), but can shuffle along.  Also if you are worried about inbreeding, the male rabbit leave the family hutch and shuffle along the line past others until they find a suitably unrelated mate.  Why we would be worried about inbreeding in immortal rabbits living on a line escapes me!

Anyway we start with one pair of children.  Lets put a c.  After a month they become adults, a.  Another month passes and they now have a pair of children, but are still there themselves.  We therfore have the original pair and a pair of children: ac.  Next month the adults have another pair of children and the children become adults: aca.  Can you see how this will work?  Each month the children become adults so we replace every c with and a, each pair of adults has a new pair of children but stays as adults, so we replace every a with ac.  We can continue to get longer and longer sequences of rabbits on this line:

aca  to acaac to acaacaca to acaacacaacaac….

Now some puzzles.  Given a line with 21 adult pairs and 13 child pairs, how many pairs of adults and children would there be after one month?

Given p adults and q children how many adults and children will there be after one month?

Finally a more difficult one.  How will the ratio of adults to children behave month on month?  Will it

a) Get closer and closer to a particular number?

b) Keep on changing without pattern?

In either case can you say more?

You have a sack of coins of three types: brass, silver and gold. You know that the majority of the coins are gold, though they’ve been painted and partially hollowed so that you can’t actually determine the type of a particular coin. Fortunately, you have a machine into which you can insert two coins and the machine will tell you whether the two coins are of the same type or different types.

Your task is to locate a gold coin.

You will compare the coins in “passes” with each coin not being compared more than once in a pass. In a pass, every coin can be a member of a comparison or not, but any particular coin can’t be part of more than one comparison during the pass. Your goal is to minimize the number of passes required to be sure that you locate a single gold coin.

You should be able to describe how many passes (at most) your solution will require rather than the number of passes increasing arbitrarily with the number of coins that turn out to be in the bag.

Is it true that if you give me any number you please (say 3, for example), I can come up with a new number (in this case 3/2) so that the sum and products of our numbers are the same? ( 3+ 3/2 = 3 x 3/2 = 9/2 )

A couple more examples: if you give me 0, I’ll give you 0, and 0+0 = 0 x 0.

If you give me 2, I’ll give you 2 and 2 + 2 = 2 x 2.

If you give me -1, I’ll give you 1/2 since -1 + 1/2 = -1 x 1/2.

Can I always give a response to your number, or is there a way to stump me?