Code puzzle thread +_+

[Kyriakos]

Might be a better way indeed of solving it, but it seems straightforward enough.

Spoiler :

ABC divisible by 3,4,5 => divisible by 60 => C = 0
BCD divisible by 5,8 => divisible by 40 => D = 0
=> "BCD" also divisible by 100 => divisible by 200
=> B E {2, 4, 6, 8} => B = 2 since the other digits have been explicitly excluded
=> AB a multiple of 6 which ends with "2" => AB E {12,42,72}
=> A = 7 since 1 and 4 have been explicitly excluded

The code is 7 2 0 0

Again, what I did was far worse. After the initial factor plan with unknown x,y I actually did the various possible x,y to exclude with the known false digits (and the remainder for BCD is not that little...)

Still, I think that there should be a way to do this with polynomial division and/including a two-variable equation
What follows isn't that, however, it's just a rewriting of what you did, combining it with the not that bad parts of my own method. Of course I'd had utilized c,d=0 if I had seen it.

Spoiler :

 

[Thrasybulos]

A huge pie is divided among 100 guests.
The first guest gets 1% of the pie.
The second guest gets 2% of the remaining part.
The third guest gets 3% of the rest.
And so on, the last guest getting 100% of the last part.

Who gets the biggest piece?

 

[Kyriakos]

A huge pie is divided among 100 guests.
The first guest gets 1% of the pie.
The second guest gets 2% of the remaining part.
The third guest gets 3% of the rest.
And so on, the last guest getting 100% of the last part.

Who gets the biggest piece?

I suppose this formula describes it? (I know it's not the solution; I may well have written the Σ notation wrongly too)

Spoiler :

And I have to assume there's a clever way to go around it. Or you may indeed be asking for something of calculus level (which I'd need to refresh in my memory...)

 

[Thrasybulos]

Spoiler :

And I have to assume there's a clever way to go around it.

There's a way indeed to limit yourself to answering the precise question: you're asked which is the biggest, not what size the biggest piece is.

Of course, answering that second question would yield the answer to the first, so it's a natural reflex to try that first. But if it proves too unwieldly... just remember it isn't asked.

 

[Kyriakos]

Spoiler :

There's a way indeed to limit yourself to answering the precise question: you're asked which is the biggest, not what size the biggest piece is.

Of course, answering that second question would yield the answer to the first, so it's a natural reflex to try that first. But if it proves too unwieldly... just remember it isn't asked.

Ok, I got that there's another way.

Spoiler :

Anyway the corresponding x for max of f(x) would just tell us who (in the progression) gets the largest piece; f(x) of that x would present the value. I will give it more thought

 

[Kyriakos]

@Thrasybulos can you give a clue? (other than the one about arriving at position without something that gives value). Because currently I am very much stuck Since I can't establish this as an ab with b not being defined recursively.

 

[Thrasybulos]

It's a bit hard to give a clue that's both helpful while not giving the whole thing away.
Let me try...

Spoiler Clue 1 :

You shouldn't try to express an equation: an inequation is what you're after.

Spoiler Clue 2 :

Compare what any person gets with what the next person gets.

 

[Kyriakos]

@Thrasybulos there is a minor chance I did arrive at a correct answer, but there very well may be some error with the algebra and the Sums. Fwiw, I am getting a result x=9, as in my solution attempt below, but this may well be wrong anyway or at least incomplete.
I am looking forward to your solution!

My own:

Spoiler :

If this is correct, I take it to mean that the next number will always be bigger than the one before it, as long as you don't go over an integer value of 9=>from 10 onwards you get a smaller piece of the pie=> x E N of max of f(x) is 9.

I am pretty certain it is wrong anyway (tried the resulting formula for 10 slices and it doesn't give a correct result), just don't know what I botched in the algebra (I suspect the inequality division is a part of it). Please help with that if you can...!

 

[Thrasybulos]

Well, you essentially got it.

Spoiler :

You got to the right inequation, so I think you just forgot at the end what it exactly represented, which leaves you to be off by 1.

Person n gets handed piece p.
That person takes p x n / 100 out of it.

They then hand over to the next person p x (1 - n / 100) = p x [(100 - n)/100].
That next person takes p x [(100 - n)/100] x (n + 1)/100 out of it.
So person n+1 gets a bigger share than person n if:
p x [(100 - n)/100] x (n + 1)/100 > p x n / 100
which simplifies to:
[(100 - n)/100] x (n + 1) > n
(100 - n)(n + 1) > 100n
100n + 100 - n² - n > 100n
100 - n² - n > 0

Which is true for n < 10.
So until the 9th person, the next person gets a bigger piece.
Then, from the 10th person onwards, then next person keeps getting a smaller piece.

Which means the 10th person gets the bigger piece.

 

[Kyriakos]

I suck
So I suppose it was just that to answer I should have calculated x+1, with the x I got.
But this is actually good, because I was worrying it was all wrong and therefore I suck worse.

For the case of 10 slices and 1/10 (1), 2/10 (1-1/10) etc, this leads analogously to a n^2+n-10<0=>n<2.7 and consequently <3 =>the third person takes the largest slice.

I am quite enthusiastic about connotations of this for understanding the 2nd degree single-variable polynomial, tbh.

As for my inequality division, it worked, but I have to check why. In general it won't work...

(edit: I redid my calculation for slices=10, and yes, I managed to calculate wrongly; it is indeed f(3)>f(4) ^^ )

Very cool problem, anyway. I will see if I can find something to post next.

 

[Thrasybulos]

I'm out of town this week-end, with limited internet access, so no hurry.

 

[Kyriakos]

Fwiw I tried to ask elsewhere (including Reddit) and it may well be that my method doesn't work with inequalities at all. Maybe (since the right equation is gotten in the end, which can't be just luck) it works with equalities (ie establishing when the function will reach the point where f(x) is the same as f(x+1) for real x) and then using analysis to show that in this type of function indeed this was the max (in x+1). Ie just x^2+x-100=0 being where f(x)=f(x+1), and since before that it was ever rising, after that it has to be falling=>that was the max.
Can't answer that myself, as for the time being I am not as fluent in analysis as needed. But perhaps f(9.5) is indeed the same as f(10.5) and then it can be shown that the max is their halfpoint or something close enough to work for integer 10.

 

[Kyriakos]

This is easy (contains a basic trick, though), but posting because currently I can't find anything better There may still be different starting approaches (I will post mine later).

Just find x. For the record, in the presentation it spoke of x being real.

 

[Thrasybulos]

Spoiler :

Solved it in a very naive observation => intuition => confirmation way.

m1 + m2 + m3 + m4 = 4 => hmm... m1 = m2 = m3 = m4 = 1 ?
2001, 2003, 2005, 2007 => looks like years... the kind of puzzle sites like posting near New Year's day? So the answer should be a recent year...
2001 + 19 = 2020, 2003 + 17 = 2020, 2005 + 15, 2007 + 13 = 2020 => ok, the answer has to be 2020.
Check : (2020 - 19) / 2001 = 1, (2020 - 17)/2003 = 1, etc.
Bingo. x = 2020

So yeah, once you notice each member is expressed as (x - a1)/b1, (x - a2)/b2, etc... with a + b being constant, it's easy to get that ((a + b) - a) / b = 1, so x = a + b.

 

[Kyriakos]

Correct

Spoiler :

It was always going to be difficult for this to not be that subset of the reals. Personally I first noticed that they all have to be at least roughly equal and also very near 1, but there is no room for divergence from the first (if it was over 1, so would the next be, and if under, likewise), so they all are equal and 1=>x=2020.

 

[Thrasybulos]

Those were supposed to be "code puzzles", we've strayed a bit.

Let's get back to the original theme :

Spoiler :

The puzzle is considered solved as long as you decrypt the final message.

 

[Arakhor]

GCHQ often do incredibly difficult challenges simply as a recruiting policy.

The best I've managed to work out so far is that 1 is TIME, 4 is TREE and 5 is STOCKING.

 

[Thrasybulos]

I remember solving it last december, although I didn't get everything.

Spoiler :

For instance, for 5, I gathered it had to be "STOCKING", but I had only solved the "ST" part. I didn't get the other 3 letter groups.

 

[Arakhor]

In part 5, the bottom clue has to be NG because the other six letters are spelling out WRAPPI... I didn't get the middle two either.

 

[a pen-dragon]

for part 5:

Spoiler :

sequence number 3: looking at the position in the alphabet of each letter one obtains: WU = 23,21 -- SQ = 19,17 -- OM = 15,13.
The natural continuation is 11,9, which happens to be KI.

part 6:

Spoiler :

All hints in the text point to binary code. This is a 5 digit encoding, so numbers up to 31, so I assume positions in the alphabet. Either the full or empty notes are 1. Trying out with full = 1, this results in 3,1,18,15,12 or in words CAROL. If empty=1 then this can not represent positions in the alphabet directly due to numbers greater 26, so I disregard that option.

part 7:

Spoiler :

The occuring 4 is very suspicious. Assuming a letter substitution code, the following words have the same structure as 4-letter word. Starting from there the code is easy to solve. My result:
"First solve the code.
Identify a 4-letter word.
Look everywhere.
Maybe it's very obvious."
The way I interpret this is that the code may be any 4-letter word on this card (everywhere), so maybe GIFT (very obvious?) or CARD.