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1. Problem 358*

Find all positive integers such that the product of all positive divisors of is .
2. (Original post by Felix Felicis)
Problem 385*

Find all positive integers such that the product of all positive divisors of is .
I am using Dalek's method of periphrastic divergences and Euler-taradiddles to say that;

We have

If we take the base, and multiply it by we have .

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

is now our new exponent after considering the periphrastic divergence.

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

.
3. (Original post by Zakee)
I am using Dalek's method of periphrastic divergences and Euler-taradiddles to say that;

We have

If we take the base, and multiply it by we have .

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

is now our new exponent after considering the periphrastic divergence.

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

.

You've forgotten the lateral solutions though!

.

Simples.
4. (Original post by Felix Felicis)
Problem 385*

Find all positive integers such that the product of all positive divisors of is .
Solution 385*

The product of all positive divisors of is , where is the divisor function.

Thus, if satisfies this condition, then must be a power of .
Thus, , where and , for some integer .
Thus, , and .
So
Thus,
By the multiplicativity of , since , and if for some prime , then , we have that .

If we let , then is strictly increasing on the naturals, and thus has at most one solution. is that solution.
5. (Original post by Felix Felicis)

You've forgotten the lateral solutions though!

.

Simples.

Oh, I'm truly sorry. I haven't been doing much Mathematics recently, especially Dalekian Mathematics. I'll learn more some day.
6. Solution 355

Let be such that , the claim is that if then does not have property . Let for so , then by the pigeonhole principle there exist such that for some . And of course, has property

Solution 356

Have I completely and utterly missed something here? If for any we have we may group the population into pairs and call the set of these . Then obviously the number of clubs . Equality is only possible if we allow for some .
7. (Original post by Lord of the Flies)
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Solution 355

Let be such that , the claim is that if then does not have property . Let for so , then by the pigeonhole principle there exist such that for some . And of course, has property

Solution 356

Have I completely and utterly missed something here? If for any we have we may group the population into pairs and call the set of these . Then obviously the number of clubs . Equality is only possible if we allow for some .
It's fine.

Problem 359**

In a town there are people and clubs - . If is odd for all and is even for all , then . Is the bound attainable?

Problem 360**

Let , be such that for the inequality holds. Show that .

Problem 361**

Let be integers, - a positive integer. Prove that the numbers form a complete residue system if and only if .
8. Did anyone ever get an answer to that horrible integral a while back? I don't recall seeing one.
9. 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 represents the length of the rectangular section and represents the radius of the semicircle.
Determine the length, , 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

Problem 363*

Given that the sum of the angles , and of a triangle is radians, show that =

Problem 364***

Determine the extreme values of

=

where = , = , and
10. (Original post by Arieisit)
Problem 363*

Given that the sum of the angles , and of a triangle is radians, show that =
Spoiler:
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Triangle, so c+B=180-A, for any x, =
11. (Original post by The nameless one)
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Triangle, so c+B=180-A, for any x, =
Is that all?

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12. (Original post by Arieisit)
Is that all?

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well, almost
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I am known for being lazy, and that is why i just skipped the previous question, cannot be bothered to read it
13. (Original post by Arieisit)

Problem 364***

Determine the extreme values of

=

where = , = , and
Spoiler:
Show
Is Lagrange the way to go here or am I barking up the wrong tree?
14. (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?
15. (Original post by CD315)
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Is Lagrange the way to go here or am I barking up the wrong tree?
You may be on to something.

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

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16. (Original post by 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 represents the length of the rectangular section and represents the radius of the semicircle.
Determine the length, , that maximises the area enclosed by the track.

I must have misunderstood the question.

The area enclosed is this is maximised with respect to x when you differentiate with respect to and equate it to 0.
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 , and of a triangle is radians, show that =

This is very simple I think.
eq 1
eq 2

so:
eq 3
sub eq 3 into eq 1:

Therefore:

So:

Sin is odd so:

We have written this equivalently to eq 2.
Thus:

Problem 364***

Determine the extreme values of

=

where = , = , and
I liked that middle question can't be bothered with the last one.
17. I've been generalising random expressions and thought you lot might like this one.

Problem 365
**/***

Find a closed form expression for in terms of elementary functions where .

Be sure to consider the different cases that arise according to the values of a, b and c.
18. (Original post by Arieisit)
Problem 363*

Given that the sum of the angles , and of a triangle is radians, show that =
Solution 363

My first solution of the thread!
19. (Original post by Jkn)
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?
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.
20. (Original post by k9markiii)
I liked that middle question can't be bothered with the last one.
I absolutely love your solution for question 363! When I did it, it was in no way similar to your version. Nice perspective!

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