/sci/ - Science & Math

>Can you solve the unabomber's math trick?
You don't solve tricks, you use tricks to solve problems.

>>10016167
He calls it a trick himself if you read it

>>10016158
>Armand
>Bartholomew
>Claudius
such based names from based Ted

>>10016158
I tried to work it out and got multiple solutions where they can all start with 50 cents but there is only 1 solution where only 1 has 50 cents to start and its CIaudius, is this the correct answer?

>>10016158
Where did you get that jpg, OP?

So I tried solving it out, and I ended up with an expression for what Bartholomew and Claudius started out with in terms of Armand's starting value. It turned out to be
[eqn]B=\frac{5}{3}A[/eqn]
[eqn]C=\frac{4}{3}A[/eqn]
where A is free.

Then I tried setting A, B, and C equal to 50 cents. The only one that worked was Bartholomew because the others resulted in fractions of cents, and I guess that's the trick.

>>10016158
2a, b-a, c
2a, 2(b-a), c-b,
2a - c + b, 2(b-a), 2(c-b)
This is the representation of exchange of money at each step.

Now they all have the same amount of money so

2a - c + b = 2(b-a) = 2(c-b)

Therefore, 4a = 3b = 3c

Clearly, there is no integer solution for a,b, and or c starting with 50 cents.

>>10016230
no you made a mistake

the equation is:
a+b-c = -2a + 2b = 2a - 2b + 2c
which you can solve as:
c = (4/3)a
b=(5/3)a

which implies it had to be bartholomew
because
>if claudius started with 50 pennies, then armand started with (4/3)*50 pennies, which is not an integer number of pennies
>if armand started with 50 pennies, then bartholemew started with (5/3)*5 pennies

however if b started with 50 pennies, then armand started with 30 pennies and claudius started with 40 pennies

btw at the end of the game they all had 40 pennies

>>10016260
oops that second greentext should read

>>if armand started with 50 pennies, then bartholemew started with (5/3)*50 pennies, again not an integer number of pennies

>>10016260
Seems reasonable but can you show your steps to get to the equation? I am confident you didnt distribute one of the b terms correctly.

>>10016260
seems like you got the same result as mine (>>10016229)
yay, -1 brainlet points for me

>>10016270
so A B and C start with:
a, b, c
then after the first game
a+a, b-a, c
then after the second game
a+a, 2b-2a, c-b+a
then after the final game
a+b-c, -2a+2b, 2a-2b+2c

>>10016158
>number theory
he really was a wackjob

>>10016377
it's not really number theory, it's just algebra over the integers. not deep at all imo

the problem is a tough algebra problem though, as you can see by it having flummoxed at least one /sci/ anon

>>10016286
I disagree with Claudia's winnings after game 2. That makes Armand and claudia different than what I had.

B obviously, this is completely basic.

hold on what context was this letter writen
this sounds extremely friendly, why would he be helping some kid(?) with math

>>10016617

>>10016617
See

>>10016623
How powerful is he?

>>10016192
Playing a nasty trick by giving a difficult problem is different from creating a "math trick"

>>10016561
okay, maybe i made a mistake. show me where i made the mistake, or at least show me why my solution is incorrect

i think i got it right honestly. and the fact that i'm wasting time on the unabomber's math problems instead of writing my phd thesis is a bit embarassing

when were the most recent photos of him taken and where can I find them
76 is pretty old, none of the pictures I can find look 76

>>10016623
No wonder why I get "how" all the time

>>10016158
can you actually write mail to him?

>>10016628
well he’s a genius; he failed to bring down industrial society and return humans to hunter-gatherer culture and got put in a cell... personally back to the trees sounds dumb but he has a point about losing our dignity and environment to the industrial machine

>>10016623
>4/7 for getting the right answer but using notation wrong
god math pisses me off
ive had math teachers like this
piss off its the right fucking answer isn't it, i can implement it on a computer and do something real and practical with it so I'd say that counts as getting it right

>>10017338
>4/7 for implementing the right algorithm but using the brackets wrong
god programming pisses me off
ive had programming teachers like this
piss off its the right fucking algorithm isn't it, i can write it in pseudocode and prove it works so I'd say that count as getting it right

>>10016731
it'll probably get you on a watch list

>>10017402
what if you buy his books from amazon
where the hell is the money going anyway, he sure as fuck can't use it and why would he want his family to get any of it

>>10016158
it was only impossible because it was about money. Rewrite it as a math problem and it works. I know because I was there when it happened, although it didn't look like it did and I wasn't the mathematician in the room and I don't know who was

I presume the Unabomber was hiding.

I wrote the final quantities in terms of starting quantities and solved the matrix. However I just guessed how many pennies they ended up with starting with 50 beacause why not. That leads to A and B having fractions of pennies, I tried some other numbers until 40 in that case:
>a0=30, b0=50, c0=40
But I don't really know how could I prove this is the only possible answer where someone starts with 50 pennies tho.

>>10017472
R0. 30,50,40
R1. 60,20,40
R2. 60,40,20
R3. 40,40,40

It checks out. Therefore I was wrong and anon was right. His solution is in the thread.

2A - (C - (B - A)) = A - (C - B) = A - C + B
2(B - A) = 2B - 2A
2(C - (B - A)) = 2(C - B + A) = 2C - 2B + 2A

A - C + B = 2B - 2A
3A - C = B
C = 3A - B

A - C + B = 2C - 2B + 2A
3B = 3C + A
C = (3B - A)/3

2B - 2A = 2C - 2B + 2A
4B - 4A = 2C
C = 2B - 2A

3A - B = (3B - A)/3
9A - 3B = 3B - A
10A = 6B
A = 6B/10
B = 10A/6

A cannot be 50.

C = 18B/10 - B
C = 8B/10
B = 10C/8

C cannot be 50.
B must be 50.

If B is 50:
A = 6(50)/10 = 30
C = 8(50)/10 = 40

Trying A=30, B=50, C=40: 30 50 40 -> 60 20 40 -> 60 40 20 -> 40 40 40

>>10016158
Too much reading, don't care.

>>10017536
Reading is for losers xd op is such a fucking nerd lol

Lol this is fucking great.
I was actually planning to send him a letter myself.

>>10018195
it looks like he would just say "thanks for the letter read my book"

>>10018219
This, he's to busy and prison is constantly trying to fuck with his mail.

Answer: S starts with 50 cents.

The proof is as follows:

Let A,B, and C denote the amount of pennies Andrew, Bartholomew, and Claudius initilly possess respectively. Then, by round:

R1: (A,B,C)
R2: (2A,B-A,C)
R3: (2A,2(B-A),C+A-B)
R4: (A+B-C,2 (B-A,2(C+A-B)).

The condition that R1=R4 implies that:

1: A+B-C=A; B=C
2: 2(B-A)=B; B=2A.

Hence, in terms of A, R4=(A,2A,2A); if only one person starts with 50 cents, it must logically be A since B=C whenever R4=R1.

Thus, A begins with 50 cents.

QED

Feel free to input A=50 into both R1 and R4 to convince yourself of this proof if you've yet to be persuaded.

>>10018380
>The condition that R1=R4 implies...
The condition is that in R4 they all have the same amount of pennies, not that they end up with the same amount of pennies they initially had.

>>10018428
Fine then

B is the answer.

Proof:

Let R1=(A,B,C) and R4=(A+B-C,2(B-A),2(C-(B-A)))=(A4,B4,C4) Moreover, fix A4=B4=C4.

Then the condition that B4=C4 implies that C=2(B-A). One obtains upon inputting this into A4 and equating with B4 the expression:
A4=3A-B
A=3B/5.

This further implies that R1=(3B/5,B,4B/5) and R4=(4B/5,4B/5,4B/5).

The constraint that each element A1,B1, and C1 of R1 are natural numbers implies that B mod 5=0 or B=5n where n is an element of the natural numbers. Substituting, we have:

R1: (3n,5n,4n)
R4: (4n,4n,4n).

Naturally, any n will produce an integer value for all associated elements of R1 an R4; this satisfies our first constraint. Our next constraint is that A, B, or C =50.

Looking at R1, we see that neither A nor C can satisfy this constraint because neither 3n=50 or 4n=50 produces a natural number.

Hence we conclude that n=10 and B=50 with the initial round:

R1: (30,50,40).

This completes the proof.

>>10018380
honestly the question has already been solved by a few anons, the answer is bartholomew, and even the anon who disagreed later admitted he made a mistake

plus, wtf is S? friggin retarded

>>10018485
I would call into question your own intellect if you can't imagine a situation in which S was accidentally hit rather than A.

Additionally, thoroughness and clarity are paramount when writing proofs. It's not enough to say the answer is B; it must be proven that it can't be anything else.

>>10016158

theres 3 solutions in form (A,B,C)

A=Arnold, B=Bartholomew, C=Claudius ; Total that they all end up with = "Total"

---------------------------------------------------------

Solution 1: (50,83.3,66.7)
Total: 66.6 cents a person

Solution 2: (30,40,50)
Total: 40 cents per person

Solution 3: (37.5,62.5,50)
Total: 50 cents per person

Check em.

>>10018723
yeah but pennies are not divisible (duh that’s why he said pennies) so solution 2 is the only solution, as posted earlier in this thread

>>10018723
Only solution 2 is valid; pennies don't have physical subdenominations

>>10018723

***** for solution #2, switch 50 & 40

why did ted quit his mathematics professor position?

>>10018484
Yes, that's more like it.

>>10018765
to intelligent too work

>>10016158
a b c
30 50 40
60 20 40
60 40 20
40 40 40

if this dude was so smart then what did he think blowing up a building could reasonably accomplish?

>>10019836
Do you think voting accomplishes anything?

>>10019836
Committing extreme acts worthy of national news headlines is the only way for a proponent of fringe-issues to get his/her word out to the population at large. Do you honestly think that any common folk - or bright people, for that matter - would´ve reflected on these issues by seeking out his manifesto?

>>10019836
he didnt blow up buildings. he killed people who were in tech jobs.