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1. Question 1 from an old IMO,

Problem 318*

Find all 3-digit numbers such that, if you divide them by 11, you get the sum of the squares of the original numbers' digits.
2. (Original post by MW24595)
Question 1 from an old IMO,

Problem 318*

Find all 3-digit numbers such that, if you divide them by 11, you get the sum of the squares of the original numbers' digits.
Solution 318
Incomplete.
Spoiler:
Show

Let N be one such 3-digit number. Obviously N is divisible by 11, so

, with and .

Let

.

.

.

Now, check each case:

using the inequalities, a must be positive, so

Which has no real solutions.

Again, no real solutions, so the only N for is

.
3. (Original post by henpen)
Solution 318
I'm not sure whether I've interpreted the question correctly, is the wording of the question the same as 'Find all 3-digit numbers which upon division by 11 yield the sum of the squares of their 3 digits'?
Your reasoning is sadly faulty in that it ignores the case that a+b > 9.
4. (Original post by henpen)
Problem 319**
I'm working on this at the moment, but I think it may be interesting in itself.

Where is the cyclotomic polynomial in , prove that

??

Aren't there more restrictions on p and/or n?
5. (Original post by Jkn)
Problem 161**/***

Let k be an integer greater than 1. Suppose and for

Evaluate
Roflol, sorry for the delay. My Laptop is broken, and I wasn't about to type this up using the phone!

Solution 161

Notice that for , we essentially have the recurrence for the Harmonic Numbers (well, almost; its just a matter of starting point), and it is obvious that this is divergent. Further note that for , the increments are only greater, and so, the series would diverge faster than the Harmonic Series.
Either way, we have as .

Now, by the Cesaro-Stolz,

Substituting our expression for in,

For the sake of neatness, let .
So, we have, from the binomial theorem,

As as

So, we are left with,

6. There are lots of people on.here better at maths than me, so I was wondering if you could explain something to me. On page 92 of this PDF: http://www.jmilne.org/math/CourseNotes/FT.pdf, it says at the end of the second paragraph 'one sees easily...' Can any of you flesh out why it's a unique topology? In particular why can't we perform a shift on the topology to generate a new one? It's probably something obvious but I've been staring at it for half an hour and I can't for the life of me see it.

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7. (Original post by MW24595)
Roflol, sorry for the delay. My Laptop is broken, and I wasn't about to type this up using the phone!

Solution 161
Very nice! Whilst you have used the exact same techniques that I have, my solution is not identical - it's essentially a more direct application followed by a rather unexpected deduction. I will type it up in an hour or so when I've got more time!
8. Not too tough, uses some basic double integral knowledge

Problem 319**
where R is the parallelogram defined by the lines:

Problem 320**
Find the volume underneath the surface bounded by and over the xy-plane
9. (Original post by henpen)
Solution 318
Incomplete.
Spoiler:
Show

Let N be one such 3-digit number. Obviously N is divisible by 11, so

, with and .

Let

.

.

.

Now, check each case:

using the inequalities, a must be positive, so

Which has no real solutions.

Again, no real solutions, so the only N for is

.
Nicely done, so far.
But yes, as Smaug pointed out, it is incomplete.
10. Question 321*:Requires only A level knowledge... I hope this provides a challenge

The function f is defined by f(x) = x⁴ - 9x³ + 28x² - 36x + 16, x ∈ R; and u = x + 4/x , x ≠ 0

(i) Express u² in terms of x

(ii) Using the result from (i) show that if f(x) = 0, then u² - 9u + 20 = 0

(iii) Hence, determine the values of x ∈ R for which f(x) = 0

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11. Solution 321*

i.

ii.

iii.

12. Did it even challenge you to prove part ii?

To the OP, sorry for not using Latex, I only found out after... I'm a noob

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13. (Original post by Arieisit)
Did it even challenge you to prove part ii?
Not really, I spotted it almost immediately. There aren't many problems of that difficulty on this thread so I thought I'd have a shot .
14. (Original post by TheMagicMan)
Spoiler:
Show
There are lots of people on.here better at maths than me, so I was wondering if you could explain something to me. On page 92 of this PDF: http://www.jmilne.org/math/CourseNotes/FT.pdf, it says at the end of the second paragraph 'one sees easily...' Can any of you flesh out why it's a unique topology? In particular why can't we perform a shift on the topology to generate a new one? It's probably something obvious but I've been staring at it for half an hour and I can't for the life of me see it.

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If you are still interested...

We know that from we can recover the neighborhood filter of the identity. Now, just note that for each , and coincide with the neighborhood filter of (since the maps and are homeomorphisms and , , , i.e. condition ); hence the topology is uniquely determined.

Solution 318

Let . Note that and . Thus, .
Consider two cases:

If , we have , which gives and , for .

If , we get . Taking modulo 4, we see , and so and , by casework.

Solution 318

Let . Note that and . Thus, .
...
Nice solution. Undoubtedly I'm being thick, but what happened in the 'thus' logical procedure?
16. (Original post by henpen)
Nice solution. Undoubtedly I'm being thick, but what happened in the 'thus' logical procedure?
It can't possibly be 22, because the most it can be is a_0+a_2 <= 18.
If you are still interested...

We know that from we can recover the neighborhood filter of the identity. Now, just note that for each , and coincide with the neighborhood filter of (since the maps and are homeomorphisms and , , , i.e. condition ); hence the topology is uniquely determined.

.
Thanks. I did actually get it in the end when I came back to it. I'm not really sure why I couldn't see it first time up.

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18. (Original post by Smaug123)
It can't possibly be 22, because the most it can be is a_0+a_2 <= 18.
Of course, thanks.
19. Problem 322**/***

Given

Find a closed form for
20. (Original post by FireGarden)
Problem 322**/***

Given

Find a closed form for
Solution 322

Differentiate both sides to get

This is a simple seperateable differential equation, leaving the details out the solution is

Using the original equation we get

Put this with the above to get

This gives us that all solutions must be of the form

Plugging this back into the original equation shows that all of these are indeed solutions.

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