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The Proof is Trivial!

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Problem 358*

Find all positive integers

n

such that the product of all positive divisors of

n

24^{240}

(edited 10 years ago)

Reply 2301

Zakee

Original post by Felix Felicis

Problem 385*

Find all positive integers

n

such that the product of all positive divisors of

n

24^{240}

I am using Dalek's method of periphrastic divergences and Euler-taradiddles to say that;

We have

24^{240}

If we take the base,

24

and multiply it by

2

we have

48

.

Now do the power divided by the base multiplied by two:

\frac{240}{48} = 5

5

is now our new exponent after considering the periphrastic divergence.

Therefore, the new exponent is the value computed above by the Euler-taraddidle method:

n = 24^5

(edited 10 years ago)

Reply 2302

Felix Felicis

Original post by Zakee

I am using Dalek's method of periphrastic divergences and Euler-taradiddles to say that;

We have

24^{240}

If we take the base,

24

and multiply it by

2

we have

48

.

Now do the power divided by the base multiplied by two:

\frac{240}{48} = 5

5

is now our new exponent after considering the periphrastic divergence.

Therefore, the new exponent is the value computed above by the Euler-taraddidle method:

n = 24^5

You've forgotten the lateral solutions though! :shock:

24^5 = 2\cdot4^5 = 244444

.

Simples.

Reply 2303

BlueSam3

Original post by Felix Felicis

Problem 385*

Find all positive integers

n

such that the product of all positive divisors of

n

24^{240}

Solution 385*

The product

\pi(n)

of all positive divisors of

n

\displaystyle n^{\frac{\sigma (n)}{2}}

, where

\sigma (n)

is the divisor function.

Thus, if

n

satisfies this condition, then

24^{240} = 2^{720} \cdot 3^{240}

must be a power of

n

.
Thus,

n = 2^{a} \cdot 3^{b}

, where

ak = 720

and

bk = 240

, for some integer

k

.
Thus,

a = 3b

n = 24^b

and

b|240 = 2^4 \cdot 3 \cdot 5

.
So

b \in \{ 1, 2, 3, 4, 5, 6, 8, 10, 12, 15, 16, 20, 24, 40, 48, 60, 80, 120, 240 \}

Thus,

n \in \{ 24, 24^2, 24^3, 24^4, 24^5, 24^6, 24^8, 24^{40}, 24^{48}, 24^{60}, 24^{80}, 24^{120}, 24^{240} \} =: A

By the multiplicativity of

\sigma

, since

24^n = 2^{3n} \cdot 3^n

, and if

m=p^k

for some prime

p

, then

\sigma(m) = k + 1

, we have that

\sigma(24^n) = (3n+1)(n+1)

.

If we let

f(n) = \pi(24^n) = 24^{\frac{n}{2}(3n+1)(n+1)}

, then

f

is strictly increasing on the naturals, and thus

f(n) = 24^{240}

has at most one solution.

n = 5

is that solution.

Reply 2304

Zakee

Original post by Felix Felicis

You've forgotten the lateral solutions though! :shock:

24^5 = 2\cdot4^5 = 244444

.

Simples.

Oh, I'm truly sorry. I haven't been doing much Mathematics recently, especially Dalekian Mathematics. I'll learn more some day. :smile:

Reply 2305

Lord of the Flies

Solution 355

Let

Q_m \subseteq S

be such that

|Q_m|=m

, the claim is that if

m>n

then

Q_m

does not have property

P

. Let

A_k=S\cap \{2^q(2k-1):\;q\in\mathbb{N}_0\}

for

1\leq k\leq n

(

\bigcup A_k = S)

, then by the pigeonhole principle there exist

p,q\in Q_{m}

such that

\{p,q\}\in A_k

for some

k

. And of course,

Q_n=\{n,n+1,\cdots 2n-1\}

has property

P

Solution 356

Have I completely and utterly missed something here? If for any

i,j

we have

|A_i|\equiv |A_i\cap A_j|\equiv 0\pod{2}

we may group the population into pairs and call the set of these

T

. Then obviously the number of clubs

\leq \mathcal{P}(T)=2^n

. Equality is only possible if we allow

A_i=\varnothing

for some

i

(edited 10 years ago)

Reply 2306

Mladenov

Original post by Lord of the Flies

Spoiler

It's fine.

Problem 359**

In a town there are

n

people and

m

clubs -

A_{1},\cdots, A_{m}

. If

|A_{i}|

is odd for all

i

and

|A_{i} \cap A_{j}|

is even for all

i,j \in \{1,\cdots,m\}

, then

m \le n

. Is the bound attainable?

Problem 360**

Let

A_{i} \subseteq \{1,\cdots,n \}

i \in \{1,\cdots,m \}

be such that for

i \not= j

the inequality

|A_{i} \cap A_{j}| \le 1

holds. Show that

\displaystyle m \le n+1+ \dbinom{n}{2}

.

Problem 361**

Let

a,A,B

be integers,

M

- a positive integer. Prove that the numbers

Aa^{m}+Bm \pmod M

form a complete residue system

\pmod M

if and only if

\gcd(B,M)=1

Reply 2307

james22

Did anyone ever get an answer to that horrible integral a while back? I don't recall seeing one.

Reply 2308

Arieisit

Problem 362*

This is actually very similar to an A level question but without the added guidance that they usually provide.

A sports association is planning to construct a running track in the shape of a rectangle surmounted by a semicircle. Let the letter

x

represents the length of the rectangular section and

r

represents the radius of the semicircle.
Determine the length,

x

, that maximises the area enclosed by the track.

EDIT

The perimeter of the track must be 600 metres.

I completely forgot to put that in. Sorry :frown:

Problem 363*

Given that the sum of the angles

A

B

and

C

of a triangle is

\pi

radians, show that

\sin(A)

\sin(B+C)

Problem 364***

Determine the extreme values of

f(x,y,z,u,v,w)

\frac{1}{1+x+u}

+

\frac{1}{1+y+v}

+

\frac{1}{1+z+w}

where

xyz

a^3

uvw

b^3

, and

x,y,z,u,v,w>0.

(edited 10 years ago)

Reply 2309

The nameless one

Original post by Arieisit

Problem 363*

Given that the sum of the angles

A

B

and

C

of a triangle is

\pi

radians, show that

\sin(A)

\sin(B+C)

Spoiler

Reply 2310

Arieisit

Original post by The nameless one

Spoiler

Is that all? :holmes:

Posted from TSR Mobile

Reply 2311

The nameless one

Original post by Arieisit

Is that all? :holmes:

Posted from TSR Mobile

well, almost

Spoiler

Reply 2312

31541

Original post by Arieisit

Problem 364***

Determine the extreme values of

f(x,y,z,u,v,w)

\frac{1}{1+x+u}

+

\frac{1}{1+y+v}

+

\frac{1}{1+z+w}

where

xyz

a^3

uvw

b^3

, and

x,y,z,u,v,w>0.

Spoiler

Reply 2313

Jkn

Original post by james22

Did anyone ever get an answer to that horrible integral a while back? I don't recall seeing one.

Do you mean the cos(tan(x))/x one? (or something like that)

If so, no solution was posted. I feel I made a tiny bit of progress though hit a wall pretty quickly. I can post my "progress" if you want to try and get the ball rolling again? :tongue:

Reply 2314

Arieisit

Original post by CD315

Spoiler

You may be on to something.

I don't like giving hints if I didn't see any progress.

Posted from TSR Mobile

Reply 2315

k9markiii

Original post by Arieisit

x

represents the length of the rectangular section and

r

represents the radius of the semicircle.
Determine the length,

x

, that maximises the area enclosed by the track.

I must have misunderstood the question.

The area enclosed is

A = 2rx+\pi r^2

this is maximised with respect to x when you differentiate with respect to

x

and equate it to 0.

0 = 2r

As this only gives a value of the minimum, this implies the maximum will increase as x increases.

Problem 363*

Given that the sum of the angles

A

B

and

C

of a triangle is

\pi

radians, show that

\sin(A)

\sin(B+C)

This is very simple I think.

\sin(A) = Im(e^{iA})

eq 1

\sin(B+C) = Im(e^{i(B+C)})

eq 2

A+B+C=\pi

so:

A=\pi - B - C

eq 3
sub eq 3 into eq 1:

\sin(A) = Im(e^{i(\pi -B - C)})

Therefore:

\sin(A) = Im(e^{i\pi}e^{-i(B + C)})

e^{i\pi} = -1

So:

\sin(A) = Im(-e^{-i(B + C)})

Sin is odd so:

\sin(A) = Im(e^{i(B + C)})

We have written this equivalently to eq 2.
Thus:

\sin(A) = \sin(B+C)

Problem 364***

Determine the extreme values of

f(x,y,z,u,v,w)

\frac{1}{1+x+u}

+

\frac{1}{1+y+v}

+

\frac{1}{1+z+w}

where

xyz

a^3

uvw

b^3

, and

x,y,z,u,v,w>0.

I liked that middle question can't be bothered with the last one.

Reply 2316

Jkn

I've been generalising random expressions and thought you lot might like this one.

Problem 365**/***

Find a closed form expression for

\displaystyle \int_0^{\infty} \frac{x^a}{(b+x^c)^d} \ dx

in terms of elementary functions where

d \in \mathbb{Z}

.

Be sure to consider the different cases that arise according to the values of a, b and c.

Reply 2317

user2020user

Original post by Arieisit

Problem 363*

Given that the sum of the angles

A

B

and

C

of a triangle is

\pi

radians, show that

\sin(A)

\sin(B+C)

Solution 363

A + B + C = \pi

\Rightarrow B + C = \pi - A

\begin{aligned} \Rightarrow \sin (B+C) & = \sin (\pi - A) \\ & = \sin \pi \cos A - \sin A \cos \pi \\ & = - \sin (A) \times (-1) \\ & = \sin (A) \end{aligned}

My first solution of the thread! :woo:

(edited 10 years ago)

Reply 2318

james22

Original post by Jkn

That's the one I meant, I'm curious what the solution is because I could not find any way to make simpler. I managed to convert it to several different integrals but they didn't look any nicer.