Well?

Well?

Other urls found in this thread:

en.wikipedia.org/wiki/Irrational_number
en.wikipedia.org/wiki/Cantor's_diagonal_argument
en.wikipedia.org/wiki/E_(mathematical_constant)
en.wikipedia.org/wiki/Fractal
twitter.com/SFWRedditGifs

Harsh roll mein friend

You're fat.

21

15+5+1+1=22

21

Threads over kill yourself faggot nerd

This

kys

Finally, some non retarded anons.

Got three wrappers left m8
Have yourself an extra one

You're right

15+5 = 20 you retards.

Please tell me you're only pretending to be this stupid

right after I posted 21 I realized it was actually 22

...

if 1 chocolate is 1 usd and you have 15 usd that is 15 chocolates. if 3 wrappers is 1 chocolate then thats 15 divided by 3. you dumb cunt.

can't you ameritards count?

15 chocs = 15 wrappers.
15 wrappers = 5 chocs.
5 chocs = 5 wrappers.
5 wrappers = 2 wrappers and 1 choc.
1 choc = 1 wrapper
2 + 1 = 3
Those 3 wrappers turn into one chocolate. You just got 22 chocolates.

How many wrappers do you get from those 5 extra candy bars user? and How many wrappers from them?

That leaves you with five wrappers
Three of wich you exchange for a chocolate
Eat it and that leaves you with 5-3 = 2 and add the bar you just ate 2+1 = 3
Exchange the wrappers for a bar and that gives you the last one
Hence 22

Again stop pretending you're this stupid user

69 xD

15 + 5 + 1 + 1 = 22

15 is the first batch. You end up with 15 wrappers and you can replace every 3 with a chocolate bar:
15/3 = 5

Exchange 3 of the wrappers with a chocolate bar:
5-3 = 2 wrappers left.

Now you have 1 extra wrapper:
2+1 = 3
3/3 = 1

forigve my drunken math

22 chocolates total, 1 left over wrapper

>69 xD

It's 20, you choco-cucks.

29.

chocolate consumers can't solve this.
17 but i don't buy it

Over 9000

chocolate consumers can't solve this.
22 but i don't buy it

15. No one said the chocolates came pre-wrapped

lel

(0,0) -> (15,15) -> (5,20) -> (2,21) -> (3,22)

22 chocolates.

Plus you have 1 wrapper left

...

so you proceed and team up with somebody and you get your 1/3 split ?

22 consumers can't solve this.
chocolate but i don't buy it

Easily solvable in a general way: n = (x mod 2) / 2 + 1.5x - 1, where x is your amount of money. With x = 15, that gives n = 22.

Wrong. Assuming 8% sales tax you can only buy a max if 13. This will yield 4 on the initial wrapper exchange and 1 on the second exchange.

WHAT KIND OF RETARDED STORE WHO GIVES CHOCOLATE FOR FREE FOR 3 WRAPPERS LOL

But then you would each have 1/3 of a wrapper which you could pair with 8 more people for another chocolate. So the answer is infinite chocolates.

who says the $1 isn't incl. g.s.t?

Diabetes. Because you just ate 21 chocolate bars on the spot just to get the 22ed bar.

15

OP here. I'm currently entering a sugar coma. Please stop sending chocolates.

22 chocolates can't buy this.
But I don't solve.

Nevermind.
Delete this.

ThisAlso, in the US, food is not taxed.

Here's a helpful computer program:
Prelude> let solve m w = if w >= 3 then 1 + solve m (w - 2) else if m > 0 then 1 + solve (m - 1) (w + 1) else 0
Prelude> solve 15 0
22

Not even 22 chocolates can solve this, and i am not buying this.

What about whoring yourself out to the guy who's selling the chocolate?

>in the US, food is not taxed

Ok then, guessing you are not an economics major. Or someone who wins at board games a lot.

Or if that program was too helpful,
Prelude> let y f = f (y f) in y (\f w m -> if w >= 3 then 1 + f (w - 2) m else if m > 0 then 1 + f (w + 1) (m - 1) else 0) 0 15
22

first pass: get 15
second: get 5
third: get 1 2/3
fourth get 5/9
fifth: get 5/27
sixth: get 5/81
seventh: get 5/243
eights: get 5/729
ninth: get 5/2187
...
without rounding you can get an irrational number of chocolate for $ 15

>series of rational numbers
>you can get an irrational result
Oh well.

JEWS

it is an infinite series but after this point you need a microscope to measure amounts

>buy one chocolate
>divide chocolate bar infinitely
>have infinite chocolates
>have $14
>profit

An infinite sum of rationals is also rational. We know this because the sum of two rationals is rational, and at no point in the infinite sum will that stop being true.

Has anyone answered "seven" yet?

you are saying that there are no irrational numbers? en.wikipedia.org/wiki/Irrational_number

$15 gets you 15 chocolates

15 wrappers get you 5 more chocolates

5 wrappers get you 1 more chocolate with 2 wrappers left over

1 new wrapper plus the 2 leftover wrappers gets you 1 last chocolate

15 + 5 + 1 + 1 = 22

No, I'm saying that I have a rudimentary understanding of set theory. For example, en.wikipedia.org/wiki/Cantor's_diagonal_argument

How do you represent a sum of an infinite series as the ratio of integers?

Well, if it's convergent, then the number it converges to, and if it's divergent, then as positive or negative infinity. Have you taken calculus?

en.wikipedia.org/wiki/E_(mathematical_constant)

go back to school and learn more then
en.wikipedia.org/wiki/Fractal

15 original chocolates
15 wrappers=5 extra chocolates
5 extra wrappers=1.6 chocolates
1 extra extra wrapper+2 leftover extra wrappers=1 chocolate
22 chocolates with one wrapper left over

15