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7:11 AM
I found something interesting. Let's represent your equations as $A+B=C$. Let $A\| B$ mean "A concatenate B" and $d(x)$ be the "digits of some assigned $x$". Also, let $n$ be the number that corresponds to the position of each equation ($n=1$ for first equation, $n=2$ for second equation, etc) then for some value $D$, if $$D=\frac{A\| B-d(A)\cdot d(B)}{\frac{1}{2}|A - B|+2-n}$$ then $D$ is an integer, and close to $C$ :) — Mr Pie 24 mins ago
@MrPie ^^ how does that work with 5 + 19 = 97?
@TheSimpliFire we get 79
97 :D
from what I gather $$D=\frac{519-5\cdot(1+9)}{\frac12|5-19|+2-3}$$ not an integer
Oh..... wait a sec, lemme try again
How do you define d(x)?
sum of digits?
@TheSimpliFire no
Just the digits
7:14 AM
so d(x) = x?
So we wouldn't do $5\times (1+9)$ we would do $5\cdot 1\cdot 9$
that makes sense
let's try another one
@TheSimpliFire d(x) was just the set of its digits. Yeah, saying "d(x) = digits of x" is a bit vague, my bad
D = (753 - 7*5*3)/(0.5(53-7)+2-1)
I think we should get 27
7:16 AM
How did you come up with that?
@TheSimpliFire trial and error :)
There should be a reason why D is an integer
(heh, me being a nerd)
idk really. I just go with my gut, or my instinct, I guess...
We know that either A or B is single digit
and the other one is a double-digit
Consider the case where A=10x+y and B=z
Ah yes
7:18 AM
Then $$D=\frac{100x+10y+z-xyz}{\frac12|10x+y-z|+2-n}$$
geez, that was quick. Substituted and wrote out the latex??
I think while I type in MathJax ;)
2 years of experience
@TheSimpliFire I have to do separately or else I stuff up the mathjax, from my experience. Otherwise I'd be constantly editing my answers and the question will be on the front page for hours xD
so now we need to solve for n
7:21 AM
so firstly y-z is even
if y-z is odd then the other 1/2 term must come from the other fraction
can we factorise $100x+10y+z-xyz$? I've been trying, and can't
@MrPie The weird thing is that for some value of n your equation does produce an integer
E.g. 86 + 7
@TheSimpliFire I sense a conjecture coming...
n=12 solves the problem
@MrPie Apparently $$\frac{531}{\frac{79}2+2-12}=18$$
faceslap i made a mistake
7:27 AM
haha 9-5-1 perhaps :P
@TheSimpliFire will there always be an $n$ for three random numbers?
Two random numbers with one being single digit and the other being double
@TheSimpliFire yeah, I was just gonna add that
$$\frac{(123-1\times 2\times 3)}{\frac{23-1}{2}+2-\color{red}{0}}=9$$
Is $n$ always positive?
It will not work if $100x+y+z-xyz$ is a prime number @MrPie
@TheSimpliFire if we can factorise the equation then we can see when it can be prime or if it is never prime, but I can't. (Imma be lazy and go on wolfram alpha and see)
7:32 AM
let's find a counterexample
z must be odd
That means $y$ is odd
Well $y-z$ is even
9 mins ago, by TheSimpliFire
this worked
even though we had 79/2
so I had an initially wrong observation
@TheSimpliFire Ok, let $D>0$, and $n\geq 0$. That limits our choices for $n$.
7:34 AM
at least one of x,y must be even
431 - 4.3.1 = 419 is prime
Forcing n=22
So there will always be a solution @MrPie
Set $\frac12|A-B|+2-n=1$
@TheSimpliFire for primes
Not only for primes
Composites are divisible by 1 too ;)
$n=\frac12|A-B|+1$ will always work
@TheSimpliFire -_- but now it's less interesting...
If we don't do that then we have found a counterexample
Just make the top prime
@TheSimpliFire or, we can just make some restrictions and say the numerator can't be prime.
And now we can say that there might be at least two values of $n$
Where one of them is $\frac12|A-B|+1$
7:38 AM
We can separate two cases
A-B is odd, A-B is even
Note that we are still doing the case where A=10x+y and B=z @MrPie. The case A = x, B = 10y+z is different
@TheSimpliFire ohhh, I thought that didn't make a difference because of abs
...clumsy me
A>B in this case
If A-B is odd, we can do the following
For every increment +1 of n, the denominator reduces by 2
Therefore at some point it will reach 1, where we just left off
Whether the numerator is prime is currently irrelevant
If A-B is even = 2k, we can do the following
But now $k$ needs to be an integer, right? Oh wait, it will be because $A$ and $B$ are integers
@MrPie yes
we have $n\in(0,k+2)$
Since A-B is even, A and B must be of the same parity
If both A and B are even, then the numerator is divisible by 2, and we can set n = k to make D an integer
It is trickier when both are odd
If they are odd then that means $k$ and $n$ have an opposite parity, but this is obvious trivial (better word). Trying to look for some more depth here ;)
7:46 AM
For example, the case with A=43 and B=1 yields no solutions apart from $k+2-n=1$ which is trivial.
Hang on
If both are odd, and at least one of x,y is even, then D has no solutions at all!
Let $A=10x+y$ and $B=z$ as you said... oh YEAH
Thus for any possibility of a solution, all of x,y,z must be odd
xD I have found basically nothing on my own except the equation itself
That's why A=43 and B=1 doesn't work as 4 is even
What about $A=16$ and $B=6$?
7:49 AM
5 mins ago, by TheSimpliFire
If both A and B are even, then the numerator is divisible by 2, and we can set n = k to make D an integer
If $x=1$ then the rest are even then that's a solution
@TheSimpliFire ik, but I thought you meant $x,y,z$ must always be odd. Sorry, I read too hastily
@MrPie They must always be odd if both A and B are odd (better phrasing)
Let x=2a+1, y=2b+1, z=2c+1
Actually, since $10$ is even then if $y$ and $z$ are even, we are good to go.... which sense because if $y$ is even then. Again, obvious, but maybe this even trait more specifically could help us?
Or I'm just barking up the wrong tree since we already covered this ;)
man, I've lost it XD
THat might be useful
@MrPie Now we need to find out when D is an integer and when it isn't
Note that k+2-n is divisible by 2 so we're good
@MrPie screw that, I'm writing a program in PARI/GP :)
$$D=\frac{5(13+2a+2b)+2c - 4abc-2ab-2ac-2bc-a-b-c)+65}{k+2-n}$$
7:57 AM
much quicker
@MrPie -2(4abc...) not + :)
and also -1 at the end
and are you sure it's not 8abc :P
@TheSimpliFire no (I don't think) because we are multiplying all by $2$, but on the previous note, yeah you were right since we are subtracting
ah yes
note 110/2=55
Nooo! Bloody timer!!
Pretend there is no $65$ in the numerator
@OmegaKrypton Some crazy formula MrPie made up here
8:00 AM
oh this one... i saw that... CRAZY
@MrPie 5*11=55 and 5*13=65
@TheSimpliFire yeah. I put $13$ and not $11$
.....don't make me check this all again ;) (nah, I'm not that mean)
I get this
you can cancel some a's, b's and c's
I think ours are equivalent save the slight typo at 13
@TheSimpliFire yeah, I just went straight to factoring for sake of parity, but you just got the like terms together and then factorised, which I should have done....
The main question. What next :)
8:04 AM
Let $n=k$?
I mean, whatever $a,b,c$ is (likewise, whatever $A$ and $B$ is), the numerator is even
@MrPie You solved it
So we can formulate the entire thing as follows
@TheSimpliFire nah, I think I can tell that you knew before I did, considering that you put a little :) at the end ;)
shhh :P
but eh, thanks for the tick haha
Oh, and one last thing for you:
@MrPie is that another answer for the simple arithmetic #8
8:07 AM
if so post it as an answer, it'll get +1s lol
:) :) :)
@TheSimpliFire wait.... did we prove it?
Because $413-4\times 1\times 3$ is not even
$\Large\color{blue}{\stackrel{\bullet\bullet}{\smile}} \Large\color{blue}{\stackrel{\bullet\bullet}{\smile}} \Large\color{blue}{\stackrel{\bullet\bullet}{\smile}}$
@MrPie Did you forget that 4 is even so the trick doesn't hold
8:09 AM
puts his tick on a crossbow and shoots it out the window
considering that T sounds like D amidst laughter, sometimes.
In conclusion, for $A=10x+y$ and $B=z$ with $x,y,z\in\Bbb N_{\le 10}$, the value of $D$ will always be an integer. If $A$ and $B$ are odd then set $n=\frac12(A-B)+1$. If $A$ and $B$ are even then set $n=\frac12(A-B)$.
:D @MrPie
@TheSimpliFire Dammit!
oops did i reveal something @MrPie
Q: A Smuggled List of S'S

Omega KryptonSmuggled this strange list from [redacted] 's place. Maybe he is making a puzzle. Who is the victim? Scarlet covers Swindeler imprecation Sour disordered equipment Smash journal Sinful America Surgeon oath Storms carport Working for Generalist Countdown - an...

Yes i do, but i won't delete it. you guys would have a little head-start. shhh...
@OmegaKrypton hehee
that was a very delayed ping
ok, to the S'S!!!!!!!
8:14 AM
yay :)
sinful america might be related to wild west
not sorry :P
typo at swindeler :P
oh thanks
@OmegaKrypton doesn't that change the anagram of that?
Wait.... who said I should anagram the words on the list? ;)
and swindler should be plural, edited
@OmegaKrypton ok, well thanks to two more specific tags, I think I know the first step to take, perhaps.... I have 4 hours spare :P
hehe, just realised I'm in the top 11 in terms of rep on OR ... and yet I have no experience in it XD
8:21 AM
To my surprise, there isn't which I would have expected coming from one of the other tags. Hmmm...
minimal, so didnt add it
If I'm being honest, I joined OR because there was a lot of maths notation and not because of the programming behind it
anyway, im sure you have that piece of little knowledge
If I'm being honest, I joined OR because @TheSimpliFire joined it and I wanted to get a small id number :P
@OmegaKrypton Coincidentally, I am the 123rd user of OR
8:23 AM
@TheSimpliFire @OmegaKrypton fun fact: in the centre of the word "America" with letters "America" I can make the word crime. MWAHAHAHA Cue sound of thunderstorm and flash of lightning
@TheSimpliFire i know
You're not wrong @MrPie
@MrPie Core of America = Messy crime :P
@TheSimpliFire and this is why Trump is the president anagrams are fun
For a moment I thought that was an anagram @OmegaKrypton until I saw the double s
8:24 AM
@TheSimpliFire what was an anagram?
the title?
The title is not a hint
Core of america ~
@MrPie Cue sound of thunderstorm and flash of lightning - should be the other way round :P
@OmegaKrypton YOU ARE A GENIUS
no im not
(for the lightning and thunderstorm thing)
8:28 AM
in OK's 屋企(home), Jun 21 at 7:19, by Mr Pie
I, user 'Omega Krypton' = OK, a top merry genius!
you see, YOU made this
"Swindlers imprecation"
"Swindlers imprecation"
"Swindlers imprecation"
stop that
That is either the best coincidence to date or just very awesome phrasing - either way, it's still very awesome ;)
8:30 AM
@MrPie coincidence :P
I do this face a lot: ;)
My right eye is always winking
my left eye cant wink
@MrPie @TheSimpliFire you see what i had removed?
We can see
8:32 AM
good shhh...
you guys just saw something from me that you should see. i removed them already
"Swindlers imprecation"
^this is the easiest one, to you
hello @Haran talking bout puzzling stuff
I am no swindler!!
Hello @Haran
8:36 AM
@Haran Am revisiting this
Q: Does there exist polynomials $P$ s.t. $P(k)\mid k!$ holds for only finitely many $k\in\Bbb N$?

Tianlalu Question: Does there exist polynomials $P$ with integer coefficients such that $$P(k)\mid k!\tag{*}$$ holds for only finitely many $k\in\Bbb N$? I think such polynomial doesn't exist (don't know how to prove it). For some specific polynomials, we can prove there exists infinitely many $k\in...

@MrPie @TheSimpliFire ping me at my home if you have any ideas ;)
I put at +50 bounty on yesterday
@OmegaKrypton by the way, I've always known for "imprecation" to mean "a swear word" (thanks to my humanities teacher). But searching up the definition for a clarification, it says "a spoken curse". I don't know what it means by this. Something like "hakuna matata you're now a banana" or a curse/swear word?
Hint usefulness level 1 (will post to public later): one for one
8:40 AM
@OmegaKrypton ahh, gotcha ;)
Honestly I have no idea what's going on ;)
@MrPie tell him, if you wish... i wont repeat [evil smile]
@TheSimpliFire it is basically the equivalent of Spongebob's "Holy Mother of Neptune!"
well yeah, I know what imprecation means
Sometimes I'm naughty and sometimes I'm not. Sometimes I delete messages and sometimes I don't :)
8:42 AM
Or Marty's "Sugar Honey Ice Tea!"
I can read them but I can't be bothered atm :P
Or my [censored]
kicked from the chat due to indecent language
Current situation: @TheSimpliFire is either doing whatever is bothering him/her. @MrPie is working hard on my puzzle but gets distracted. @Haran has no idea of what we are talking about. I am laughing at @MrPie 's puns and anticipating for some intelligent guy/ gal's answer
Haran has left :P
8:47 AM
@OmegaKrypton nah, I'm looking. I've just got separate tabs open (one for the puzzle, one for this).
just kidding for the purpose of kidding ;)
@OmegaKrypton I'm writing a dissertation so I'm kinda busy...
I'm here to procrastinate
wow keep it up!
Q: The Puzzling Times

Rubio$$\bbox[orange]{\begin{array}{rcl}\\\hline\huge\ \ \star\ \ \star\ \ The&\huge{Puzzling}&\huge{Times\ \ \star\ \ \star\ \ }\\\hline\\\end{array}}$$   Vol. 4, No. 1 $\raise 2pt \tt{\large{\ \ \ \ \ \ \ \ MONDAY,\ JANUARY\ 9^{th},\ 2017\ \ \ \ \ \ \ \ }}$ Price: One      $\raise 55pt\begin{arra...

@OmegaKrypton or maybe I am. I saw "The Puzzling Times" on related questions.... it looks amazing
an applause for Rubio! :)
I saw that!
8:48 AM
I bookmarked it. REAL impressive.
They must've spent weeks creating that
And it's all in MathJax
I have no idea what to anagram. I believe we are supposed to use the words and find synonyms or other substitutes of some sort - then anagram. I tried to also look between the letters and find something else (because if the S thing was deliberate, some other things might be deliberate)
no comment
@OmegaKrypton I did find something interesting, maybe/maybe-not related. Am I allowed to say something, maybe say but then remove it, or make no comment on it?
8:52 AM
When you align the S's ("S's" from the title), it makes a K. Omega Krypton??
nope. coincidence :P
no visual/ steganography tag
@OmegaKrypton instead of "swindler's imprecation" with an apostrophe, it is written without one
8:55 AM
doesnt matter
will add as a red herring
SHHH... :P
Btw, @MrPie, have you seen the game that Mathphile made himself?
2 days ago, by Mathphile
user image
so realistic
It's starred (*2) on the right tab
*3 now
8:58 AM
@OmegaKrypton wait, was it actually a red herring? I just saw the edit on the puzzle ;)
yes it really was
just to mislead others but not you
(cancelled to avoid duplicates :P)
@OmegaKrypton aw thanks :P
@OmegaKrypton haha, maybe you should start capitalising random letters to create the illusion that they are important
Emm.. what should i add?
this is what we say "self-important"
@TheSimpliFire were OK emphasised? OK
9:01 AM
@OmegaKrypton ok, so another word for "imprecation" as it is meant, is "swear" I think. I can make the word "swear" from "a swindler"
It will be funny if OmegaKrypton was the victim
Maybe it's jafe
Maybe it's you, @TheSimpliFire!
Or you
@TheSimpliFire nope
nope to all :P
9:02 AM
I'm guessing one of the top100 users
Maybe it's Bass? Deusovi? Bob the Bui---ok, this isn't getting anywhere
... lets use the right way to get the answer... an exhastive search is so complex, you all know :P
@MrPie you are no longer on the right track
@OmegaKrypton ok, I am fully guessing now that we find synonyms, substitute with what we are given, get an anagram. I have tried to avoid the anagram part and find other things first so I don't miss anything, but now this is slowly becoming more of an excuse of procrastination
right track, there's more, but you'll know later on
@OmegaKrypton firstly, "scarlet" is a red colour. Also, splitting the word in two, since it is a compound word, we get "scar let" and scars are red, too. So I will substitute RED... however.... I think the S's are important.
9:09 AM
why split?
and i realiSed i am So SucceSSful
the very firSt time
I dunno.... "coverlet" is also a word (seen on the anagram software)????
no wordplay tag
The trouble with "covers" is that this word has so many meanings... but I believe it is a noun (considering we have "scarlet" as a possible adjective in the same way as "sinful america").
hey, "red" is also the prefix for "redacted" so maybe that's the right word?
even though wordplay is not involved... hm
coincidence, but right\ word :P
@OmegaKrypton Oh yes! Awesome.... but I do have probably another coincidence.... I can make the word "cost" from "Scarlet covers" and the next line has "swindlers" in it which revolves around costs and money; also, I can make the word "US" from "Smash journal" and the next line has "America" which is the US.
agh crap, just cracked my neck and got whiplash ;[
@OmegaKrypton aha! Replace "scarlet" like so, and "covers" with "coats" and I can make the word "Redcoats" (not even a real anagram) but also, I can make "So redact!" ....... enigmatic puzzles are hard. I won't bore you with my attempts, but I'll try my best to figure this out ;)
9:26 AM
1. not bored, partial ans.s welcome!
2. sorry but wrong
9:47 AM
There are so many words to choose from!!
10:00 AM
Oh my God, I have been messing around with words and anagrams, trying to look at profile pictures of people's usernames. Man, I don't know where to go. I'm just running in all directions, and can't tell if an anagram is or isn't meaningless.
And considering the question has been up for 4 hours with no comments, no answers (though 42 views.... should be more!) then perhaps it is too difficult.
I have been sitting down for 2 hours. I need to stand up xD. Don't take my messages too personally, though (I did sound a bit blunt, my bad). It is technically a compliment (difficult puzzle, so an awesome puzzle-maker you are!) but hey, difficulty is relative :)
10:53 AM
@MrPie and sorry i mean red is correct but not coats
2 hours later…
1:02 PM
I've lived in the UK for 10+ years and have never heard of connexion
3 hours later…
4:13 PM
@Haran Hi.
@Grešnik hello
We are trying to get some results on this, do you have any ideas?

Nice problem
By the way @TheSimpliFire I worked for a few minutes on the $P(x) \mid x!$ problem. I was able to solve the problem for $\mathrm{deg}(P)=1$ i.e. all linear polynomials.
@Grešnik The heuristics answer provided is pretty good and gives a good idea of what the idea would be. If you want to prove anything, it would be proving the non-existence of power summers of order $\infty$.
4:30 PM
Yes of course, I also believe that there is none of them.
@Haran I would be shocked if there would be an "infinite power-summer" , but a proof that there is none would, of course , be nice.
@Peter I'll see what I can do @Peter
I really do not think much about that problem, my "focus" is now mainly on set-theoretic issues.
5:03 PM
@Grešnik hello
I have a question of trigonometry. Can you help me ?
I do not know, I haven´t done trigonometry for some time, what is the question?
How do I resolve sinh (2ln5)?
hyperbolic sine
ln5 = 5
But what do I do with 2?
Hyperbolic sine is linear combination of exponential functions, just plug 2ln5 into the expression, 2ln5 equals ln 5^2=ln25
@Grešnik Thank you!
Where can I find out more about this?
About what? Logarithm and exponential function?
5:09 PM
You have here some rules for the logarithm function:

Thank you!!!
For the question about hyperbolic sine just see what are the basic rules for logarithm function and for the composition of logarithm and exponential function.
And see how hyperbolic sine is defined as linear combination of exponential functions.
2 hours later…
7:41 PM
Who in their right mind decided to define non-increasing as decreasing and non-decreasing as increasing!!!!
I spent a good hour with someone who spent a good two hours on a question until someone pointed out that non-increasing does not mean not increasing!
okay I'm done ranting thanks for having me
7:57 PM
Q: Are these continued fractions of integrals known?

TheSimpliFireMotivated by this paper on polynomial continued fractions (Bowman, 2000), I thought about various extensions to current definitions of these fractions. What if we defined a continued fraction such that the numerator of each fraction is the integral of the previous one? That is, define th...

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