On Friday I posted this puzzle….

What date comes next in the sequence?

April 1st
February 3rd
March 4th
If you have not tried to solve it, have a go now.  For everyone else the answer is after the break.
The 1st letter of April is A, applying the same rule you get B, then C and so the next date in the sequence is the 1st December!
Did you solve it?
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. Apologies if I’m missing something here, but I don’t understand why those months were used. They seem fairly arbitrary, why not jAnuary, feBruary, marCh. Or even April, feBruary, oCtober? Is there some logic to using April, Feb, March that I’m missing?

    1. No real reason for choosing them apart from their closeness adds to their ability to mislead down some blind alleys (eg 01/04, 03/02, 04/03 …) without leading to false answers (as Peter points out above)

    1. April 1st – the first kletter of April is A
      February 3rd – the third letter of February is B
      March 4th – the fourth letter of March is C
      where can I get D from – the first letter of December – hence the answer December 1st

  2. Thank you very much, simple actually. If my brain is that bad it does not bode well for the rest of the day, week, month…………………….

    1. Don’t worry
      They are all simple once you have seen the answer
      They are not so easy when you solving them
      That is my experience anyway

  3. Hmm, I had the A-1 connection, but couldn’t make F fit with 3 nor M with 4, so I gave up.

    Was alright, I’m not a big fan of the series puzzles though 😛

  4. In response to “Slow Learner” challenging a solution that results in December 25th, I offer up these two 3rd order polynomial equations:

    d = 1 + 9.5 * x – 11 * x^2 + 3.5 * x^3
    m = 7.3442 – 4.2135 * d + 0.8972 * d^2 – 0.02884 * d^3

    where x is the sequence number (starting at zero), d is the rounded integer day result and m is the rounded integer month result. Math FTW!

    So, like both Paul and I said last week, these sorts of puzzles *can* be little more than an exercise in numerology. Certainly Richard’s answer is more “clever”, but make no mistake that it is mainly an challenge in guessing what he is thinking the solution should be, and not that there is only one possible solution to the sequence.

    1. Besides Occam’s Razor, I’m not convinced these formulae give the right answer as they miss out the third item 4th March jumping from 3/2 to 25/12. Good try though 😉

    2. Again, the point is not that this is the best answer, just that there is more than one *possible* answer. I did it because it was an fairly easy exercise in mathematics with the tools that came with my Mac. There may be many even easier ways to get December 25th (or other answers) based on some human-level reasoning, but reverse engineering the puzzle was not worth the effort. If you want that, you’re gonna have to pay me.

      And I don’t know what you’re talking about ChrisR:

      x = 2 (third item)
      d = 1 + 9.5 * 2 – 11 * 2^2 + 3.5 * 2^3 = 4 (day)
      m = 7.3442 – 4.2135 * 4 + 0.8972 * 4^2 – 0.02884 * 4^3 = 2.99964 (March)

      Back to class for you, I guess! 😛

    1. Impossibly (I can call you impossibly, I trust?) thanks for this.
      I have no idea whether your solution is workable or not, but, in common with the Occam’s Razor fans, I do not find it as elegant as Richard’s answer.
      Anyway, my challenge was to Paul Durrant, who made the assertion on Friday, not you. He may have a more elegant solution.
      Over to you Mr Durrant.
      BTW ctj I have not forgotten your bold assertions from a couple of weeks ago.

    2. I made the assertion earlier than Paul. He’s welcome to come up with another answer but, from a mathematical standpoint, it’s a solved problem.

      Also note that Richard did *not* offer an elegant solution, or any solution at all, to your December 25th challenge. Nor did I offer any other solution, elegant or not, to match Richard’s December 1st answer. Comparing the two solutions doesn’t make sense. These are clearly non-parity sequences and, like I said, there are likely many other rules that can generate many other sequences with those 3 starting data points.

    3. Slow Learner, could you remind me of my bold assertions? i only recall making assertions in normal type, or at worst italics.

  5. chiefs, simpler answer should be January 2

    chronological months; ? FEB, MAR, APR

    corresponding to

    consecutive numbers; 1, ?, 3, 4

    therefore January 2


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