Soon Friday I set this puzzle…

The day before yesterday Edward was 17.  Next year he will be 20.  How can that be the case?

If you have not tried to answer it, have a go now.  For everyone else the answer is after the break.

This sentence was made on 1st January.  Edward’s birthday is the 31st December.  He was 17 the day before yesterday.  He was 18 yesterday.  He will be 19 this year, and 20 next year!

I have produced an ebook containing 101 of the previous Friday Puzzles! It is called PUZZLED and is available for the Kindle (UK here and USA here) and on the iBookstore (UK here in the USA here). You can try 101 of the puzzles for free here.


    1. May I modestly draw attention to my previously published insights to a version of this puzzle?

      I believe my contribution was somewhat overlooked – perhaps too many of you have been at drunken parties to celebrate the new Dr Who over the weekend to pay attention to the subtlties of this puzzle.

      The solution is as always obvious to those with a deep intuitive sense of the depths of our universe. And is already causing distress and confusion among those who do not – hence the “baby in a pram without a dummy” comments about it being a repeat puzzle. Distracting yourself with trivia is a sign of failure.

      It is not a repeat – how could it be? And if it were it would be because of the failure to answer it properly the first time. A good teacher knows when to set the same examination question again.

      I wish to be humble yet: study my works. Humbly they are for you and you will learn much. It is the least I can do.

    2. This doesn’t sound like Barry Goddard at all. (Perhaps it’s Dave or one of his neighbours/relatives). Will the real Barry Goddard please come back.

    3. It’s not Barry Goddard above. It’s better than the impersonation of him last week but it’s still a bit Bobby Davro in terms of quality.

  1. Self-plagiarism or not, here’s the relevant bit of what I posted last week.

    I remembered it from a few months earlier but I couldn’t recall the solution, so I cheated. I found the answer on a website of puzzles submitted by students at Woodlands Junior School. So, a 12 year-old might have posted this puzzle (different ages but same puzzle) and I still couldn’t get it.

  2. And here I was holding out a faint hope that, this time around, the answer that Richard would give would be different. Disappointing to have this one recycled so soon, especially when Richard is probably aware of at least 101 puzzles (or perhaps some of the ones in his ebook are repeats, too?).

  3. I thought that this puzzle sounded familiar, too, so I went with a different answer:

    Edward’s age is given in Base 8.

  4. The real question is whether this was an inadvertent repeat of a puzzle or a devious social experiment sprung on us all? Guinea pigs unite!

    1. If we are just guinea pigs, the puzzle should have come with a cavy-at.

      I’ll be here all night…

  5. Richard, fuck’n’eh. Just put up a blog post that says, “I’m really busy right now. Between promoting my existing books, doing videos, the fringe festival and fuck all, I’ve got no time to think of something original for the Friday puzzle, so just bear with me and stay tuned for when I can actually serve up something I haven’t written up 100 times before.” Your fans will understand. The hangers on will drop off, and that’s okay.

    1. In the absence of any real discussion of this puzzle I am prepared to host a Q&A (Question and answer) session on the question of problem solving.

      My success in solving the types of problems set on this blog speaks for itself and needs no further introduction. My methods are more widely applicable and we could all learn from them.

      As an astrology student I bring my own divinatory ability to the task. You too can do so too.

      Please ask and I will endeavour to illuminate. Who knows? Perhaps next wee you will be able to solve the puzzle using my methods.

    2. @eddy

      You may ask here. If we outgrow this forum I am sure one of my many supporters will help provide other facilities.

      I have many years experience in teaching as a tutor. A move into online teaching is a natural step forward for me.

    3. @Edie

      Thank you for your support and comments.

      I am glad you have opened the class by asking about irony. It is a much misunderstood brain function. Like Lorde who sang about it raining on her birthday being hyperviligiant to irony is a misapplied sensory unneccesary.

      I have deeper senses that that. I have a sense of the iconic which allows me to see deeper into human nature than many people can comfortably imagine.

      We live in a quantum universe which means that observations are primary in its ongoing creation. Thus even when posting on a web site be sure to choose to make observations that have depth and are meaningful.

      This is my first lesson.

    4. It’s not the ‘real’ Barry Goddard, but this was a good post and it made me laugh. ‘Edie’ was a good touch.

  6. In the absence of one from Richard, here’s a classic bit of lateral thinking that’s been doing the rounds for years but still has the potential to confuse:

    Imagine you’re escaping from a castle / dungeon / whatever. At the end of a corridor, you face two doors, in front of which are two trolls / guards whatever. Above them is a sign which reads “One door leads to freedom, the other to certain death. One of us always tells the truth, the other always lies. You are allowed to ask one of us, one question, then make your choice.”

    Needless to say, there’s no indication of which door is which, or which troll / guard is which.

    What question do you ask to guarantee your route to freedom?

    1. @mittfh
      Your presentation of this puzzle is far better than the version of it that I heard many years ago.

    2. If you haven’t encountered this one before:

      You ask one of the trolls / guards: “If I asked [the other troll/guard, e.g. ‘your colleague here’] which door led to freedom, what would he say?”

      Then take the other door.

      Let’s label the trolls / guards T (true) and F (false), and the exits D (death) and E (escape).

      If you asked T which door to take, he’d say E.
      If you asked F which door to take, he’d say D.

      If you asked T which door F would advise you take, he’d truthfully say D.
      If you asked F which door T would advise you take, he’d lie and say D.

      So by redirecting the question as above, whichever troll / guard you ask, you’ll always get the same answer – namely, which of the two doors you should avoid (unless they have the personality of Marvin from H2G2…)

      Hmm… I wonder if it would be possible to design a variant which also incorporates aspects of the Monty Hall Problem? 🙂

  7. Here’s an old one, but I don’t think very many people know it…
    Draw a grid of squares 4 wide and 3 high, omitting (or erasing) the four corner squares, leaving 8 squares (a row of two centered above a row of four and a row of two centered below that). Now place the letters A B C D E F G H in the squares so that no two alphabetically consecutive letters touch each other, even diagonally.

    Example of an attempt:
    — G D —
    A E H B
    — C F —
    But this fails because E and F touch, as well as G and H.

    Or, prove it impossible.

    1. Got to be impossible or you wouldn’t have added the last bit.

      How to prove, though, another matter. Some Goddardian intuition may be required.

    2. I think I can prove it’s impossible, using symmetry and the three only possible first letter starting positions.

    3. It is possible.
      Took me about 5 minutes (but I did have a hint from noted tax-dodger Paul David Hewson).
      His lordship was impressed.

    4. Been away. Don’t know if you were wanting the sol’n to be posted but it took me about 30 sec to find one. I’m sure that there are a few others that are not simply rotations/reflections.

    5. Solution:

      Unique, except for reflections and rotations. The insight is to notice that the two center squares are adjacent to every other square except one. So, the letters A and H must go there, since all the other letters are “adjacent” to two other letters alphabetically.

  8. Or a different one altogether.

    The band U2 has to cross a bridge at night.

    The first problem is, there is only one torch and the bridge is only strong enough to support two people crossing at once.

    The second problem is, everyone walks at different speeds. Adam takes 10 minutes to cross. Bono takes 5 minutes. Larry takes 2 minutes. Edge takes 1 minute. When two people cross together, they must obviously walk at the speed of the slower of the two in order to keep together with the torch.

    So, it has to be two people go over. One comes back. Two people go over. One comes back etc.

    The third problem is, their concert starts on the other side of the bridge in 17 minutes. How do you get them all over in 17 minutes.

    (There are no tricks. No piggybacks etc.)

    Hope this is a bit new to some people.

    1. They either start the concert 2 minutes late, leave someone behind or someone feels their way across in the dark…

    2. I don’t care how late their concert starts. The later the better as far as I am concerned. For preference it would better if the hypocritical tax-dodger Bono and his tone deaf cronies fell into the river.

    1. @MathMiles. You’ve left E on the original side in that case.

      @Chuck Berry. Substitute U2 with the members of your favourite band or string quartet. I don’t like U2 much either, to be honest.

      @Barry Goddard. It’s good to have your implied endorsement. I am indeed trained in psychiatry.

    2. @Izzy – well, I was only giving the first two steps and leaving the rest as an exercise to the reader – but it does solve the problem.

    3. If you want me to spell it out:
      L + E – 2 min
      E – 1 min
      A + B – 10 min
      L – 2 min
      E + L – 2 min
      TOTAL 17 min

  9. There are too many here making demands on Mr Wiseman. If he does not want to set an immediate puzzle please allow him some time to recover from Glasgow.

    As it says in the very first line of the post “Soon Friday I set this puzzle…”.

    Soon Friday he set puzzle. Hard to understand not: when Friday come he puzzle set. Is not a science of rocket to understand.

    Ah yes but what Friday? Maybe that the puzzle is?

    Yet the less we have people who are not trained in the public understanding of psychiatry trying to set their own puzzles in the comments. That makes no sense at all and could all end in tears. A puzzle that is not psychiatrically sound could lead to mental distress among the less mentally endowed among us.

    So please: leave the puzzle setting to the professionals.

    And I will continue to provide quality solutions and analysis to those puzzles to help the mentally disenfranchised understand how to solve without invoking nightmares of school arithmetic lessons.

    1. You may be able to help the “mentally disenfranchised” El Bazzo, but your comments are of no use to those of us that leave our cars in the regular parking spaces at Tesco.

    2. @DAVe. Are you making fun of the physically disabled? I am not sure how else to read your comment. Please advise.

    3. @Ernie Becclestone
      No I wasn’t intending to make fun of the “physically disabled” (to use your phrase).
      Why did you think I was – please advise?
      I was intending to make fun of Barry Goddard – clearly I missed the mark.
      Nonetheless, I thought we were meant to treat all people equally now. This would imply that the “physically disabled” are just as legitimate targets for the butt of my jokes as bankers, holocaust deniers, smokers, Japanese, paedos, racists, Alex Salmond, Jade Goody or Arabs.
      If I am wrong please advise – I would hate to go against the prevailing orthodoxy.

    4. @DAVe. Thanks for your reply. It was your reference to the other spaces in Tesco’s car park that made me think you were making fun of physically disabled people. It’s a relief to know that you weren’t.

      Who you makes jokes about is a matter between you and your conscience. It’s none of my business. Making fun of people, however, is a different matter, bordering on bigotry. I’m glad I interpreted your comment wrong.

  10. A child is holding a helium-filled balloon on a string in a car so that it floats just below the ceiling with the string under tension. All vents and windows closed. What happens when the car accelerates forward? Does the balloon go forward backward or sideways relative to the car?

    1. Just seen this on youtube – fascinating, but obvious when you think about it – which I only did after seeing it.

      It’s rather like, what happens if you have a bicycle standing vertically (get a friend to hold it up) and you press backwards on a pedal when it is at the bottom of its circle?

  11. Here is a nice easy one …..

    In a finite, undirected, connected graph, an integer variable v(n) is associated with each node n. One node is distinguished as the anchor. An operation OP(n) is defined on nodes:

    if node n is the anchor, then do nothing, else set v(n) to the value 1 + min{v(m)}, where m ranges over all neighbors of n that are distinct from n.

    An infinite sequence of operations is executed, the node arguments
    n, m, … for the operations being chosen arbitrarily and not necessarily fairly.

    Show that eventually all v(n) stabilize. That is, that after some finite prefix of the infinite sequence of operations, no further operation changes v(n) for any node n.

    1. Is this a collection of points forming a closed manifold, such as those of a topologically closed surface or an analogue of this in three or more dimensions?

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