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Maths problems

Could you give me some solution for those questions?

1. prove 'n^3 + 5n is divisible by 6' without using induction

2. let m, n be coprime positive integers, with m is greater than 1.
Prove that if log(base m)n is rational then it is 0.

3. there are six towns, such that between each pair of towns there is either a train or bus service (but not both). Prove that there are three towns that can be visited in a loop (going via no other towns) using only on mode of transport.
1. Factorise
2. Maybe raise both sides to a power?
3. Draw a diagram to start with I'd say
Reply 2
cyh910907
Could you give me some solution for those questions?

1. prove 'n^3 + 5n is divisible by 6' without using induction

2. let m, n be coprime positive integers, with m is greater than 1.
Prove that if log(base m)n is rational then it is 0.

3. there are six towns, such that between each pair of towns there is either a train or bus service (but not both). Prove that there are three towns that can be visited in a loop (going via no other towns) using only on mode of transport.


For the first one, n is either of the form 6k, 6k+1, 6k+2, 6k+3, 6k+4, 6k+5 where k is a non-negative integer.
You can factorise the expression and then simply try each possibility. Obviously, 6k+2=2(3k+1), 6k+3=3(2k+1), 6k+4=2(3k+2)..


For the second, then you can raise it to the powers and use some sort of contradiction. If m^(p/q)=n, where (m,n)=1 and (p,q)=1 ...i.e. it's in its lowest form...
Reply 3
cyh910907
3. there are six towns, such that between each pair of towns there is either a train or bus service (but not both). Prove that there are three towns that can be visited in a loop (going via no other towns) using only on mode of transport.
Again, I don't really see how to hint at how to answer this. A very broad hint would be to pick an arbitrary town to start with, and think about how it connects with the other 5 towns, but I don't know how much that will help.

More generally, you've posted 6 questions in the course of an evening. You should spend time on these questions yourself before asking for help. And when you do ask for help, let us know what you've tried already, and what your thoughts are.

I would also say that questions you're asking have a fairly wide range of difficulty - where are you getting them from? The first two here are pretty routine, but the 3rd one is pretty hard. (My guess would be more people know how to answer it because they've seen it before than worked it out for themselves).
Reply 4
if i am not wrong the 3rd question involves the pigeonhole principle, if ur trying qns 3 u gotta be patient becuase it will involves a lot of reasoning.
Reply 5
OCC++
if i am not wrong the 3rd question involves the pigeonhole principle, if ur trying qns 3 u gotta be patient becuase it will involves a lot of reasoning.
It's the same kind of reasoning I'd say, but the solution I have in mind doesn't use pigeonhole (I think).
Reply 6
hmm using a graph theoretical approach?
Reply 7
OCC++
hmm using a graph theoretical approach?
Well it's a standard result from graph theory, but the proof I know (which seems to be the standard one) is really just a few lines of logical reasoning.
Well, there Is a modified version of the pigeonhole principle involved. If there are 5 pigeons that go into 2 holes, 1 hole will receive at least ⌈52⌉=3\lceil \frac{5}{2} \rceil=3 pigeons. If this does not make sense, try to fill those holes while keeping all of them with less than 3 pigeons.
Reply 9
Cookie The Blue Monster
Well, there Is a modified version of the pigeonhole principle involved. If there are 5 pigeons that go into 2 holes, 1 hole will receive at least ⌈52⌉=3\lceil \frac{5}{2} \rceil=3 pigeons. If this does not make sense, try to fill those holes while keeping all of them with less than 3 pigeons.
I s'pose. Though when you only have two holes, the need to appeal to a principle seems a little moot...

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