help-28

[sturdy valve]

+ve --> while measuring it should be in same direction as incident ray

and all distances are measured from pole

over here pole is ur A or B

[buoyant urchin]

where is this question from, feels like I've seen it somewhere before

[night socket]

jee my friend.

[buoyant urchin]

previous year?

[night socket]

idk how to do sign conventio

[night socket]

yes

[buoyant urchin]

ah

[night socket]

i struggle with sign convention for radius of curvature

for lenses

please help

[buoyant urchin]

is not a plano concave lens btw

[sturdy valve]

its just a concave surface

[sturdy valve]

draw a line from pole say (A) towards ur object and tell me dirn of that line (either to the left or to the right)

[night socket]

im generally saying

i face difficulty

[night socket]

left.

for the right hand

concave surface

[sturdy valve]

bruh i said for A

but ok

yes for B its correct

[sturdy valve]

now if u draw an incident ray towards B can u tell me the direction?

[night socket]

right

[sturdy valve]

and i meant towards the centre of curvature here not the object

[night socket]

incident ray goes towards right

[sturdy valve]

yes correct

[sturdy valve]

so dirn of incident ray is opposite of the line which we drew for radius of curvature (this one)

so R is -ve

in general also u can remember ig convex lens R>0 and concave R<0

[night socket]

nah nah

that cant be applied here yk

[sturdy valve]

why not?

[night socket]

bro if u dont mind

yk the curved surface of a sphere

has some circle

to it

with that can u explain

[sturdy valve]

i gtg now..but its the same concept just imagine a sphere curved at that surface

[night socket]

ahhh

ok thanks

.closed

.close

[full forum]

Channel closed

Closed by @night socket

Use .reopen if this was a mistake.

[full forum]

Available help channel!

Send your question here to claim the channel.

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[calm rapids]

Find all integers $(n,m)$ such that $n^m = m^n$ with m \ne n.

[glossy valve]

#PhenomenalPerson
Compile Error! Click the reaction for more information.
(You may edit your message to recompile.)

[calm rapids]

could someone solve this with like pure olympiad math and no calculus involved? thx

[delicate torrent]

!status

[full forum]

What step are you on?
1. I don't know where to begin.
2. I have begun but got stuck midway.
3. I got an answer but I was told that it's wrong.
4. I got an answer and would like my work checked.
5. I have a question about someone else's work/solution.
6. I have completed the problem and don't need help anymore. Thank you.
7. None of the above

[delicate torrent]

This looks familiar to me ngl

[delicate torrent]

!noans

[full forum]

The purpose of this server is to help you learn; please don't ask for direct answers. Ask for guidance, explanations, or feedback instead.

[delicate torrent]

!da2a

[full forum]

Asking the actual question right away is more likely to get responses.

Asking "Can I ask...?" or "Does anyone know about...?" doesn't give people enough information to decide whether they can help, and answering can feel like a promise to help with the actual question, which they might find themselves unable to.

[calm rapids]

uhh I got an answer of (2,4), (4,2), (-2,-4), (-4,-2) but like couldnt really prove why they are the only solutions

[calm rapids]

alright

[silk bridge]

take ln

[delicate torrent]

I remembered my teacher once helped us with this before

No logs was needed

[silk bridge]

hm..

[calm rapids]

interesting

[calm rapids]

what is that solution though

[delicate torrent]

Im trying to remember

Or i'd just do it myself

[delicate torrent]

So im thinking of this

Consider $m \mid n$

[glossy valve]

1 divided by 0 equals Infinity

[delicate torrent]

That means $\exists k \in \mathbb Z, n = km$

[glossy valve]

1 divided by 0 equals Infinity

[delicate torrent]

Substitute in the original equation, we got $(km)^m = m^n$

[glossy valve]

1 divided by 0 equals Infinity

[delicate torrent]

Then we can solve for $k^m$

[glossy valve]

1 divided by 0 equals Infinity

[delicate torrent]

That didn't go well

[calm rapids]

k^m = m^n / (m^m)

ok

[delicate torrent]

Any ideas?

[silk bridge]

why k belongs to z

right m|n you specified

[silk bridge]

(km)^m = m^km

then raiseto 1/m

km = m^k

k= m^k-1

[calm rapids]

they are all integers

[silk bridge]

n = km = m^k wtf

[calm rapids]

alr 15 mins is up so <@&286206848099549185>

[silk bridge]

hm.. idk with that one make cases

but lets try ln

but you say no calculus

by taking ln we are just going to see when the graph of lnx/x repeats values for x belongs to integers

[calm rapids]

well theres a quote that you can solve every IMO q or maths olympiad q without calculus

[calm rapids]

alr

[silk bridge]

avg olymp type shit let's see

[delicate torrent]

ofc lol

[calm rapids]

n ln m = m ln n

so ln m / m = ln n / n

[delicate torrent]

here is what i came up with rn

(assuming that $m$ and $n$ are positive)

[glossy valve]

1 divided by 0 equals Infinity

[calm rapids]

nice

[delicate torrent]

now just consider $d = 2$ and prove that there's no root for $d \geq 3$

[glossy valve]

1 divided by 0 equals Infinity

[calm rapids]

alr just explain why gcd(m,n) = 1 results to be a contradiction

[delicate torrent]

$m \mid RHS$ but $LHS = RHS$ so $m \mid LHS$

[glossy valve]

1 divided by 0 equals Infinity

[delicate torrent]

Using the fact that $\gcd(m, n) = 1$ to continue

[glossy valve]

1 divided by 0 equals Infinity

[calm rapids]

oh alr thx

so then if l = d^(l-1) we consider: now just consider $d = 2$ and prove that there's no root for $d \geq 3$

[glossy valve]

#PhenomenalPerson
Compile Error! Click the reaction for more information.
(You may edit your message to recompile.)

[delicate torrent]

Yep

[calm rapids]

now just consider $d = 2$ and prove that there's no root for $d \geq 3$

[glossy valve]

#PhenomenalPerson

[delicate torrent]

Yes

So what happens at $d = 2$?

[glossy valve]

1 divided by 0 equals Infinity

[calm rapids]

just saying if gcd (k,l) = 1 always works here right ?

[delicate torrent]

We used the gcd(k, l) = 1 fact in here

[calm rapids]

oh wait it does

cuz m/n can always be reduced to a simplest fraction form

[delicate torrent]

For $d = 2$ then consider $l = 2$ and prove $l = 2^a$ ($a > 1$) is impossible

[glossy valve]

1 divided by 0 equals Infinity

[delicate torrent]

Cool

[calm rapids]

l = d^{(l-1)}

wait why is texit not responding

$l = d^{(l-1)}$

[glossy valve]

#PhenomenalPerson

[calm rapids]

there we go

[delicate torrent]

$l = 2$ works well

[glossy valve]

1 divided by 0 equals Infinity

[calm rapids]

if d = 2 then l = 2

[delicate torrent]

l = 2 then k has to be 1 (since k < l)

[calm rapids]

now for $d \geq 3$

[glossy valve]

#PhenomenalPerson

[delicate torrent]

You need to prove it's impossible

[calm rapids]

wait if d = 2, then doesnt l = 1 work as well though

[delicate torrent]

But note that $k < l$

[glossy valve]

1 divided by 0 equals Infinity

[calm rapids]

oh yeah

[delicate torrent]

$l = 1$ then we can't find a value for $k$

[glossy valve]

1 divided by 0 equals Infinity

[delicate torrent]

.

[calm rapids]

so l is 2 ^ a but if a greater than or equal to 2, why doesnt it work just clarifying

[delicate torrent]

Wdym?

[calm rapids]

so l is 2 ^ a but if a greater than or equal to 2, why doesnt it work just clarifying

for case d = 2

[delicate torrent]

For a >= 2 then RHS grows faster than LHS

[calm rapids]

yeah

[calm rapids]

ok now its just this

$l = d^{(l-1)}$ and $d \geq 3$

[glossy valve]

#PhenomenalPerson

[calm rapids]

its a bit hard to prove its impossible not gonna lie

[silk bridge]

where are we

[calm rapids]

this is like literally all I need

oh wait we haven't considered m and n being negative yet

oh well

[calm rapids]

just prove this then

[calm rapids]

@delicate torrent are you still here ?

[atomic valve]

it's the same argument except it doesn't work earlier

[calm rapids]

wdym

[atomic valve]

you have d and d^l, so one side is d^(d^l) and the other is d^(dl)

and you know l > 1

do you see the problem

[calm rapids]

uhhh which equation are you refering to

[atomic valve]

m^n = n^m

[calm rapids]

@delicate torrent @silk bridge @atomic valve thanks for your help today but I gtg now so I'll post this problem tomorrow again

you guys can stay back for a while on this q

[atomic valve]

it has been solved though

[calm rapids]

actually if you explain this

[delicate torrent]

Use the same argument i told you earlier

Some side grows faster than other side

[calm rapids]

ohhh

wait I get it now

thanks a lot

but like how do you express all these in a formal maths competition though

[atomic valve]

you would write it out

“getting it” doesn't mean having a vague idea of what goes wrong, it means clearing the smoke and crystallising what exactly happens

[calm rapids]

ok

[atomic valve]

by earlier work you know that of m and n, one is the power of the other

convince yourself that this is true

[calm rapids]

like I actually have to gtg

alr I'll leave it here

thanks guys anways

bye

maybe I'll post this tomorrow here

[atomic valve]

i'm not sure what you want, like the full solution written down?

[calm rapids]

sure

I'll see it tomorrow

if you can

alr bye

[atomic valve]

We are solving for (positive) integer solutions for
\[ n^m = m^n. \]
Write $d \coloneqq \gcd(n,m)$ and define $k,l$ such that $m = dk$ and $n = dl$. This gives
\[ (dl)^{dk} = (dk)^{dl}. \]
Assume without loss of generalisation that $k < l$. We can rearrange to see
\[ l^k = d^{l-k}k^l \]
such that $k^l = 1$ by coprimality. Hence $k = 1$, and $m = d$, $n = d^l$.

Our equation is thus $d^{dl} = d^{d^l}$.
We know $d > 1$ such that this is equivalently asking for 
\[ l = d^{l-1}. \]
But also $l > 1$, so $l$ is at least $d$. If $d = 2$ then $l = 2$ is indeed a solution, and by the time $l = 3$ the right hand side is already bigger, hence nothing else works. If $d > 2$ then already $l = d$ has right hand side bigger, so nothing works.

[full forum]

@calm rapids Has your question been resolved?

[atomic valve]

if you really wanted to formalise the rhs lhs business you would define f(d, l) = d^(l-1)/l on ℤ_≥2 × ℤ_≥2 and say that f is monotonically increasing with respect to d and l

[zenith kernel]

ln(x)/x strictly decreases for x>1

Oh, no calculus…

[atomic valve]

they don't want calculus

[full forum]

Channel closed

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[full forum]

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• Do not immediately ping people or roles. After 15 minutes, feel free to ping <@&286206848099549185> once.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[pseudo roost]

Question 14

[pseudo roost]

I wanted to know if I’m correct

[warped frost]

this step is iffy

the top line is right, but the step going from the top line to the bottom line is wrong

[pseudo roost]

Oh

Can you help me then

[warped frost]

hint: there's a common factor on the right

[pseudo roost]

Wdym?

I through r the two cancel

[warped frost]

consider factoring out something from the right side first

[pseudo roost]

Oh

[warped frost]

then divide through both sides by the thing you factored out

[silk bridge]

$2^{a+b} = 2^a\cdot2^b$

[glossy valve]

scoob

[warped frost]

that works, but is not necessary here

[warped frost]

though would be nice as an alternative

well I suppose you're using scoob's method then

[silk bridge]

[warped frost]

I'll let scoob take over if he wants to

[pseudo roost]

I don’t understand what to factor our

[silk bridge]

nah go on

[warped spire]

I would pay particular attention to the 2^6 on the right

[pseudo roost]

Is that what u mean?

[pseudo roost]

For factorising?

[warped spire]

yes

try factoring out a 2^6 from the RHS

[pseudo roost]

What’s rhs

[warped spire]

right-hand side

similarly, LHS = left-hand side

[pseudo roost]

Oh ok ok

Like this?

But when do Ik when to factor out

[warped spire]

⚠️

[delicate torrent]

Problemo problemo

[warped spire]

oh infinity is here

suppose I'll let him take over

[delicate torrent]

😭

-# why lol

[warped spire]

not a helpful discussion for this channel

[silk bridge]

yo infinity

[delicate torrent]

?

[silk bridge]

-# let's get back to OP

@warped spire

[delicate torrent]

In this case, you should subtract $2^{\frac{x}{2}}$ on both sides and then factor

[glossy valve]

1 divided by 0 equals Infinity

[pseudo roost]

I don’t think understand

[delicate torrent]

Hold up

[silk bridge]

side note when you have x^a = x^b + x^z you cannot just compare powers

[pseudo roost]

Oh

What do you mean compare

Pls

[split hatch]

isn't a substitution the quickest here?

[delicate torrent]

From here, you can't say a = b + z

[warped spire]

$$2^a = 2^b \Rightarrow a = b$$ OK
$$ 2^a = 2^b + 2^c \Rightarrow a = b + c$$ NG

[glossy valve]

fox(x, y): ℝ² -> ℂ (Seia)

[delicate torrent]

Golden rule no. 1: you can only compare powers when your equation is something like this: $x^a = x^b$ where $x \not \in {0, 1, -1}$

[glossy valve]

1 divided by 0 equals Infinity

[warped spire]

I would... but that would be confusing atp

[split hatch]

wait I might be genuinely cooked because I can't think of another way

mfw you can't do basic maths

[silk bridge]

-# i missed hanako was trying to form quadratic ig

[delicate torrent]

Hold up

[pseudo roost]

Oh ic

[delicate torrent]

Let's try to redo this again

And we'll tell you where you're wrong

Step by step

[pseudo roost]

Ok

[delicate torrent]

So $2^x = (\sqrt{2})^{x + 8} - 64$

[glossy valve]

1 divided by 0 equals Infinity

[delicate torrent]

What do you do here

[delicate torrent]

?

[pseudo roost]

Change the 64 to prime base

[delicate torrent]

Cool

Golden rule no. 2: you should decide your bases to work with first before you put your pen down

In this case, the base you work with is 2

[delicate torrent]

Cool

[pseudo roost]

ok

[delicate torrent]

So $2^x = (\sqrt{2})^{x + 8} - 2^6$

[glossy valve]

1 divided by 0 equals Infinity

[full hazel]

Is this a good form for the equation?

[void nova]

That's correct, yeah

[full hazel]

Can it be solved without graphing?

[void nova]

Now you can make a substitution, t = (√2)^x

[pseudo roost]

Do I multiply Bith sides by square

[void nova]

Wdym?

[warped spire]

yes

[pseudo roost]

Bith

Both

[delicate torrent]

OP's trying to redo the question

[void nova]

Yeah sure but I didn't understand what they meant by multiplying by square

[pseudo roost]

To get rid of the root

[void nova]

Oh ok, so you meant squaring both sides

[delicate torrent]

You can turn the root into 2^0.5

[void nova]

But no, that's not useful

Yeah exactly

[delicate torrent]

Hold up

[void nova]

Too much spoiler 😬

[warped spire]

let the OP try for herself, please.

[delicate torrent]

Please don't interupt the OP

[full hazel]

Mb

[delicate torrent]

Anyways @pseudo roost

Golden rule no. 3: for these type of questions, try to convert the most into the base that you had decided

[pseudo roost]

Can I change their to 2^1/2

[pseudo roost]

Oki

[delicate torrent]

That's what you should do

[delicate torrent]

Okay

[delicate torrent]

Try to simplify the exponents

[pseudo roost]

Like this?

[delicate torrent]

Yes!

[delicate torrent]

Golden rule no. 4: try to make as many constants as you can

[pseudo roost]

Ok

What are constantly

[delicate torrent]

Constants are basically numbers

[pseudo roost]

Oh ok

[delicate torrent]

For instance: 5, 3, 6

[pseudo roost]

Can leave the fraction as x +8

Bc the denominator is the same

[delicate torrent]

Try to use the exponent rules

On $2^{\frac x2 + \frac 82}$ to see what you can do

[glossy valve]

1 divided by 0 equals Infinity

[pseudo roost]

Cancel out the two’s?

[delicate torrent]

Try to use the rules in here

[pseudo roost]

Multiply

Huh

[delicate torrent]

Hint: $a^{b+c} = a^b \cdot a^c$

[glossy valve]

1 divided by 0 equals Infinity

[delicate torrent]

Yes!

[pseudo roost]

Can I factor out?

[delicate torrent]

you should move the ones with $x$ on the exponent to the LHS

[glossy valve]

1 divided by 0 equals Infinity

[delicate torrent]

and all other stuffs to the RHS

[pseudo roost]

Oh

So grouping

[delicate torrent]

hm

yeah

[pseudo roost]

Do I have to multiply the 2^x/2

By two

[delicate torrent]

?

wdym?

[pseudo roost]

To leave x

[delicate torrent]

uhh

no need to

[delicate torrent]

you can turn $2^x = \left( 2^{\frac{x}{2}} \right)^2$

[glossy valve]

1 divided by 0 equals Infinity

[delicate torrent]

this will help you form a quadratic

[pseudo roost]

Wait wht

[delicate torrent]

so in here right, you turn $2^x$ into what you just said

[glossy valve]

1 divided by 0 equals Infinity

[pseudo roost]

So multiply by 1/2?

But why do we do this

[delicate torrent]

you can see a form of a quadratic equation in here

[delicate torrent]

do you see it?

[pseudo roost]

Oh the k method

What golden rule goes this apply too pls

[delicate torrent]

lol

well

i do golden rules to different type of questions

it's more formally known as the substitution method

you can search that up

[pseudo roost]

How do you know when to use the k method

[delicate torrent]

spot the pattern

• practice = key

[pseudo roost]

Ic

Tyy

[delicate torrent]

np

anymore questions to ask?

for some more golden rules

[pseudo roost]

I don’t think so for now

But do I use the quadratic formula here bc the normal factorization doesn’t work

[delicate torrent]

yeah ofc lol

i mean

it's a quadratic

[pseudo roost]

Says undefined

[delicate torrent]

hold up

[delicate torrent]

it has no roots right?

since it has no roots means no values of $k$ satisfies

[glossy valve]

1 divided by 0 equals Infinity

[pseudo roost]

Oh

Ic

Tysm

I gotta go

[delicate torrent]

and since no $k$ satisfies then no $x$ satisfies

[glossy valve]

1 divided by 0 equals Infinity

[delicate torrent]

!done

[full forum]

If you are done with this channel, please mark your problem as solved by typing .close

[pseudo roost]

Ty everyone

.close

[full forum]

Channel closed

Closed by @pseudo roost

Use .reopen if this was a mistake.

[delicate torrent]

welp no more golden rules lmao

[full hazel]

Was it that same question?

It does have a real solution no?

[delicate torrent]

oh fuck

[delicate torrent]

FUCK

[full hazel]

Here, factor out 2^4

Then substitute 2^x with your favourite letter of the alphabet

You get a quadratic equation that factorizes completely

[delicate torrent]

@pseudo roost

[full hazel]

Ideally, turn 2^x into √2 ^2x, to avoid dealing with fractional powers

Then, substitute (√2)^2x with, say, t or v or u or y or p or–

[pseudo roost]

.reopen

[full forum]

✅ Original question: #help-28 message

[pseudo roost]

But why pls

Could u explain

[delicate torrent]

-# i messed up real bad, but you can factor out a 2^4

[pseudo roost]

Wait lemme do it again

[delicate torrent]

hold up, leave until here

erase everything else

[pseudo roost]

Oki

[full hazel]

Okay so look at

Two raised to x
And
Two raised to x/2

It is technically a quadratic equation, but not in a very good form

[pseudo roost]

Oh

[full hazel]

It's easier to see and work through when the powers are not fractional

[pseudo roost]

Ic

[pseudo roost]

So I make the power non fractional

[full hazel]

Yes

[pseudo roost]

Oh I was thinking of something different

[delicate torrent]

but still

remember the golden rules

-# sorry for messing the work up

-# but the golden rules shall apply

[pseudo roost]

It’s ok

[full hazel]

Yes

As @delicate torrent said previously, you should decide on a base
I just chose a base that makes the exponents "cleaner"

[pseudo roost]

The only common base is 2

[delicate torrent]

$\sqrt{2}$ is cleaner

[glossy valve]

1 divided by 0 equals Infinity

[delicate torrent]

but most people (including me would go for 2 at the first instinct)

[pseudo roost]

So I can just root it

[pseudo roost]

Could u explain the second one

[full hazel]

Root and square, yes

'2' becomes "the square of square root of two"

[full hazel]

Sure

[delicate torrent]

or if you wanted a base of 2

then you already have the quadratic

use the k method as you said

formally known as the substitution method

[pseudo roost]

But there aren’t two terms that sre thr same

[delicate torrent]

?

well

remember what you converted before

[delicate torrent]

this

[pseudo roost]

Yes

[full hazel]

You're talking about grouping
And there is

2 raised to 8/2 is 16
2 raised to 6 is 64

16 is a common term that you can separate

[pseudo roost]

Oh

What idk use the powers

[delicate torrent]

i think you factored it wrong

[delicate torrent]

-# ||complete the square or quad formula solves the job||

[full hazel]

Yes, but since they were factoring it anyway

[pseudo roost]

I really don’t understand what your saying

[full hazel]

Honestly you have a good quadratic equation here

[delicate torrent]

-# ||the expression OP has here can be rewritten into a clean square and no more||

[full hazel]

Plug the values into quadratic formula

[delicate torrent]

hold up

hold up

[delicate torrent]

OP is currently in this step

[full hazel]

They're the same step
I rewrote what they wrote

[pseudo roost]

But there’s no square

[delicate torrent]

so as what @full hazel said, you can try to use the quadratic formula

hold up

[delicate torrent]

use this

you found that out yourself and it's correct

and then use the k method that you mentioned before

[pseudo roost]

I have to start again bc I e forgotten

[full hazel]

Honestly, yes

Fresh slate will do you good

[delicate torrent]

🔥

[full hazel]

Also, @pseudo roost
Genuine question: have you done exponential and logarithmic functions previously?

This is not something you can get the intuition for unless you have an idea of power, quadratic and exponential functions

It's a very nice question that combines all of those things

[pseudo roost]

I have done the first one but I’d what the second is

And I just stated doing this kind of exponents

[full hazel]

Oh okay

[delicate torrent]

okay

you're fine

[full hazel]

No nvm logarithms
Mb

[delicate torrent]

2nd one is basically just helps for getting rid of the base

[full hazel]

Start over and we can work through it easier this time

[pseudo roost]

I’m confused about the rhs

[delicate torrent]

wait

turn $2^x = \left( 2^{\frac x2} \right)^2$

[glossy valve]

1 divided by 0 equals Infinity

[delicate torrent]

it should be clearer to use the $k$ method (formally substitution)

[glossy valve]

1 divided by 0 equals Infinity

[silk bridge]

is this the same question?

[delicate torrent]

yes

[silk bridge]

where are we stuck ?

[full hazel]

You made a mistake in step 4
You squared LHS, but not the entire RHS

And you don't have to square anything at all

[delicate torrent]

@pseudo roost

that should be better to handle

[pseudo roost]

Yes thank you

[pseudo roost]

Oh

I’m confused

[full hazel]

Just do this from the beginning

[pseudo roost]

But how do I do the right hand side

[silk bridge]

but do what infinity said

[delicate torrent]

hell yeah

[pseudo roost]

The 2^8/2

[delicate torrent]

it's better

[delicate torrent]

well it's a constant

and a constant is just a number

[pseudo roost]

Idk if it’s positive or negeative bc it’s multiplying the 2^x/2

[delicate torrent]

don't mind that for now

replace every $2^{\frac x2}$ with $k$

[glossy valve]

1 divided by 0 equals Infinity

[delicate torrent]

and then worry about that later

[full hazel]

k^2 = k • 2^4 - 2^6

[pseudo roost]

I got k^2 -k +48

[delicate torrent]

no no no no

-# you made the mistake that i previously made to help you

[pseudo roost]

Oh

[delicate torrent]

do what @full hazel said

[delicate torrent]

also i'd recommend searching up the substitution method on the internet

there are many resources on that

[pseudo roost]

I will

I’ll drop this for now

Thank you everyone

.close

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[delicate torrent]

idk how when i searched for it, they used it for system of equations

😭

but the main idea is to define a new variable to make the problem easier

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[muted token]

Is the x in $(-x)^{-n}$ positive when n is even, and negative when its odd?

[glossy valve]

Vortac

[sand escarp]

try some values of n

[muted token]

I did

$1 \div (-2) \div (-2) = \frac{1}{4}$

[glossy valve]

Vortac

[muted token]

$1 \div (-2) \div (-2) \div (-2) = -\frac{1}{8}$

[glossy valve]

Vortac

[sonic stratus]

Consider that, if n is even, then -n also is.

and thats about it.

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[granite shadow]

This came from a video of deriving the mirror equation for a convex mirror, why tf is PG=0

[sand escarp]

approximately 0

if look at how they wrote PG = 0

not sure if that's the case tho

[granite shadow]

Approximately 0
I mean that only happens when the convex mirror doesn't have 2mm as diameter💔

Isn't PG closer to 0 if the convex mirror is more "unround" like having longer diameter or centre of curvature length, but it the center of curvature length is low, the mirror will be way "rounder" effectively making PG closer to the nearest hot moms instead of 0 (i might get banned for writing this but I'm not wrong, yes?)

But doesn't PG being close to 0 also affected by height?

I mean height of the object

The higher the object is the further G will be from P (btw if it's not apparent, EGC is an right angle)

I mean the less flat the mirror is, and the higher the object is, instantly make PG far from 0

Huh

Imma just let autoclose, maybe if somebody appears and looking to help

My conclusion for now is,
Whether PG is close to 0 or not is based from

1. Height of the object
2. Length of curvature
   The higher the object the farther PG is from 0
   The longer the length of the curvature, the closer PG is to 0

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[quiet gate]

wy did i get 3 and what is wrong with this

[quiet gate]

why did i get 3 ad what is wrong with this

[prime pier]

im not sure what you mean by "getting 3", but this solution is correct

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[hoary oyster]

for 128 i’m NOT getting the same volume or SA and chat gpt is getting something completely different, can someone help

[night socket]

yk heron's formula?

[bright bronze]

did you take the base as an isosceles triangle or smthn? coz I get the same answer

[bright bronze]

not needed tbh, the base is a right angled triangle

[night socket]

no.1 rule is to never ask chatgpt.

[night socket]

works.

mb

[night socket]

u added 3 rectangles right

[hoary oyster]

The formula it gives is 1/3 B*H

for the volume

n it doesn’t give me the formula for the SA

also wouldn’t the base be 12*10?

do we view it kinda on its side

man i don’t even know what shape this is 😭

[queen crater]

Just look at the right angles

[hoary oyster]

Ohh so that tells me where base is

[queen crater]

It's a prism with right triangle bases

[hoary oyster]

so the formula for the volume would be b*h??

[queen crater]

Yea

[hoary oyster]

i think i’m doing sm rly wrong cus isn’t that js 10*12

[queen crater]

You think the base is 12cm² ?

[hoary oyster]

no

[queen crater]

Then why 10*12?

[hoary oyster]

wait

what did u multiple

?

to get the base

[queen crater]

Do you not see the right triangles?

You have the hypotenuse and one side

[hoary oyster]

Yes

oh

OH

is it 9

cus pathagreum theorem

[queen crater]

What is

[hoary oyster]

i don’t understand 🥹

i see the right triangle

what is the base and height that u got

[queen crater]

You said "is it 9", I just don't know what that "it" is referring to

[hoary oyster]

the base

of the

right triangle

[queen crater]

Not sure what a base of a right triangle is, so how about you tell me all the sides of the triangle?

[hoary oyster]

I honestly don’t even know which way to look at the shape

[queen crater]

Just list the sides for me

[hoary oyster]

12,10,15

are given

[queen crater]

Again, what are all the side lengths of the triangle(s)?

[hoary oyster]

12 and 10

[queen crater]

I'm about to give up on you

Can you highlight which segments correspond to the given lengths?

[queen crater]

Why do you think all those green segments are 12cm?

[hoary oyster]

Idk cus there’s a 90 degree angle

[queen crater]

A 90º angle doesn't automatically give you isosceles triangles

[hoary oyster]

oh okay so then what do i do?

do i find the height by using trig

[queen crater]

You have a right angle and two side lengths, surely Pythagoras is easier

[hoary oyster]

Alr i’ll do that then

[hoary oyster]

Also i don’t recommend giving up so easily next time you help someone 🙂

just needed answers

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[tranquil hemlock]

Determine the height of the plane.

[tranquil hemlock]

Im really confused as to why tan is used to answer this question since i cant determine what sides will be opposite/adjacent etc

[atomic valve]

there are two right triangles you should draw, with angles 32° and 56° respectively. the length of their sides are related—how?

[tranquil hemlock]

couldnt tell u

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[vale flax]

21 students in a class, teacher sends 4 to library every day. The teacher chooses randomly for the first 4 days from the list of students who havent gone. If i havent went from day 1-3, whats the chances i go on day 4\

[vale flax]

my first thought was this

denom is ways to choose 21c4

and numerator will be 9c4

but uh thats not working so idk

[lime ether]

why?

how many people are left to choose from on the start of day 4

[vale flax]

9

[vale flax]

uhhh i was thinking denom is saying total ways to pick 4 from 21 but numerator is on day 4

[lime ether]

not sure how you got that

the numerator should be the number of ways to choose 4 people where you are one of them

on day 4

the denominator is the number of ways to choose 4 people (with or without you) on day 4

[vale flax]

ohhh

so numerastor is uh

8c3

denom is 9c4

last question is this one

wait second to last

" a recent survey showed that 60% of students play video games, 10% read, and 8% do both" Draw a venn diagram for this

[lime ether]

yep

[lime ether]

whats confusing

[vale flax]

i did this at first right

but like its wrong idk

[lime ether]

you need a box for the total set of students

what about those who don't read or play video games

also

what is the 60 and 10 supposed to represent in your diagram

seems like you're saying 60% play video games but don't read

and 10% read but don't play video games

which is false

[vale flax]

uhm

[vale flax]

how would i fix it

do i subtract 8%

[lime ether]

yes

[vale flax]

from each bc rn its saying 68% and 18%?

[lime ether]

yea

[vale flax]

okay so if im given either, and both%, i subtract the both% from both to solve it

[lime ether]

yea

but as i said before

you also need to incorporate the students who do neither

since they don't make up the full 100% here

[vale flax]

do i just do 1-.52-.02

[lime ether]

no that isnt all of the students who do at least one

you've only subtracted the students who do only one

what about those who do both

[vale flax]

oh so to get neither i do 1-.52-.02-.8?

just subtract every %

so boom 38%

[lime ether]

nice

[vale flax]

okay last one for sure

30 students in my class, teacher chooses 5 students at random to make a group, probability that me and 2 friends will be in a group

denom is just total possible ways to pick groups of 5 from 30 kids

so 30c5

[lime ether]

yep

[vale flax]

now numerator is im not too sure

for this do i assume we are like 1 thing?

MeF1F2, Random, random

[lime ether]

wait this is supposed to be one group right?

or are you saying 6 groups of 5

[vale flax]

just 1

[lime ether]

ok

so you're just choosing the two other people to be y'alls friends

[vale flax]

uh yes

ohh

wait

so if i need 2 more people

and us 3 are already in

is it just 27c2?

yo thats kinda hard

when do i know if a question will require this like subtraction

like what we did with the first question

[lime ether]

we didn't do complementary counting?

wdym subtraction

[vale flax]

like how we subtract our own group from total

[lime ether]

like how the total number of people we choose from is smaller than you thought?

[vale flax]

yeah i guess

[lime ether]

this just comes down to understanding what you're selecting

[vale flax]

oh okay

is there words that give it away

[lime ether]

i don't think so

its context dependent

[vale flax]

oh okay

i think i kinda get it

thank you for the help

.close

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[lime ether]

you're welcome

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[knotty grail]

tangent equation?? cubic in y?? no clue really

[knotty grail]

checking

[charred carbon]

Lmao I just took an exam that has a similar problem

There's an approach, if the circle touches the parabola at a point M then tangent of the parabola would also be tangent of the circle at that point

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@knotty grail Has your question been resolved?

[zenith kernel]

I solved it by direct calculation, so I don’t know how I can give you a hint. No clever methods used. Only one step I twisted a little bit: (x-a)^2+(y-b)^2=r^2, at the step y(a-x-4)=32 I multiplied by y both sides, make it 8x(a-x-4)=32y. Other than that, pure computation. ||(x-16)^2+(y+8)^2=320|| I got

[charred carbon]

<@&268886789983436800>

[silk bridge]

i feel cubic is inevitable

[zenith kernel]

I didn’t use any cubic

Nothing higher than 2

[void nova]

How? I'm curious, since I was also trying but got stuck

[zenith kernel]

But it’s pure computation. You sure you are fine i show you my result?

[zenith kernel]

If I used anything clever I might say what clever thing I used as a hint. But it’s just computation. So I can only show you the full steps

[silk bridge]

spit

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[charred carbon]

Awwww no one tries my approach

[night socket]

who pinged.

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[knotty grail]

sorry back

[knotty grail]

reading chat

hold on

[zenith kernel]

Step 1: the circle passes (0,0) and (0,-16), that means:
||a^2+b^2=r^2
a^2+(b+16)^2=r^2
Together give you b=-8
and x^2-2ax+y^2+16y=0||
Step 2, consider at (x,y) two tangent are equal, which tells you
||-(x-a)/(y-b)=4/y , that gives you
y(a-x-4)=32||
Step 3. Now we do a little trick:
||Multiply by y both sides
8x(a-x-4)=32y, which is
x^2+(4-a)x+4y=0.||
Now you have everything
Step 4, just continue to calculate:
||Together with x^2+(8-2a)x+16y
=x^2-2ax+y^2+16y=0, gives you
12y=(a-4)x.||
Step 5, let t=12/(a-4), then ||x=8t^2, y=8t
back to y(a-x-4)=32, t(12/t -8t^2)‎ = 4
t=1 or -1, t=-1 won’t work. t=1
a=16, (x-16)^2+(y+8)^2‎ = 320||, which is the result.

[knotty grail]

hum

[zenith kernel]

Someone told me to split. So you can reveal one step at a time

[silk bridge]

i said spit it out, nvm

[zenith kernel]

Oh spit, i thought split

[silk bridge]

that is clever

[knotty grail]

um how did you go from\
$x^2+(4-a)x+4y=0$\
to
$12y=(a-4)x$

[knotty grail]

like wouldnt there be a y^2 somewhere in the latter step

[zenith kernel]

This x^2+(4-a)x+4y=0, combine with

x^2+(8-2a)x+16y=0 we obtained from step 1

Those two together give us 12y=(a-4)x, or say x=ty

[knotty grail]

x^2+(8-2a)x+16y=0 we obtained from step 1
well yeah but wasnt that
x^2-2ax+y^2+16y=0

oh

got it

[zenith kernel]

y^2=8x

Okay

[void nova]

I would have never thought of multiplying by x

[knotty grail]

quite the convoluted equation solving

[zenith kernel]

Yeah only that step is not brainless computation. But everything else is just procedure

[silk bridge]

everywhere you look the soln says find repeated roots, by discriminant

[knotty grail]

huh

i wasnt able to find any solutions in the first place

[silk bridge]

eh where did you get this from

[knotty grail]

aakash test series

sometimes a bit too crazy

[silk bridge]

oh wierd test series ahh moment

i hope no shit like this comes

[zenith kernel]

I heard this kind of computation problems are common in places like India, and South Korea

[knotty grail]

and/or redundant trivia

yes

[zenith kernel]

Yes, foreign language, the most annoying part is a/an to me

[knotty grail]

what

uh ok hold on i need to read through this once again

[sturdy valve]

oy purple universe check dms

[charred carbon]

Okay that's also what I did lol

The computation part is still pain in the a** tho

[zenith kernel]

It’s okay. Whenever you find the calculation becomes complicated then you know you need to discard that path. Since it’s high school problem, surely no too complicated computations show up

[sturdy valve]

😂😂😂

[charred carbon]

Actually no I did a slightly different way and got -y^4/64+8y=0

Eh but it's also kinda the same thing

[charred carbon]

Not in my school tho, smth the equation here is just so fuc*in complicated that the only way is to use a calculator, and ofc we're allowed to use

Tho I still hate it

[zenith kernel]

I hate calculation too, not very mathematical

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@knotty grail Has your question been resolved?

[knotty grail]

so i think one way to put it is that we solve a system of three equations in h, k and a (where (h, k) is the point of tangency)\
$h^2+k^2-2ah+16k=0$\
$k^2=8h$\\
$\frac{4}{k} = \frac{-(h-a)}{k+8}$

[knotty grail]

okay i think i can live with that

thanks

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[frigid carbon]

I just realized this is similar to writing parametric form of parabola, finding normal equation then centere lies on normal,
Equate distance of P from cente to radius then we get t=+- 4 , hence x coordinate of centre

[frigid carbon]

Uh

Sorry

.close

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[frigid carbon]

Or is it just the same

[silk bridge]

similar

[knotty grail]

i think that circles back to cubic somehow (pun not intended)

or idk i need to try that method again

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[dusky garden]

Questions and my workings are in the above 3 images.
But they seem to be incorrect as all the interval calculators online get the same answer, which is different from my own

Can anyone give me a lil help 🥺

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@dusky garden Has your question been resolved?

[silk bridge]

it doesn't try it

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@dusky garden Has your question been resolved?

[full forum]

@dusky garden Has your question been resolved?

[dusky garden]

<@&286206848099549185>

[gentle echo]

?

[dusky garden]

Hey my question is here and my workings are above it

[dusky garden]

Trying to calculate 95% interval

[gentle echo]

Lemme see

[dusky garden]

Thank you my goat <3

I went wrong somewhere, but I'm not exactly sure where

[gentle echo]

Step 1: Compute means
Mean payload: x̄ = 193.77 KB
Mean I/O time: ȳ = 75.21 ms
Step 2: Compute regression coefficients
Sums:
Sxx = Σ(x_i - x̄)²
Sxy = Σ(x_i - x̄)(y_i - ȳ)
Slope:
a = Sxy / Sxx ≈ 0.3786
Intercept:
b = ȳ - a * x̄ ≈ 2.13
Regression equation:
y = 0.3786 x + 2.13
Step 3: Predicted values
y_pred = ax + b
Step 4: Residuals
residuals = y_i - y_pred
Step 5: Standard error of estimate
Se = sqrt(Σ(residuals²)/(n-2)) ≈ 0.9541
Step 6: 95% confidence interval for predicted y
t-value (df=33) ≈ 2.034
CI formula:
y ± t * Se * sqrt(1/n + (x0 - x̄)² / Sxx)
Example: for observation 1 (x=1024):
y_pred = 0.37861024 + 2.13 ≈ 388.85 ms
95% CI = [388.77, 390.93] ms
For all observations, the 95% confidence intervals are:
Observation | Lower CI | Upper CI
1 | 388.77 | 390.93
2 | 10.83 | 11.61
3 | 25.99 | 26.73
4 | 239.28 | 240.54
5 | 147.50 | 148.31
6 | 35.47 | 36.18
7 | 7.41 | 8.20
8 | 9.31 | 10.09
9 | 8.17 | 8.96
10 | 14.24 | 15.01
11 | 22.20 | 22.95
12 | 173.19 | 174.11
13 | 77.53 | 78.18
14 | 125.57 | 126.31
15 | 6.65 | 7.45
16 | 30.17 | 30.89
17 | 16.52 | 17.28
18 | 48.36 | 49.04
19 | 66.17 | 66.83
20 | 5.14 | 5.94
21 | 41.16 | 41.85
22 | 4.38 | 5.18
23 | 19.17 | 19.92
24 | 3.24 | 4.05
25 | 3.62 | 4.43
26 | 330.64 | 332.44
27 | 4.76 | 5.56
28 | 5.90 | 6.69
29 | 281.19 | 282.69
30 | 203.41 | 204.47
31 | 56.70 | 57.36
32 | 91.15 | 91.82
33 | 12.34 | 13.12
34 | 4.00 | 4.80
35 | 107.04 | 107.73
Step 7: Interpretation
a = 0.3786: Each additional KB of payload increases I/O time by ~0.38 ms
b = 2.13: When payload = 0 KB, expected I/O time ≈ 2.13 ms
Step 8: Residual analysis
Residuals are small and randomly distributed
Standard error Se = 0.9541 ms
This shows the regression fits the data well

@dusky garden

It's correct right?

???

Ping me after you're done

Am I correct or not

@dusky garden

[gritty rose]

did you just copy paste this from some ai

[dusky garden]

I'm pretty sure confidence level is supposed to be two values, and its supposed to be very close to the mean

So I don't think that is correct

[dusky garden]

Do you have any idea on how to solve it, sorry to ping you if you do not like pings

[torn jolt]

HOLY

wy cant i find the maths of my lvel 😭

[dusky garden]

I've been trying to get is solved for like 5hrs

[dusky garden]

?

[torn jolt]

nthing-

[dusky garden]

<@&286206848099549185> Does anyone know how to solve this, I've been trying to figure it out for hours my workings are above

My answer is there (184.46, 184.78) but it is incorrect

[silk bridge]

what is this stats ?

[dusky garden]

I think it should be like statistical inference or something

Its something I'm studying as part of Computer Science

[silk bridge]

sorry, no idea but maybe try ug channels

[full forum]

@dusky garden Has your question been resolved?

[full forum]

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[slate glacier]

Remainder when (77)^(77) is divided by 8, is
(Can anybody explain various methods here, I saw one which used binomial and one which used modulo, also what would be the quickest method for this)

[turbid badge]

modulo would be the quickest

[slate glacier]

I see, let me see if I can share how they solved it

[gritty rose]

Oh they did both

[slate glacier]

I don't get what they did here

[gritty rose]

First line is at least two steps. Binomial expansion and modulo 8 the powers of 77 for all but two of the terms

[slate glacier]

yup, okay so the thing is, I'm kinda unfamiliar with this stuff, could you please make it a bit more simple

[turbid badge]

u know the binomial formula?

[gritty rose]

What is "this stuff" exactly

[slate glacier]

yes, the combination one right?

[turbid badge]

yea

all the terms they took modulo 8, the power of 76 is at least 2

[slate glacier]

modulo entirely, and binomial theorem needs proper revision

[turbid badge]

and 76 has 4 as a factor, so all those terms divide 8

[gritty rose]

Oh yes I was mistaken. It's 76 not 77. Ty Green for the correction

It could be done with 77 but that'd be a little harder tracking powers of -1

[gritty rose]

Read https://brilliant.org/wiki/modular-arithmetic/

There's a similar problem with 17^17 mod 10

[slate glacier]

okay so wait

77^77 mod(8) yes?

[turbid badge]

👍

[slate glacier]

and we make it (76 + 1)^(77) because 76 has even divisors?

[turbid badge]

yeah and it's convenient to have 1

[slate glacier]

okay, now (76 + 1)^(77) mod(8), what do we do now?

[turbid badge]

okay we have to use the binomial formula, but we do not directly expand it

we have nC_k a^(n-k) b^k

as k goes from 0 to n

u know that 76 has 4 as a factor

[slate glacier]

yes

[turbid badge]

now every term upto k=n-2, do you agree the power of 76 will be at least 2, if u take a as 76?

[slate glacier]

yes

[turbid badge]

then we also have at least two 4s, so we have a 16 in all those terms, which 8 divides

[slate glacier]

oh yeah

that makes sense

[turbid badge]

so all those terms divide 8 and would be 0 mod 8

[slate glacier]

yes

[turbid badge]

the sum of those terms also divides 8, and you can write it as 8k, which is what your solution did

now you still have two terms remaining

when k= n-1 and k=n

[slate glacier]

yes

[turbid badge]

can you see how they were dealt with?

[slate glacier]

yes, damn

It's easier than I thought

[turbid badge]

so the remainder in the end is 5

you can also do it directly using modulo without binomial expansion

[white karma]

|| one way to do this is to prove and then use(or maybe just use) (a+b)^p\equiv a^p+b^p mod p||

||This can be generalized so that one has (a_1+...+a_n)^p\equiv a_1^p+...+a_n^p mod p||

||Now write the 77 of the base as 1+1+..1 77 times and note that the exponent 77 is 7*11||

The way I provided is silly but it's a way to do it

you could also use Euler's theorem

That would be another way to solve the question

[slate glacier]

what's that?

[white karma]

Do you know Euler's totient function φ

[slate glacier]

not at all unfortunately

[white karma]

np, alright so let's start from there

euler's totient function is a function φ:N*->N* defined as follows: for any n in N*, φ(n) is the number of elements =<n which are relatively prime to n

so for example there are 2 numbers =<4 and relatively prime to 4, namely 1 and 3. Hence φ(4)=2

Actually there is a nice closed form that gives you φ(n) for any given n

[slate glacier]

wait wait, but why do we include 1 in primes here?

[white karma]

well we aren't talking about prime numbers

we are talking about relatively prime numbers

ie numbers with no common divisors >1

or in other words, numbers whose gcd is 1

[slate glacier]

Okay, so 1 is never a prime number but always a relative prime number right?

[white karma]

We want the number of elements k=<n such that gcd(k,n)=1

[white karma]

Right, 1 is relatively prime with any (non-zero) integer

[slate glacier]

okay got it

[white karma]

ok so now we have the definition of Euler's totient function

[slate glacier]

Just confirming, φ(15) = 12?

[white karma]

No

it's 8

[slate glacier]

{1, 2, 4, 7, 8, 11, 13, 14}?

[white karma]

You only removed the divisors which are >1 here I suppose.

But that's not really everything

[slate glacier]

So not even the multiples of the primes divisible by it?

[white karma]

Because for example 9 has a common divisor with 15, namely 3

[slate glacier]

I mean I removed 1, 3 and 5

[white karma]

You don't want any common factors other than 1

[white karma]

Well you shouldn't remove 1

you should remove 15 for example

[slate glacier]

yeah sorry I actually removed 15, IDK why I said 1

[white karma]

you should remove any number =<15 which has any common factor with 15 than 1

[white karma]

np

[slate glacier]

So is this set fine?

[white karma]

Yeah this sounds correct to me

Nice

ok so in fact there is a way to always find φ(n) without having to check every number =<n

[slate glacier]

okay

[white karma]

If $n=\prod_{k=1}^m p_k^{a_k}$, it can be proven that $\varphi(n)=n\prod_{k=1}^m(1-\frac 1{p_k})=\prod_{k=1}^mp_k^{k-1}(p_k-1)$

[glossy valve]

ali yassine

[white karma]

So if you are given a natural number n, you find it's prime factorization and then you can compute φ(n)

for example 15=3x5, so φ(15)=φ(3)φ(5)=2x4=8

[slate glacier]

Oh wow

And I see even powers won't be a problem here

[white karma]

Right

So now that we know how to compute φ(n) in a better way than just checking whether each number =<n has a common factor with n other than 1, we are ready to state and use Euler's theorem.

Euler's theorem says the following: given a positive integer $n>1$ and a positive integer $a$ which is relatively prime to $n$, ie such that $\gcd(a,n)=1$, the following holds: $a^{\varphi(n)}\equiv 1\pmod n$

I assumed n>1 just to avoid something like having mod 1

But it should also work if n=1

[slate glacier]

wait, but what's m?

[glossy valve]

ali yassine

[white karma]

Mb it's supposed to be n

a special case of this is when n is a prime number, in this case the theorem is also called fermat's little theorem

[slate glacier]

okay and also can we take an example before we apply it in that question

[white karma]

Yes sure

[slate glacier]

Oh so that's what that theorem is, I had heard of it

[white karma]

Yeah, but the theorem I presented is a generalisation of fermat's little theorem to any positive integer modulus

[slate glacier]

okay, so like a branch of it?

[white karma]

if by branch of it you mean that fermat's little theorem is a special case of Euler's then yes.

[slate glacier]

yes yes

[white karma]

Ok so let's take an example (or more if you want) before using it to solve the original question

[slate glacier]

okay

[white karma]

choose a positive integer

[slate glacier]

131

[white karma]

We will take this as our modulus

choose a positive integer relatively prime to 131

[slate glacier]

oh god, for modulus let's keep a smaller one

like 7

[white karma]

Alright

[slate glacier]

or wait, 11

[white karma]

Ok 11

Now choose a positive integer relatively prime to 11

[slate glacier]

7

[white karma]

ok, let's try to compute 7^102 mod 11

[slate glacier]

🙏

okay

[white karma]

so what's the first step?

[slate glacier]

that function of 102?

number of relative primes of 102

[white karma]

Here is the theorem

what's a and what's n in our case?

[slate glacier]

a = 7 and n = 11