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?
Richard Wiseman 
Clever! I was nowhere near solving that.
Same here. Good puzzle. One feels one perhaps should have stuck at it a little longer!
I wasn’t even close. Very good.
YES! I actually got it! (and quickly!). Will now spend the rest of the day celebrating!
Is it me? Would it not be more chronological if the puzzle started Jan 2, Feb 3, March 4? The answer would still be December 1st.
But then Apr 5 would be a perfectly good answer.
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?
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)
thanks ChrisR, seems a bit rough to me but oh well.
I must be being really thick, but I can’t even understand the answer. Can someone help me please?
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
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…………………….
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
i was very away from it .good one
Success! And woah, did I try solving this out of the box before linguistically. When I finally tried a linguistic approach, 12 minutes had passed. So kudos to you who tried that approach first! And congratulations to everyone who solved it!
I did not solve this one 😦
April 4th liked this puzzle.
April 4th December 1st April 4th October 1st October 3rd 🙂
u said idiot
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 😛
How do you get the 1st? Sure it is December but does it matter what day I’m December it is?
The assumed day-number is the letter-index, Anonymous.
btt: Great Puzzle – I had no idea at all…
It took me a while, but I finally figured it out. Boy, I’m really getting dumber in my old age.
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.
I think Occam’s Razor applies here
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 😉
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! 😛
Whoops! I must learn how to use Excel better. (Slinks quietly away)
so clever!
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.
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.
Slow Learner, could you remind me of my bold assertions? i only recall making assertions in normal type, or at worst italics.
It took me a couple minutes to even understand the answer. Too convoluted to be a plausible puzzle.
chiefs, simpler answer should be January 2
chronological months; ? FEB, MAR, APR
corresponding to
consecutive numbers; 1, ?, 3, 4
therefore January 2
peace.