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**wintersolstice**- Replies: 0

This is actually one of 3 things (main things) I've been studying/investigating recently, (maybe I'll post the other 2 at a later date)

It's a sequence I created and I wanted to submit it to the OEIS (Online Encyclopaedia if Integer Sequences) the only problem is, to submit anything you need to register an account and unfortunately, since I don't have any official credentials (I don't work in fields of maths or anything and don't know any who is a member) I can't register. (btw I did search the OEIS for my sequence and far as I know it wasn't there) so I thought I'd try sharing it here. to get some opinions but also to get help studying and generating it. I don't know anything about computer programming, my only tool is an Excel Spreadsheet which is limited to what it can do and it can't offer much help to generate the sequence (I think I generated about 40 terms by brute force, but can't find where I saved them, so I'll only be able to post a few terms.

Any to describe the sequence I need to define a "super-block"

a super-block, is a set of consecutive terms with the following property:

you have a repeated block that appears multiple times in the "super-block" which is seperated by gaps of equal length. Below is a few examples

1)

In this case the repeated block is

and it has a gap of length 3. which are:2)

the repeated block is

and the gaps are

now there are two values

gap count: the number of gaps (1 less than the number repeated blocks)

gap length: how many terms (not digits) are in each gap

the gaps don't have to be the same, only the same length

the repeated blocks have to be identical (same terms, same order, same abundance)

a super block is illegal if its gap count is greater than it gap length

to explain consider

repeated block is

it has 1 gap of length 0 (yes that's still considered a gap)

so it's illegal

same repeated block, this now has 1 gap of length 1 so it's allowed

same repeated block, this now has 2 gaps of length 1 so it's not allowed

also the super block can start anywhere in the sequence

now to the sequence

you start by asking "can I put 1" if so do, if not try 2 then 3 etc

so

(no problemsrepeating block

1 gap of length 0 (this will happen if two consecutive terms are the same

so

no problem

repeating block

1 gap of length 1 no problem

we can't use 1 (because of above and 2 would be another 1 gap of length 0 repeated block of

so

next we might try

2 gaps of length 1 so not allowed so

and so on

but that's a block of

**wintersolstice**- Replies: 0

There's a function (Ackermann Function) (though there are related functions)

Note: I had to improvise a bit with notation, plus I reused a few symbols (letters for variables) sorry if it's long and complicated, I also hope I've done it all correctly

defined as follows

rule 1

i.e.

rule 2

i.e.

andotherwise

rule 3

I think I've come up with another way to express it

take

andand use the different levels of arithmetic

level 0

(successor function)level 1

(addition with B copies of 1)level 2

(multiplication with B copies of A)level 3

(exponentation with B copies of A)level 4

(tetration with B copies of A)and so on...

means level Y so means A to B level Y of arithmeticso based on this

with level 0 you do

Proof

first start with M=0

which is

then take M=1

according to rule 2according to rule 1 according to rule 3 according to rule 1so you get the sequence

etccompared to

etcso

and

according to rule 2according to the previous case according to the previous case

creating this sequence

compared to

which means

and

m=3

according to rule 2according to previous caseso with a first term of 5 and the operation multiply by 2 and then add 3

the formula will be of the form

multiply by 2 then add 3

and

so

and

so

so

now if L is used to mean a given level (where we know the formual works, and I've shown it works up to level 3) and LL means the next level

lets assume it works for a value of

first assumption according to rule 2and given the above rule

since

for any levelalso given that

substitute

according to first assumptionto generalise (is a second assumption, though it works for 0 and 1)

so

substituting

and

with (P+3) 2's)

with (P+4) 2's)

given that

and so on

**wintersolstice**- Replies: 0

Statement: The sum of the angle defects of a Polyhedron (that doesn't intersect itself) is equal to 360 (degrees) multiplied by the Euler characteristic

Proof:

the Euler characteristic is

F (Faces)

E (Edges)

V (Vertices)

lets say that F, E and V are all we know

the angle defects can be represented as

The sum of the angles around each Vertex are

now

and so on....

Add them up

now we know V so we need

This represents the sum of all the angles in all the faces

The angles in an N sided polygon are:

or

If the number of sides in each face are given by

then the the sum of the angles are:

and so on...

Adding

is the sum of the number of edges in all the faces, and since there are two faces to an edgeso substituting that in

and therefore

which simplifies to

rearranging

I figured out whats wrong: (and its not me that wrong)

**wintersolstice**- Replies: 14

I read a puzzle on the main page (under the logic puzzles) and I came up with an answer and a reasoning behind it, however when I checked the answer, it was wrong, today I attempted a brute force search to try and establish which answer was wrong (the answer on the site or my answer. and I still came to the same answer and can't find a flaw in it, also I have a question regarding the given answer

first here is the link

http://www.mathsisfun.com/puzzles/mrs-goodsworthys-hats.html

basically if my solution is wrong then I need to know:

1) whats the flaw in my reasoning? and

2) the other question that was in my hidden text

or is my solution right and the answer needs editing?

thanks

**wintersolstice**- Replies: 1

In regards to this puzzle:

https://www.mathsisfun.com/puzzles/cars-across-the-desert.html

I've found a slightly different solution which I'm sure is valid it also uses the same number of "extra" cars

**wintersolstice**- Replies: 0

sorry if this has been posted.

to differentiate the following

start with the following

then take natural logs of both sides

diferentiate both sides

NOTE 1: since I could seem to write the derivative of f(x) and g(x) as f'(x) and g'(x) I had to change the to h(x) (for g'(x)) and i(x) (for f(x)

NOTE 2 this derivative was done using the chain rule and the product rule

multiplying by y

and sustituting

Hopefully this is right.

To solve:

Define the following 8 variables A, B, C, D, E, F, G and H.

then the roots are:

bobbym wrote:

Hi;

I doubt it would look like mine but post it if you like.

I've said it doesn't look like yours lol

anyway thanks I'll post it later

bobbym wrote:

I'm wondering if I should post mine using the various variables substitution? btw its different from yours

btw my solution involves 8 variables to be defined (like in yours)

bobbym wrote:

What are you having a problem with?

I have solution expressed in terms of other variables which in turn are composed of others and if I do all the substitution it will be big and I want to simplify but am no good at that

**wintersolstice**- Replies: 10

Linear

Quadratic

Cubic

will get quartic soon

**wintersolstice**- Replies: 2

If you spot any mistakes please let me know so I can edit, though I'm hoping there won't be any

where

This is the formula for the area of a triangle whose sides are a,b and c

Proof:

(NOTE: this proof uses Pythagoras' Theorem so in (dia 3) there's a simple proof of that)

(here are the diagrams)

start with the following (dia 1)

now

multiplying by b

substituting that into the area formula

squaring both sides

according to (dia 2)

and according to the cosine rule

whose proof is as follows:

again (dia 1)

(left triangle)and

(right triangle)rearranging the first equation

substituting into the second

expanding the bracket

canceling the X squared and a little rearrangement

from the left triangle

which means

substituting

rearranging

squaring

now the formula derived from (dia 2)

rearranging

which means

multiplying both sides by

now

so

altering the first fraction on the right

so

expanding the bracket on the right (stage 1)

expanding the bracket on the right (stage 2)

simplifying and rearranging

multiplying by 16

adding some terms to the right (that all cancel)

rearranging

factorisation (stage 1)

factorisation (stage 2)

bit of splitting joining and adding terms

rearranging

factorisation (stage 3)

factorisation (stage 4)

bit more splitting joining and adding terms

rearranging

factorisation (stage 5)

factorisation (stage 6)

reordering

little more splitting

dividing by 16

which means

substituting

square rooting both sides

**wintersolstice**- Replies: 4

sorry I didn't know where to post this but I'm having trouble with LateX

**wintersolstice**- Replies: 11

here's an interesting puzzle (no idea what difficulty level this would be.)

the idea is to create 10 different expressions each totaling 12, using 5 of the same digit (so 5 0's, 5 1's, 5 2's, 5 6's, 5 5's, 5 7's 5 8's and 5 9's)

rules:

1)you have to use EXACTLY 5 of that number no more no less

2)you can't use any other numbers (e.g.

would add a 2 you can't do that)3) you can use any arithmetic symbols, the √ symbol (since it has no number on it) and factorial (!)

4) you can also use decimal point and recurring sign (e.g. 8 = 8/10 and 8(recurring) = 8/9

5) you can't put two(or more) numbers together (e.g. you can put two 5's together to get 55 and you can't do 55 =55/10

6) you can also use brackets

there may be multiple solutions to some of them:D

heres the first one to start you off

EDIT the following solution is invalid since it works for all digits (except 0)

D/D + ((D+D)/D)

any other such solution is also invalid

vikramhegde wrote:

Ok so the procedure is like this.

First consider a triangular number sequence [ fn = (n+1).n/2] for n = 1,2,3..

The 18th number on this sequence is 171.

And sum of all triangular numbers till 171 (the tetrahedral number) crosses 1000

Now, for each subsequent drop of the first ball (when it doesn't break) we add the previous number on the triangular sequence.

So ball number one is dropped for the first time from Floor number 171, second at (171+153)=324, third at (324+136)=460, fourth at (460+120)=580, fifth at (580+105)=685, sixth at (685+91)=776, seventh at 854, eighth at 920, ninth 975, tenth at 999. Let us call these numbers M#1, M#2, M#3....Now, if it breaks in the first try at 171, we start dropping the ball at floor numbers 18, 18+17, 18+17+16,

Similarly for any break of the first ball on floor M#x, the floor number at which we drop the second ball is given by -

{(M#x-1)+P} and if the second ball doesn't break at this, we continue on the sequence - {(M#x-1)+P+(P-1)}, {(M#x-1)+P+(P-2)}, {(M#x-1)+P+(P-3)}... {(M#x-1)+P+(P-P)} Where P is the position of {M#x - (M#x-1)} on the triangular sequence.Having broken the second ball somewhere, we go back to the last try where we didn't break it and work our way up with tries on each floor till it breaks.

We will find that 19 is the least number of tries required.

I'm sorry I'm not trained in mathematics and hence have to put it in such a round about manner. I'm not so familiar with the notation and the use of sigma functions and had to invent some notation of my own. I hope I've explained it adequately.

Also, I'd be happy if someone could explain it a more simple manner.

just to correct you on that it should be M#x-1 + p + 1 to see why imangine dropping the first ball from 172. You can still do it even if it breaks on that floor other that your solution is spot on:D

NNikolay wrote:

Wintersolstice, I have showed how your proof is wrong and have given a solution. What else can I do?

Second modification has a solution too. 2 questions with 3 possible outcomes give 9 combinations! Of course you loose some information due to Random behaving really randomly and because you don't know the words upfront. But it is enough to separate 6 cases.

Hope you will read the solution before telling it's wrong

I completely misread your last post that's all:D When you said "modification" I thought you meant you had modifiyed your original solution! and when you posted your "second modification" I thought it was describing your solution! (giving clues I mean:D)

Sorry about that!:D

anyway I've had a look at the solution but it's going to take a bit of time to get my head round it (although this has ruined for me! Nevermind there's the other puzzle)

btw have you seen my "variation" that I mentioned early (it's not that difficult really though)

NNikolay wrote:

4. You need a solution in two questions.

but that's impossible, it can't be done in two questions, look at my "proof" above it shows that two question isn't enough, if you need it explaining I'll try and explain it better for you:D

try out EVERY combination of answers you get from the two questions (that give you information) and see if that can tell you who's who, some combinations will tell you but some combinations won't.

btw I haven't looked at your solution yet but with my proof I don't think I need to.

MathsIsFun wrote:

There is an option "Never show smilies as icons for this post" in the Post Reply form (in case you really want to type )

sorry to revive this thread but even what wa said tin the quoted post deson't work!

NNikolay wrote:

Hi wintersolstice,

Your proof is based on the assumption that first of 3 questions can't give you any information except sample word from god's language. This is not correct. I mean, this is correct if you have one question only. But in our case you can get more information as a result of all 3 questions.

what I assumed is that you were using the "one question to determine yes/no followed by two questions to work out who's who (which you've proved wrong:D)

however the reason I believed that is because:

I tried the original solution on this version and realised that when you ask the first question (because you don't know any words) you can't actually find any information from the first question apart from a sample word even after the second tow questions.

here the original solution and you see if you can find information form the first question

this is just one solution but you'll find that using this solution is impossible on your version.

the only thing I can suggest is:

write down your own solution of 3+ question showing what the questions are asked to who and what in cirscumstances is each question asked(without showing anyone of course) and see if it works for all 6 possibilites (of who's who) and all possible answers and true and false questions and what is yes and no etc), this way you know it has a solution:D. That's what I did when I posted my variation of the puzzle:D

btw when I posted my proof I only said I don't "think" it's impossibe I never said I "know" it's impossible:D and it was imcomplete so hopefully I have completed it now:D

anonimnystefy wrote:

Hi wintersoltice

I've no idea what you're saying??? plase explain.

I put a question mark after my answer (-i?) to show I wasn't sure if that was right

bobbym wrote:

Hi;

I thought is was just me. I do not get that either.

I think I do get it actually:D

mathgogocart wrote:

correct i,-1,_

-i?

I don't think this puzzle is theoritcally possible! (though I could be wrong:D)