Archive for answers

GC. Another Buncha Prisoners

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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GB. Hat Strategy

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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GA. Stacking the Chips

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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Morris: World of Britain 2: Proof and Paradox

paradox-clockIn working out the proof for World of Britain I came across a paradox.  Maybe smarter Math Factorites can help me out?  My sanity could depend on it.

In the puzzle you have five different tasks.  On each day one of these tasks is given at random.  How long do you expect it to take to get all five tasks?

First consider a simple case.  Suppose some event has a probability, p, of happening on any one day.  Let’s say that E(p) is the expected number of days we have to wait for the event to happen.  For example if p=1 then the event is guaranteed to happen every day and so E(p)=1.

How can we calculate E(p)? 

Read the rest of this entry »

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Yoak: Followup to A Rather Odd Car Trip

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]

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FR. Who Wants To Be A Mathematician?

Kyle and Chaim finally get back into the studio!

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!

Who Wants To Be A Mathematician

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

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FM. Bamboopalooza One

The very first bamboo star in Fayetteville

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?)

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FL. Algebra on the Radio

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?

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FK. Twiddling Screws

It’s really rather counterintuitive, I think: when two screws are twiddled past each other, do they move closer, or move further apart, or stay the same distance from one another?

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FJ. Loyd’s Atomic Easter Eggs

This puzzle comes from the collected works of the great puzzler Sam Loyd:

How many eggs can be packed in a 6×6 crate, if no more than two can lie on any row, column, or diagonal (even a short diagonal), and an egg must be placed in each of two opposite corners?

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