Puzzle of the Week #97 - Paddocks

I usually publish a puzzle each Friday, but here's a bonus one:

Draw fences between some of the posts so that each post is at the junction of exactly THREE fences.

These fences will divide the field into several PADDOCKS; any paddock whose area is greater than a single triangle will contain a NUMBER, which will indicate the number of TRIANGLES that make up the paddock.

The boundary fence is already in place, so any post on the boundary only needs one more fence emerging from it in order to make up its full complement.

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Puzzle of the Week #96 - Self Working Trick

The following is a self-working card trick I invented a few years ago for my daughter to perform.

In the instructions below I have left out a crucial piece of information – what is the value of ‘x’? (Technically there are 3 possible answers, but I’m looking for the smallest of those).

 

Take a normal pack of 52 cards, well shuffled.

Discard the first ‘x’ cards.

Deal the remaining cards into 3 equal piles.

Ask the mark to choose one of the piles to discard.

Ask the mark to choose another pile. Show them the card on the bottom of that pile and then place that pile on top of the other remaining pile (so that the target card is roughly in the middle of the pile).

Deal all of the cards into two piles, left-right-left-right-etc.

Pick up the right pile and again deal them left-right-etc.

Continue until all but one of the cards is in the left pile. The card remaining in the right pile is the target card.

Puzzle of the Week #95 - The Photograph

A mother and daughter are sat together on the settee, flicking through a photo album, whilst a massive international event plays out live on the television.

‘See this photo?’ says the mother, ‘this was taken on New Year’s Eve 1999. What a night!’

‘How old were you in that photo Mum?’ asks the daughter.

‘Well, funny you should ask that, I was halfway between your age now and my age now, precisely to the day’.

Since their dates of birth are 12th March 1980 and 4th May 2005, what was on the TV?

Puzzle of the Week #94 - Hidden Words

Find the hidden word in each of the following sentences – they are on a related theme.

 

Are high winds or heavy rain forecast for this afternoon?

Are you a stud or a jock?

Diablo is Spanish for devil

Disco Stu, artist in residence.

I’d like to implant a gene to make me less lazy.

I’ve been on a Russian journalism course.

Is Kevin or Mandy in charge?

It’s easier to go under the prison wall, rather than over.

There’s a film version of Blankety Blank, with Matt Le Blanc as Terry Wogan.

Who’s in charge now, Mandy or Kevin?

 

 

Puzzle of the Week #92 - Nought and Crosses

This isn't a puzzle in any traditional sense, however I thought I'd share it here anyway:

Noughts and crosses (tic tac toe) is famously a game which will always end in a draw if both players are playing rationally.

However with a simple change to the rules which would appear to be both sensible and fair - namely that neither player may place their first mark in the centre square - the player who goes first can always force a win.

The puzzle then, if there is one, is merely for the reader to satisfy them self that this is true, and to determine, if you were going first, how you would force a win.

 

Puzzle of the Week #91 - Quotebreaker

I have taken a quotation, and I have replaced each of the letters with one-, two- of three-digit numbers according to the table below. Can you change it back to letters?

Be careful though, as some sequences of numbers could lead to several words, for instance 31110 could mean CAT (3,1,110), but could equally mean MAD (31,1,10).

This particular quotation is from Leo Tolstoy.

1 31132 3332 1 110203311110313210 31213011 11313023 201103 11033 12331021311110 2021103 1333130 13210 1031121 11033 202131103113012 1111211102121 313310232213213, '11033101121 21'31 1333213213 11033 33311211102 1101131132110121-122111211 31213011103 13210 110201132 10211103110 111100 13210 103301111100'.

Puzzle of the Week #86 - Paddocks

Draw fences between some of the posts so that each post is at the junction of exactly three fences.

These fences will divide the field into several paddocks; any paddock whose area is greater than a single triangle will contain a number, which will indicate the area of the paddock that contains it.

The boundary fence is already in place, so any post on the boundary only needs one more fence emerging from it in order to make up its full complement.

For example:

Here is the puzzle:

Puzzle of the Week #81 - Fixture List

In a new football league there are five teams: Allesley, Bannerbrook, Canley, Dunsmore and Earlsdon. Over the course of 10 weeks, each team must play every other team twice (once home and once away). There will be two games each week, with one team getting a rest. It is your job to organise the fixtures.

Is it possible to organise the fixtures so that each team alternates between home and away games?

If it is, find an example fixture list; if it is not, why not?

Puzzle of the Week #79 - Cumulative Unrepetitiousness

This is really quite a tough one, so I wouldn’t blame you for sitting this one out. I promise to be gentler next week!

I have invented a function of a number, let’s call it the ‘cumulative unrepetitiousness’, C(n). The function looks at all the positive whole numbers up to and including n, and splits them into two categories: into category A go all of the numbers which contain some identical consecutive digits (such as 113 or 3457335), and into category B go all of the numbers that contain no identical consecutive digits (such as 34567 of 2323). C(n) is the size of category B minus the size of category A.

The value of C(n) either goes up or down by 1 as n goes up by 1:

C(1) = 1

C(2) = 2

C(3) = 3

C(4) = 4

C(5) = 5

C(6) = 6

C(7) = 7

C(8) = 8

C(9) = 9

C(10) = 10

C(11) = 9

C(12) = 10

etc.

As you can see, for small values of n, C(n) is always positive.

When n gets big enough, C(n) is always negative.

For a brief time in the middle, C(n) crosses the zero line several times. In fact there are a total of 35 positive values of n for which C(n) = 0, before it heads off into the negative zone for ever more.

Your task is simply to find the first of these.