Crazy Hard Math Problem Week #4 [solved]

-Incognito-

Firedrake969 wrote:
1. Gave you my proof already xD
2. If A and B are consecutive, either A - 1 = B or A + 1 = B. That's what I meant.

OK then just hang tight, they should make a new theory called, -incognito's- constant after today

Firedrake969

wow, @Znapi
xD

@Incognito well, let's see… Especially since there is a proof about how you can't do it

Last edited by Firedrake969 (Sept. 6, 2015 15:57:51)

-Incognito-

Firedrake969 wrote:
wow, @Znapi
xD

@Incognito well, let's see… Especially since there is a proof about how you can't do it

so we shall!

Firedrake969

could you comment it on my profile? I'll delete it once I read it but I'm curious to see how you're solving it for small numbers if not with n^2 n + 41

Last edited by Firedrake969 (Sept. 6, 2015 15:59:33)

Znapi

ev3coolexit987654 wrote:
Znapi wrote:
What abput a simple a-b+4? It either equals 3 or 5; it isn't always the same.
nope, always gives 5, “consecutive” implies a<b

Darn. Can we use the weird if statement things?

-Incognito-

Znapi wrote:
ev3coolexit987654 wrote:
Znapi wrote:
What abput a simple a-b+4? It either equals 3 or 5; it isn't always the same.
nope, always gives 5, “consecutive” implies a<b
Darn. Can we use the weird if statement things?

as in piece wise functions?? (if they always have different outputs)

Znapi

-Incognito- wrote:
Znapi wrote:
ev3coolexit987654 wrote:
Znapi wrote:
What abput a simple a-b+4? It either equals 3 or 5; it isn't always the same.
nope, always gives 5, “consecutive” implies a<b
Darn. Can we use the weird if statement things?

as in piece wise functions?? (if they always have different outputs)

Yeah, I couldn't remember what they were called. I probably can't do anything with it though because of that rule :p

Firedrake969

http://prntscr.com/8d9zhs
assuming a true generator, being that it produces primes and only primes

Last edited by Firedrake969 (Sept. 6, 2015 16:08:07)

-Incognito-

Firedrake969 wrote:
http://prntscr.com/8d9zhs
assuming a true generator, being that it produces primes and only primes

since i actually want someone to answer the question

, i'll change the rules so that the numbers 1-4 (as in 0 and 1 , 1 and 2, 2 and 3, 3 and 4) are the only ones you have to look at for a and b

then everyone is happy

ev3coolexit987654

if <(a) < [50 ]> then
return [101] ::control
    else
return [65537] ::control
end

Znapi

Firedrake969 wrote:
http://prntscr.com/8d9zhs
assuming a true generator, being that it produces primes and only primes

But does it have to be efficient?

-Incognito-

Znapi wrote:
But does it have to be efficient?

rules changed, see the first post

Firedrake969

so a^2 + b + 41
assuming b = a + 1

because otherwise there is no such formula that would give you prime numbers… so good thing you made it solvable xD

(anyone want to try mine?

)

Last edited by Firedrake969 (Sept. 6, 2015 16:23:21)

ev3coolexit987654

Firedrake969 wrote:
so a^2 - a + 41

ftfy

WooHooBoy

WooHooBoy wrote:
Easy: 3 + 3 + 4

hard: 3 * 3 * 3 - 4 - 3 - 3

Insane!!!: f(a, b) = 3

Edit: Ninja'd by an edit

So..

Insane!!! f(a, b) = p(a * b)

p is the prime number function.

ev3coolexit987654

WooHooBoy wrote:
WooHooBoy wrote:
Easy: 3 + 3 + 4

hard: 3 * 3 * 3 - 4 - 3 - 3

Insane!!!: f(a, b) = 3

Edit: Ninja'd by an edit

So..

Insane!!! f(a, b) = p(a * b)
p is the prime number function.

Cheater!!!!!

WooHooBoy

ev3coolexit987654 wrote:
WooHooBoy wrote:
WooHooBoy wrote:
Easy: 3 + 3 + 4

hard: 3 * 3 * 3 - 4 - 3 - 3

Insane!!!: f(a, b) = 3

Edit: Ninja'd by an edit

So..

Insane!!! f(a, b) = p(a * b)
p is the prime number function.
Cheater!!!!!

It's not against the rules…

Superdoggy

AFAIK there isn't actually a method to generate primes from variables. Pretty sure that to make sure something is a prime you have to check it against most of the numbers smaller than it. If there is a formula which is like “enter two variables, apply this to them, and get this nth prime” we don't know about it yet. I think. Pls correct me if I'm wrong.

So currently the insane problem is pretty much next to impossible.

ev3coolexit987654

Insane: a+b will always equal 3,5,or 7, so ceiling of sqrt(a+b+1) would give 2 or 3. I win!

Last edited by ev3coolexit987654 (Sept. 7, 2015 00:52:02)

Superdoggy

-Incognito- wrote:
Insane!!!: Write a formula that uses two consecutive positive integers, a and b, to always equal a prime number. (Note: The formula should not always equal one number )

Edit: ^ Above only requires the formula to work until the numbers 3 and 4

Okay, now that's possible.

a * a + b * b + a * 2

1 * 1 + 2 * 2 + 1 * 2 = 7
2 * 2 + 3 * 3 + 2 * 2 = 17
3 * 3 + 4 * 4 + 3 * 2 = 31

EDIT:

ev3coolexit987654 wrote:
Insane: a+b will always equal 1,3,5,or 7, so ceiling of sqrt(a+b+1) would give 2 or 3. I win!

Oh lol that's so obvious why didn't I think of that? Now I feel silly xD

Last edited by Superdoggy (Sept. 7, 2015 00:52:43)

Discuss Scratch