Crazy Hard Math Problem Week #4 [solved]

-Incognito-

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Hi everybody, you know the gist, its another Sunday, and i have another problem for ya'll to solve!

If you weren't here the last couple of posts, at least i have some good news for you,

You've come to the right place to be challenged!

So every week something goes up in the AT's that is math related, and you guys have to be the one to solve it.

but Anyway…

Last week's Winners

Znapi
ev3coolexit987654

Last Week's Question:

#3 The Cone Problem

Feel free to rage quit….

Mathematically Yours,
-Incognito-
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#4 Number Sense

Here are the Numbers you can use

3 , 4

Easy Question:
Write the numbers above in such a way that they equal 10

(example: using the numbers 2 and 4 —- 2+4+4 + ((4 / 2 ) /2) = 11 )

Not So Easy Question:
Without using addition (+), double negatives (- -), or absolute values, Write the numbers in such a way that they equal 17

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

Last edited by -Incognito- (Sept. 6, 2015 16:30:09)

WooHooBoy

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)

Last edited by WooHooBoy (Sept. 6, 2015 13:16:01)

Superdoggy

I'm not gonna even bother with the Easy Question. xD

3x3x3-4-3-3

Okay the insane question has a loophole, sorta.

a / a + b / b + b / b –> always equals “3”

Superdoggy

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

Lol! That's almost exactly what I did (except you ninja'd me)

xD

Last edited by Superdoggy (Sept. 6, 2015 13:42:45)

-Incognito-

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)

LOL, But i found something that uses more understandable terms, (HINT: it has something to do with exponents

)
BTW Am i reading your equation correctly??? -> the function of a and b is equal to the probability of a and b

which means f(a,b) = p(a) * p(b)

f(3,4) = 3/7 * 4/7 = .244897959… (a irrational number)

because in that case….

a prime is:
a positive integer(whole number, negative or positive) that is not divisible without remainder by any integer except itself and 1, with 1 often excluded

and irrational numbers cannot be integers

Last edited by -Incognito- (Sept. 6, 2015 14:09:26)

Firedrake969

nvm

Last edited by Firedrake969 (Sept. 6, 2015 14:15:14)

liam48D

Easy: 3 * 3 + 4 / 4
Not so easy: floor(4 * 3 - 4 / 3)

As for insane..

Superdoggy wrote:
a / a + b / b + b / b –> always equals “3”

That can be improved, a / a + b / b, which always equals 2. 2 is a prime number.

Last edited by liam48D (Sept. 6, 2015 14:52:03)

ev3coolexit987654

Easy: 3+3+3+4-3
Hard: (3*4*3/sqrt(4))-(4-3) <- doesn't count as double negative right?
Insane: (will fill in later)

Last edited by ev3coolexit987654 (Sept. 6, 2015 15:06:39)

-Incognito-

liam48D wrote:
Easy: 3 * 3 + 4 / 4
Not so easy: floor(4 * 3 - 4 / 3)

As for insane..

….

a / a + b / b, which always equals 2. 2 is a prime number.

I Already wrote:
(Note: The formula should not always equal one number )

Last edited by -Incognito- (Sept. 6, 2015 15:13:16)

-Incognito-

ev3coolexit987654 wrote:
…
Hard: (3*4*3/sqrt(4))-(4-3) <- doesn't count as double negative right?

…

LOL, no that works

i Think that is one of the more creative ways to answer the hard question

Last edited by -Incognito- (Sept. 6, 2015 15:16:57)

Marsover

no this is….

Hard: (3^4)-(4^3)

you have to give me credit for doing it in only 4 numbers

Did you even try to Insane problem?? its impossible, i'm almost tempted to find something on google…

Last edited by Marsover (Sept. 6, 2015 15:25:37)

ev3coolexit987654

Marsover wrote:
no this is….

Hard: (3^4)-(4^3) you have to give me credit for doing it in only 4 numbers

I can do better:

(join(3)(4))/sqrt(4)

Marsover

ev3coolexit987654 wrote:
Marsover wrote:
no this is….

Hard: (3^4)-(4^3) you have to give me credit for doing it in only 4 numbers
I can do better:

(join(3)(4))/sqrt(4)

*claps slowly*

well done

-Incognito-

Marsover wrote:
…

Did you even try the Insane problem?? …

yeah, don't worry, i only give problems that i have already solved

Last edited by -Incognito- (Sept. 6, 2015 15:29:25)

ev3coolexit987654

Wait:

I have a list of primes

f(a, b) = the a'th item of the list

-Incognito-

This is quickly becoming the only problem that hit 100 views without being solved

-Incognito-

ev3coolexit987654 wrote:
Wait:

I have a list of primes

f(a, b) = the a'th item of the list

you're getting desperate, aren't you?

ev3coolexit987654

-Incognito- wrote:
This is quickly becoming the only problem that hit 100 views without being solved

We solved easy and hard.

-Incognito-

ev3coolexit987654 wrote:
-Incognito- wrote:
This is quickly becoming the only problem that hit 100 views without being solved
We solved easy and hard.

i guess

but insane is taking some time, XD

ev3coolexit987654

-Incognito- wrote:
ev3coolexit987654 wrote:
Wait:

I have a list of primes

f(a, b) = the a'th item of the list

you're getting desperate, aren't you?

I have an imperfect formula:
(a+b)^2-(a+b)+41

Discuss Scratch