(Original post by Remyxomatosis)
i don't know if this will be absolutely trivial to some people but oh well.
Problem 498**
Prove that if a sequence of complex numbers converges to , then .
I believe that's basically the formal epsilon delta proof of this, which I have definitely done at least once before on either an example sheet or a past paper.
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Renzhi10122
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Problem 499**
A class of students is organised into teams. Each team consists of 3 students, and no team is identical. Show that there are two teams with exactly one common student.Last edited by Renzhi10122; 22082015 at 00:12. 
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 22082015 19:50
(Original post by Renzhi10122)
Problem 499**
A class of students is organised into teams. Each team consists of 3 students, and no team is identical. Show that there are two teams with exactly one common student.
Spoiler:Show
Let us put the students into teams completely at random. Take two teams? What is the probability that they share exactly one member?
This is . To see this, note that the shared member can be any of the three on the first team. Then the second team must consist of that member and two members not on the first team; note that we cannot repeat members.
As there are n+1 teams, the probability of any two sharing exactly one member is:
Note that for , this estimate of the probability is positive and increasing. Evaluating for gives a result greater than one, so we must always have at least one pair of teams sharing exactly one member.
(Disclaimer: This is probably not handled as rigorously as would have been liked and there is probably a mistake in the precise treatment, but having seen questions of a similar sort solved before I feel like the idea, at least, is along the lines of what is required.) 
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 22082015 20:15
(Original post by DJMayes)
XLast edited by Renzhi10122; 22082015 at 20:16. 
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(Original post by Renzhi10122)
Very nice, it's an approach I would never have thought of (then again, I guess you get to know these kind of things once you're doing maths at uni). The general result is for 'more than n teams', but it's fairly obvious that if n+1 works, then so does everything greater than n+1.
What is your solution then? 
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 22082015 21:57
(Original post by DJMayes)
There's not a specific course on these sorts of questions (at least not the first 2 years) but they pop up every so often as everyone loves a bit of Erdos.
What is your solution then?Spoiler:Show
My solution is an long one.
(skip this paragraph if you want) First, we confirm that for , there are at least 2 teams with exactly one common student. By pigeonhole, one student is in at least 4 teams. Let this student be and let be in teams . Suppose that we are able to form 6 teams such that any two teams share either 0 or 2 common students. There must be a common student in teams and then two different students in those two teams to form . must contain either or both of and . Suppose contains , then the 3rd student must be a student , but then must contain since otherwise, it will have exaclty one common student with one of . However, we have 5 students, so will be identical to one of .
Now suppose that contains both such that . Then will have exactly one common student with one of , contradiction. Therefore, for , there are 2 teams with exactly one common student.
Suppose, for the purpose of induction, that for , there are at least 2 teams with exactly one common student. Note that for any students and teams, there are at least 2 teams with exactly one common student, since we may choose any teams from teams.
We show that for students and teams, there are at least 2 with one common student. Consider any teams selected from our students. Suppose that there are not 2 teams with exactly one common student. By pigeonhole, there exists at least one student '' in at least 4 teams such that Z is not in teams . Let . Then wlog, . Suppose that for some , does not contain . Then but then cannot be in any other , contradiction, therefore is in all . Thus, all are in the form . Suppose that is in one of . Then must contain since it must have 2 common students with all , but cannot be in , contradiction. Therefore, are in all and no . Suppose for , that is in some , then one of must also be in , contradiction, therefore is in no . Therefore, there are students in teams and students in teams . However, there exists at least 2 teams with exactly one common student in teams by the induction hypothesis, so for students and teams, there exists at least 2 teams with exactly one common student.
Last edited by Renzhi10122; 22082015 at 21:59. 
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(Original post by Renzhi10122)
Hang on, why is this so? If we have 100 coins, the probability of getting at least one head is not calculated in the same way? 
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Last edited by ζ(s); 27082015 at 23:33. 
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(Original post by Smaug123)
It's a bit cruel to give one whose answer must be expressed in terms of another infinite sum. The Catalan constant isn't even known to be (ir)rational.Last edited by ζ(s); 27082015 at 23:35. 
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Problem #501*/**
(Is it ok if I don't know the answer to the questions I made up?)
Two spheres of equal radii, r, are on a smooth horizontal plane. Sphere A's center is located at (5r, 0) and sphere B's is located at (0, 5r).
Sphere A is moving towards the origin with speed u, whereas sphere B is moving towards the origin with speed u_{b}. Let t represent the time.
(i) Find the range of u_{b} so that there is a collision between the spheres.
(ii) θ is the angle between their line of centers and the xaxis. Find an expression for θ in terms of u and u_{b} at the instance of collision.
(iii) Suppose θ now decreases from 45° to 0 at a constant rate from the beginning to the collision. Find an expression for u_{b}.Last edited by simonli2575; 09092015 at 20:01. 
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 11092015 21:19
(Original post by simonli2575)
Problem #501*/**
(Is it ok if I don't know the answer to the questions I made up?)
Two spheres of equal radii, r, are on a smooth horizontal plane. Sphere A's center is located at (5r, 0) and sphere B's is located at (0, 5r).
Sphere A is moving towards the origin with speed u, whereas sphere B is moving towards the origin with speed u_{b}. Let t represent the time.
(i) Find the range of u_{b} so that there is a collision between the spheres.
(ii) θ is the angle between their line of centers and the xaxis. Find an expression for θ in terms of u and u_{b} at the instance of collision.
(iii) Suppose θ now decreases from 45° to 0 at a constant rate from the beginning to the collision. Find an expression for u_{b}.
I can show you how I began this, and what happens:
Spoiler:Show
WLOG, let r=1. The distance squared between the centres of the two spheres can be written as follows (writing v instead of u_b):
A collision occurs if the equation has a real solution. You can solve this using the above completed square; I'm not going to because it's ugly as sin.
Next two parts:
(+ signs)
Knowing the time of collision by solving the quadratic equation in your first part allows you to plug it in here to solve for the angle.
I'm pretty sure the final part is impossible to solve as the question is given. Suppose the two spheres collide with an angle of 0 at time t_0. Then:
Which then tells us that:
Unparseable or potentially dangerous latex formula. Error 4: no dvi output from LaTeX. It is likely that your formula contains syntax errors or worse.\tan(\frac{\pi}{4}  \frac{\pi}{4t_0}t) = \frac{\frac{vt5}{ut5}
If u and v are constant (which seems implied in the question, and is used earlier on) then you can perform an easy substitution to deduce that tan can be represented as a rational function. This is a very, very, very big contradiction.
Last edited by DJMayes; 11092015 at 21:21. 
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 11092015 22:38
Sorry if this is completely the wrong thread but I was sent a brain teaser today that I frustratingly can't solve.
"Johnny is at his local newsstand, looking at the various magazines on display. He picked up three magazines, A, B and C. He passes the magazines to the shopkeeper, who enters the amounts of each magazine into the cash register.
'Hang on a minute!' says Johnny. 'You just pressed the multiplication button each time between amounts instead of the addition button.' The shopkeeper smiles and replies 'It doesn't matter. Either way, it comes to £5.70.'
What were the prices of the magazines?"
Again, sorry if this is the wrong place but this has stumped me, so it's probably easier than I think it is. 
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 11092015 23:46
(Original post by Sheepmaster317)
Sorry if this is completely the wrong thread but I was sent a brain teaser today that I frustratingly can't solve.
"Johnny is at his local newsstand, looking at the various magazines on display. He picked up three magazines, A, B and C. He passes the magazines to the shopkeeper, who enters the amounts of each magazine into the cash register.
'Hang on a minute!' says Johnny. 'You just pressed the multiplication button each time between amounts instead of the addition button.' The shopkeeper smiles and replies 'It doesn't matter. Either way, it comes to £5.70.'
What were the prices of the magazines?"
Again, sorry if this is the wrong place but this has stumped me, so it's probably easier than I think it is.
Let's work with cents instead, we have and , if we decompose the latter into prime factors  we get . We need only combine these factors into three numbers which add up to , a bit of trial and error leads you to and cents for each magazine. 
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 12092015 00:05
(Original post by Zacken)
Let's work with cents instead, we have and , if we decompose the latter into prime factors  we get . We need only combine these factors into three numbers which add up to , a bit of trial and error leads you to and cents for each magazine. 
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 12092015 10:29
(Original post by Sheepmaster317)
Thanks very much. That makes sense to me, it's just been a while since I've done proper maths and it's been a much trickier one than the ones in the past. I'm intrigued if the answer they give next week is the same as yours, but that's perfectly legit as far as I'm concerned. 
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 12092015 11:58
(Original post by Zacken)
Haha, you're welcome. It's a bit of a tricky question, my initial thought was to lay it out as a system of three equations and solve that, but I wrote them down and it looked horrendous. Some number theory was all it needed. I'm pretty sure the solution is correct, although whether it's unique or not I cannot say (I'm inclined to say it is, though). Do let me know if you find another approach to it! :) 
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 15092015 12:44
(Original post by Renzhi10122)
I can reduce it down to 10 cases, which is still a lot to check: wlog, A has a prime factor of 19. Then, the largest power of 5 that can divide A is 2, since otherwise, A exceeds 570, contradiction. There are at least 3 powers of 5 left, so wlog, B is divisible by at least 25. Then, you get 10 cases to check.
Rephrase as 95a + 25b + 5c = 570 (so 19a+5b+c = 114) and abc = 2^5*3*5. We must have a < 6 by size considerations. We're down to five cases. 
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 15092015 18:24
(Original post by Smaug123)
You can do a bit better. Rephrase as 19a+25b+c=570 and abc = 2^5 * 3 * 5^3. Taken mod 5, obtain c = a from the first equation. Hence iff any 5 is present in a, then a 5 must be present in c. But 5^5 is too big for all the 5s to be in b, so there is at least one in both a and c.
Rephrase as 95a + 25b + 5c = 570 (so 19a+5b+c = 114) and abc = 2^5*3*5. We must have a < 6 by size considerations. We're down to five cases.
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