This is a quickie that came out of my computer programming work adapted as a puzzle.

I propose to stand in front of you with a bag of balls.  I will hand you the balls one at a time.  You have no idea how many balls my bag contains.  I will also provide you with a magic device such that when its button is pushed, it provides you with a random number between 0 and 1.  You may use this as often as you like.

You must hold on to exactly one ball at a time.  When I hand you a ball, you must either throw it away immediately or else retain it and throw away the ball that you were previously holding.  After you have either retained or thrown away the last ball, your goal is for there to be an exactly equal chance that you are currently holding each of the balls that I have handed you.

It may help to provide a similar problem to demonstrate how you might use the magical device to accomplish something similar.  Suppose that I told you in advance that there were ten balls.  You could then use the device to generate your random number and multiply that number by 10.  This would give you a random number between 0 and 10 uniformly distributed over that range.  You could then designate all numbers up to 1 as pointing to the first ball, numbers greater than 1 up to 2 as the second ball, etc.  Your strategy would then be to throw away balls until handed the ball corresponding to your number and retaining that one from then on.  With this strategy, you would retain each of the balls with an exactly 10% chance.

This method can’t be directly adapted to the puzzle because you wouldn’t know what multiple to use before the first draw because you don’t know how many balls will be present, so you’ll need another strategy.

This is a followup to my earlier post, A Rather Odd Car Trip.  It provides a solution so if you haven’t read that yet, you should do so first as this won’t make much sense without it.

[spoiler]

Though czarrandy and others provided the distance between the cities, I’d like to show a way that you can work the problem.

We want to work out an equation that explains the amount of time traveled each  downhill, level and uphill in terms of the distances involved.

How long does it take to travel a particular distance at a particular rate?  Take the uphill portion as an example.  At 56 mph, you travel 1 hour / 56 miles or n hours / x miles for any given distance x.  So:

1/56 = n / x

Solve for n (hours) and you get:

x/56 hours traveled for distance x.

Do something similar for the level distance (call that y) and the downhill distance (call that z) and you get this equation for the time traveled in terms of the three distances traveling from A to B in 4 hours:

x/56 + y/63 + z/72 = 4

For the trip back, where uphill becomes downhill and vice versa, we get:

z/56 + y/63 + x/72 = 4 2/3 or 14/3

If we could solve for x, y and z, we would get our answer, but generally there will not be a unique solution for two equations with three unknowns.  But in this case, what we need is the value of x+y+z , which we can attempt to extract that from the equations.

First, add the two equations together and get:

x/56 + x/72 + y/63 + y/63 + z/56 + z/72 = 26/3

To collect the terms, multiply both sides by the least common multiple of the denominators, which is 504, and we get:

9x + 7x + 8y + 8y + 9z + 7z = 4368
16x + 16y + 16z = 4368
16(x+y+z) = 4368
x+y+z = 273

for 273 miles each way.  Notice that what we did was to add the equations up to get a value for the round trip, which might have been a first intuitive step if you happened to think of it.

Now… it was very lucky that we were able to nicely factor out that 16 in order to solve for x+y+z.  This won’t always be the case!  The three speeds were chosen carefully so that this would work out.  Consider if we used some other randomly chosen values:

Uphill: 50 mph
Level: 60 mph
Downhill: 70 mph

Leave the times the same, so:

x/50 + y/60 + z/70 = 4
x/70 + y/60 + z/50 = 14/3

or x/50 + x/70 + y/60 + y/60 + z/70 + z/50 = 26/3

Multiply by the least common multiple of the denominators, 2100, and get:

42x + 30x + 35y + 35y + 30z + 42z = 18200
72x + 70y + 72z = 18200

As you can see, there isn’t going to be any way to factor out x+y+z to get a unique total distance.  You could experiment to demonstrate that you can pick values for x, y and z that satisfy the equation and for which x+y+z will differ.

The final question that interested me is how to identify the cases where there does turn out to be a solution.  From the process we just followed, you can see that you want the final coefficient of the x and z terms, or the uphill and downhill time traveled, which will always be equal to each other, to also be equal to the coefficient of the y term or the time traveled level.

Since we’ve added the two equations from the round trip, the meaning of these terms is the time taken to travel round-trip over a slanted piece of road.  So in English, in order to have a unique solution to the problem, the speeds must be specially selected such that the time it takes to travel round-trip over a slanted piece of road must be the same as the time it takes to travel over a level piece of road.

I got this problem originally from Nick’s Mathematical Puzzles. There are a lot of neat puzzles on the site to enjoy.

[/spoiler]

Kyle and Chaim finally get back into the studio!

Standard Podcast [ 11:08 ] Play Now | Play in Popup | Download

We first pose a quick question: If you drive fifty miles in fifty minutes, must there be some ten minute interval in which you drive exactly ten miles?

Of course there must — mustn’t there? Well prove it!

Our main feature this week is an interview with Michael Breen, from the American Mathematical Society, who came and hosted a game show “Who Wants To Be A Mathematician!” About a hundred high school students from all over northwest Arkansas came to cheer on their classmates; Kyle Strong of Har-Ber High in Springdale came in first, winning $1250, and Karan Batra, of Bentonville placed second with $250.

Our interview includes a few sample problems… I guess we shouldn’t list too many of them, in case Micheal wants to recycle them! Mike’s also responsible for the great series of Mathematical Moments posters— check them out!

PS: We opened with the Up To One Million Dollars In Prize Money May Be Given Away gag… Always fun!

Here’s a puzzle that sounds a little like those, “A train leaves…” questions we were all prepared for but rarely saw on the SAT, but with a twist.

You are going to take a drive from City A to City B and back, but in a rather unusual car.  When travelling uphill, the car always moves at exactly 56 miles per hour.  On level ground, it travels at 63 miles per hour and finally when travelling downhill it travels at 72 miles per hour.  Assume that it transitions from one speed to another instantaneously and all of those other “mathematically perfect” qualities that make questions like this answerable.

You find that travelling from City A to City B takes exactly 4 hours of travel time.  On the return trip, driving time sums to 4 hours and 40 minutes.

How far apart are Cities A and B?

Here’s an old puzzle, I think from Martin Gardner though I can’t immediately find a reference, adapted as a fun trick for spring break — if you have just the right sort of friends.

Invite a friend to invent a polynomial of any order, but require that it have only positive integer coefficients. Next, you explain that you have to get something of a feel for the polynomial so you provide an integer and ask them to apply the polynomial to it, telling you the value. You may have to do this more than once. You then close your eyes and dramatically tell them what their polynomial is.

What strategy would you use and how many values would you have to provide?

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.

Standard Podcast [ 5:44 ] Play Now | Play in Popup | Download

On April 29, 2004, we did a piece on the very first Bamboopalooza; (the segment BG. Bamboopalooza was a couple of years later) and Jenn Starr asks about the sequence 1 3 7 12 18 26 35 … 

(Incidentally, once you unlock the secret of the sequence, can you determine how fast, asymptotically, it will grow?)

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?

Here’s a quick puzzle that I don’t think has appeared here previously.

A man leaves at exactly 6:00 AM to climb a mountain.  He may not climb at a constant rate.  In fact, he may stop to rest, or even backtrack a bit to see something interesting.  He arrives at the summit at exactly 6:00 PM and camps for the night.

The next day, he starts down at exactly 6:00 AM, again meandering unevenly, but following exactly the same path and reaching the bottom at exactly 6:00 PM.

The question is, was there some point on the path he was sure to have visited at exactly the same time on the two trips?