Post Math Challenges/Questions

>Whats wrong here?

That you never proved that any of those things make sense and you are just basically saying
"Algebra is magic".

Take a 3 digit number A, reverse its digits to get B.
If A > B then X = A - B
If A < B then X = B - A

Reverse the digits of X (If its a 2 digit number put a zero on the front i.e. 099) to get Y

Show that X + Y always equals the same number regardless of A

log(x) is a counterexample, right?

let there be a value [math]y \in \mathbb{R} [/math] which [math]f[/math] assumes only finitely many times, let [math]x_0[/math] be the largest point such that [math]f(x_0) = y [/math]. now [math]f( (x_0,\infty) )[/math] is a connected set lying in [math]\mathbb{R} \setminus \{ y \}[/math] and WLOG is therefore fully contained in [math] (y,\infty)[/math]. this implies that [math]f([0,x_0]) [/math], a compact set, contains all [math](-\infty,y)[/math]. contradiction.

amirite?

I also first thought of log(x), but it's domain is (0,infty) not [0,infty)

No, no, no, it was supposed to be a math challenge. One of them is correct, the challenge is to find out why the other one is not. (Yes you can use a calculator but where is the fun?)

Looks correct, well done

That one's pretty cool. Only times X + Y don't equal 1,089 are when the first and last of the 3 numbers are the same

Oh yeah, I forgot that detail in the question

It's a simple enough question to state that my normie Dad even enjoyed hearing the solution

A = a100 + b10 + c10
B = c100 + b10 + a
A > B iff a > c
A < B iff a < c
suppose a > c
X = A - B = (a-c)100 + (c-a)
but (c-a) is not a digit so we have
X = (a-c)100 - (a-c)
X = (a-c)99

Now there are 9 possibilities
X = 099,198,297,396,495,594,693,792,891
Y = 990,891,792,693,594,495,396,297,198

Then notice that:

X + Y = k*99 + (11-k)99 = 99k + 11*99 - 99k = 11*99

Nice.
I am taking an elementary course this semester. Did I do well daddy?

Exactly what I did to solve it

Find a 9 digit number such that the first two digits taken together are a multiple of 2, the first 3 digits are a multiple of 3 and so on, so the whole 9 digit number is a multiple of 9. There's only one solution

If [math] x^{x^{x^{.^{.^{.}}}}} = 2 [/math] then [math] x = \sqrt{2} [/math] is a correct statement.

If [math] x^{x^{x^{.^{.^{.}}}}} = 4 [/math] then [math] x = \sqrt{2} [/math] is also a correct statement.

If [math] x^{x^{x^{.^{.^{.}}}}} = 2 [/math] AND [math] x^{x^{x^{.^{.^{.}}}}} = 4 [/math] then [math]2 = \sqrt{2}^{\sqrt{2}^{\sqrt{2}^{.^{.^{.}}}}} = 4 [/math] is also a correct statement.

The only "issue" is that [math]2 = \sqrt{2}^{\sqrt{2}^{\sqrt{2}^{.^{.^{.}}}}} = 4 [/math] is wrong but none of the calculations show otherwise.

Let this number be
qwertyuio
(, means or)
If the first two digits are a multiple of 2 then
w = 0,2,4,6,8
if the first 3 digits are a multiple of 3 then
q + w + e = 3k
r10 + e = 4k
t=0,5
y=0,2,4,6,8
r + t + y = 3k
u = y = i = 0
q+w+e+r+t+y+u+i+o = 9k

r=1,e=6
w=0
q=6,3

THE ANSWER IS:
306450963

Not gonna lie there m8. I didn't even use the answers I originally wrote out. But after I had like 6 digits secured I started trial and erroring.
Great problem m8.

Ehh that defo works, but I forgot to mention that you have you use all the digits 1,2,3,4,5,6,7,8,9. That is, no 0 allowed. With 0 allowed there are many solutions but without it there is only one

I also remember another digits based problem but I've forgotten the details so I'll find that in a bit

Found it again

Find a 5 digit number A such that when you reverse its digits to get a new number B = 4 * A

Fuck should be a 4 digit number

Ah. I don't sure if that makes it easier or harder. With that then it is guaranteed that the 5th digit is 5, for example.
I am now busier so I'll see if I revisit this later.

Here is one my prof left for homework yeasterday:
Find digits x,y,z such that 5, 9 and 11 divide 2x1642y032z

28164280320

28164230325 works too.

The equations are:
z = 0 OR z = 5
x + y + z = 7 OR x + y + z = 16 OR x + y + z = 25
x = y + z OR x + 11 = y + z

Let [math]X, Y \in \mathcal M_{n-1,1}(\mathbb R)[/math] and [eqn]M = \begin{pmatrix} 0 & X^T \\ Y & 0_{n-1}\end{pmatrix}[/eqn]
Give a necessary and sufficient condition for M to be diagonalizable.

You have four numbers: 2,5,8,9.
You have to use these numbers to get 52, using the four basic operations (+,-,*,/).
You can only use each number once and you can use as many brackets as you want.

Answer is 0.867.

My favourite one from a calculus class is:

Given the longitude and latitude of two cities and the radius of the earth, find the distance between the two cities.

Pretty simple but fun application!

...

>Give a necessary and sufficient condition for M to be diagonalizable.

A necessary and sufficient condition is that M has n linearly independent eigenvectors.

8*(9-5/2)

HAHAHAHAHAHAHAHAHA yeah we're turning into third world nations. Murica got its niggers and massive corruption /mind control/indoctrination and stupid fat people. And we've got muslims to fuck our shit up

(3+sqrt(5))/2

kek
why not just cut to the chase

>A necessary and sufficient condition for M to be diagonalizable is that M is diagonalizable

wew lads, a condition on X and Y (that can be checked in linear time if you're going to tell me that the condition was also a necessary and sufficient condition on X and Y)

Here's another one:
Let [math](a_n)[/math] and [math](b_n)[/math] be two real sequences such that [math](a_n + b_n) \rightarrow 0 [/math] and [math](e^{a_n} + e^{b_n}) \to 2[/math]. Prove that both sequences converge.

Suppose that [math](a_n)[/math] doesn't converge then one of two cases must happen:

Case 1: [math](a_n)[/math] has an unbounded subsequence [math](a_{n_k})[/math].
Without loss of generality suppose that [math] \lim_{k \to \infty} a_{n_k} = \infty [/math]. Then we must have [math] \lim_{k \to \infty} b_{n_k} = - \infty [/math] but then [math]\lim_{k \to \infty} (e^{a_{n_k}} + e^{b_{n_k}}) = \infty [/math] which is a contradiction.

Case 2: [math](a_n)[/math] has a subsequence [math](a_{n_k})[/math] that converges against a [math] c \neq 0 [/math].
Then [math] \lim_{k \to \infty} b_{n_k} = -c [/math] and [math]\lim_{k \to \infty} (e^{a_{n_k}} + e^{b_{n_k}}) = e^c + e^{-c} [/math] but [math] e^c + e^{-c} \neq 2 [/math] for [math]c \neq 0 [/math] so we have another contradiction.

Therefore out assumption must be wrong and [math](a_n)[/math] converges to [math]0 [/math]. Then [math](b_n)[/math] must naturally converge to [math]0 [/math] too.

Represent 0.dddddddd... As a fraction where d exists in {1,2,3,4,5,6,7,8,9,0}

d/9

x and y are reals.
1) Study f(x, y) = (x + y, xy).

2) We have the system (u = x + y, v = xy). Find x and as functions of u and v.

Prove it

This one is ez.

Let S be Shalaquandra's ass and B the blank space between her ass and the radius

r = r

You have a radius of convergence for the infinite power tower. It is clear that the square root of 2 is in the radius of convergence, but 4 is not.

[math]
0.ddd... = d\cdot \sum_{k=1}^{\infty}\left(\frac{1}{10}\right)^k = d\cdot \left(\frac{1}{1-\frac{1}{10}} -1 \right) = \frac{d}{9}
[/math]

That thread was stupid

Where exactly is the difficulty supposed to be?

Requires calculus

[math] \int_0^{10} \left( \int_0^{10 cos(x/10)} y dy \right) dx = 25(5sin(2)+10)[/math]

I forgot to divide by the area of the quarter of circle

Great, that works.
Here's one in algebra.
Let M be a matrix with real coefficients. What is, in terms of M, the least number of coefficients you need to change to make it invertible ?

what is the total area of the sum of all the grey circles?

HA what kind of nonsens is that! that has no aplication in the real world! dont waste your time!

Infinity.

wat

not quite

For people having a hard time
[math]
\bar{y} = (\frac{M_{xz}}{M}) = \frac{\int\int y dA}{\int\int 1 dA}= \frac{\int_{x=-10}^{x=10}\int_{y=0}^{y=\pm\sqrt{100-x^{2}}} ydydx}{\frac{\pi r^{2}}{2}}
[/math]

To all the analysts:
How can I express delta as a function of epsilon without including c? I can't seem to figure it out. Any hints?

[eqn] \delta_\varepsilon = \sup_x |g'(x)| \varepsilon [/eqn]

You can do this for all [math]C^1[/math] functions. Just look at the fundamental theorem of Calculus.

are you saying all C^1 functions are uniformly continuous? because that's not true

The ones with [math] \sup_x |g'(x)| < \infty [/math] are.

right so nowhere near all C^1 functions... not even x^2

It's actually
[eqn] \sup_x |g'(x)| \delta_\varepsilon < \varepsilon [/eqn]

[eqn] \left| g(x) - g(c) \right| = \left| \int_c^x g'(t) dt \right| \leq \|g'\|_{\infty} |x - c| < \varepsilon [/eqn]

if M is an n*n matrix, then you need n changes. or more precisely, you need k changes where k is the algebraic multiplicity of 0 as the eigenvalue of M.

I would have said geometric multiplicity,(that is, n-rk(M)), maybe that's what you meant ? Take the matrix [math]M = \begin{pmatrix}0 & 1 \\ 0 & 0 \end{pmatrix}[/math]. You need only change the bottom right coefficient to make it invertible but 0 has algebraic multiplicity 2.
In any case I expected a proof.

Given an n*n matrix, would you agree that you can make it invertible by changing all the coefficients ? (just replace each coefficient by the correspondent coefficient in the identity matrix). Now i'm asking the minimum number of changes you need to do this.

Can't you just take any value [math]a [/math] that is not in the spectrum and then subtract [math]a [/math] from the diagonal?

Yup, that means changing n coefficients (one for each on the diagonal), but you can do better than that

no, I meant algebraic, but it's wrong, should be geometric. I didn't realize that changing one entry could change more than one eigenvalues

[math].86997R^2[/math]

I get 1.758

I think a good first question might be "why is this construction even possible ?"

This is what Wolfram tells me

Show the werk son.