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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In 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)? 

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Here’s a classic from Martin Gardner:

Suppose that you have a 3″ on a side wooden cube and a buzz saw.  You wish to cut the cube into 27 smaller cubes, each 1 cubic inch. It is easy to see that you can do this with six cuts.  You simply hold the cube in its original position while making two cuts that trisect each face.

Can it be done if fewer cuts?  If so, tell us how.  If not, prove that it can’t be done.

“It was a bright, cold day in April, and the clocks were striking thirteen.”

opens George Orwell’s novel ‘Nineteen Eighty-Four’.

1.  By an amazing coincidence thirteen squared is 169 which is the number of times my clock read the right time recently in a single calander day.  Normally it only reads correctly 164 times in a calander day.  This is even more surprising as my clock has been stopped for several years.  How can this be?

My solution combines a number of different techniques.  If you can think of any way a stopped clock can read correctly more than twice a day please post in the comments.  If you can think of something I’ve missed then we may be able to get a bigger answer!

2. I have a second clock which runs slightly fast and I have no way of adjusting it.  How can I make my clock read the right time?

3.  I noticed recently that my third clock was two minutes fast.  It runs at one minute per minute.  It tells me the right time four times a day.  Why?

For inspiration you may want to listen to Peter Sellers (Bluebottle) and Spike Milligan (Eccles) discussing the stopped-clock problem way back in 1957. 

Thanks to New Scientist’s Feedback Column and it’s readers for some of the idea’s here.

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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This is a classic puzzle from Martin Gardner that also appears in Peter Winkler’s Mathematical Puzzles: A Connoisseur’s Collection.  (At least I think so.  My copy is buried in a box somewhere.)

First, a warm-up puzzle:

You’re blindfolded and I will place two cards on the table in front of you, each either face up or face down at my option.  We’ll then play a game taking turns.  On your turn, you may turn over either one or both cards.  On my turn, I may either swap the positions of the cards without changing which might be face up or not at my option.  You are unable to detect whether or not I’ve made a swap.  You win as soon as both cards are face up.  Your task is to identify a set (deterministic) strategy that you can use such that regardless of how I place and switch cards you are sure to win.

[spoiler]

The strategy that you can employ to be sure to win this version is to flip both cards on your first turn, and then flip one card on your next turn.  Then finally you flip both cards on your final turn.

There are three states the cards can start in: both up, both down and one of each.  If they are both up, you win without taking a turn.  If they are both down, your first turn causes you to win.  If one is up and one is down, you maintain this state (though you reverse the cards) in your first turn.  On your second turn, you either turn up the down card (and win) or turn down the up card.  In this latter case, your third step of flipping both cards makes you a winner.

[/spoiler]

Now, consider a tougher version of the problem.  You now have four cards, each on the corner of a square table.  The setup is the same as before.  You’re blindfolded and I place the cards faces up or down in a way that I think will stump you.  You flip 1,2,3 or all cards on your turn, and then I either leave the table alone or rotate it 90, 180 or 270 degrees.  You can’t tell if the table has turned.  You then do some more flipping, etc.

(Note that I can’t arbitrarily re-order cards — only rotate the table.  Their relative position remains constant.  I didn’t labor this point in the first puzzle as with two cards it isn’t a distinction.  You can imagine that the rules are identical and the table can only be rotated 180 degrees.)

Can you identify a pattern of steps you can take that will ensure victory regardless of my placements and turning?

I am in a total immersion game called ‘World of Britain’.

Every day you can take a daily task, if you dare.  There are five tasks and each day one of these tasks is given at random.  You could get the same task each day if you were unlucky.

The tasks are:

·         Cycling in Edinburgh; the best city in the world has plenty of cycle lanes to help you avoid the traffic, some of them are a bit surprising.

·         Cheese Rolling in Gloucestershire; can you beat the cheese? 

·         Bog Snorkelling in Wales; my favourite British sport, bar none!  Watch the action here!  

·         Gurning goes back to at least the thirteenth century.  You may think it is just about pulling funny faces, and you would be right.

·         Mud Racing  How do you know who won?

My math question is:  assuming you win everything you enter how many days would you expect it to take to win all of these competitions?

p.s Gurning is now international.  Here is a US gurner completely oblivious to the exciting belly-flop championship happening behind him.  Apparently “everyone and their butt crack is welcome”

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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I have in front of me three boxes.  One contains all red balls.  One contains all blue balls.  The third contains a mixture of red and blue balls.  The three boxes are labeled as to their contents, and the three labels correspond to possible contents of a box, but you know that the labels are all misplaced.

You may draw one ball (at random) out of a box of your chosing.  After that, you must rearrange the labels so that they correctly reflect the contents of the boxes.  How do you do this?