TT's Maths Thread

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   !!!!!!!!!!!
 

[dcc]

From the comment you left in my karma log, it is obvious that you've never read the Hitchhiker's Guide to the Galaxy.

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lol I did a thesis on it in year 10 extension english

[Over9000]

The meaning of life, if you will.

HOW DID YOU DO IT???


did u guess?

[kenhung123]

Gundam 00 is SOOOOOOOOHHHHHHHHH GOOOOOOOOOOODDDDDDDDDDDD I cleaned my room
My personal review of gundam 00: ITS FUKN GOOD
My calculator used to smell yummy, what happend to it?
Now it smells like shit and its dirty
So random!

[Over9000]

Gundam 00 is SOOOOOOOOHHHHHHHHH GOOOOOOOOOOODDDDDDDDDDDD I cleaned my room
My personal review of gundam 00: ITS FUKN GOOD
My calculator used to smell yummy, what happend to it?
Now it smells like shit and its dirty
So random!

wtf, wtf r u on?? kuong?

[Over9000]

I'm gonna do the same whole personal benefit thing too:

Let be a set consisting of n positive integers, prove that the set contains at least n elements.

QED.

[TrueTears]

wdf gone for holidays for a week and my thread is filled with spam =.=

[TrueTears]

yay thanks kamil for hint I got it

Let the set which contains numbers that can be formed by reordering their digits be called .

So for example, one type of can contain or etc. This contains a total of elements.

Another type of can contain etc which would only have elements.

Therefore to count the number of multiples of between and would be denoted by

Thus using the pigeon hole theorem, if we have many multiples of and groups of set then at least one set will contain

Thus if we can find out how many there are such that then we have answered the question.

Consider using encoding to find out .

Let denote the amount of in a digit number such that

So say for example, the number has , and the rest Now notice how

Now we have created a bijection between and a certain set which means if we can find out how many different solutions there are to then we have found out how many groups of there are (which is equal to )

Now using a similar trick Ahmad used earlier:

Consider 's.

That string above would represent the string such that

Now if we move the sign somewhere say like this:



That string above would represent the string

This means that by rearranging the signs we get a different set of solutions to

So how many ways can we rearrange the signs?

Well there's different 'slots' to place and and we need to pick slots to place a which means the slots fall naturally into place.

Therefore there are ways of rearranging the signs.

And since we created a bijection between the different rearrangement of the strings and



Subbing back in yields:



So yes there do exist at least digit numbers divisible by , all of which can be obtained from one another by a reordering of the digits

Eh just relooking at my working is there a mistake somewhere? Cause I worked out the multiples of 7 from 1 to 9,999,999,999 but the question asked for multiples of 7 with 10 digits only...?

[TrueTears]

Some new questions:

1. How many strictly increasing sequences of positive integers being with and end with ?

2. For any set prove that the number of its subsets with an even number of elements is equal to the number of subsets with an odd number of elements. For example, the set has four subsets with an even number of elements (the null set has elements which is even), and four with an odd number of elements.

3. Use a combinatorial argument to show that with



4. We are given points arranged around a circle and the chords connecting each pair of points are drawn. If no three chords meet in a point, how many points of intersection are there? For example, when , there are intersections.

[kamil9876]

question about multiples of 7: I think it assumes that say, 9 is really 000 000 000 9. I think if you included that smaller numerator then you wouldn't get a lower bound of 15456 but something smaller and hence you will not be able to conclude that it's greater than 10000.

1.)
Clue:

the stricly increasing sequence 2,4,6,8...998 (with 1 and 1000 on the ends) can be represented as CNCNCNC...NCN

where C means chosen, N means not chosen.

2.)

You can turn this into an algebra problem, ie prove that:


which is equivalent to:



4.)

Interesting Q. Try to look for some patterns. E.g: say you label your points like a clock: . Now if you connect to then for that particular chord, you have (i-2) points on the right, and (n-i) on the left, now the only chords that will cross your chord will be the ones being connected to points on opposite sides, so now you have to work how many different such chords are there for each situation. Then do this for all pairs, and try to figure out what you have overcounted and how to fix it.

[addikaye03]

http://vcenotes.com/forum/index.php/topic,20009.0.html

I asked some more Q here, if people would like to attempt them, they're interesting imo

[TrueTears]

question about multiples of 7: I think it assumes that say, 9 is really 000 000 000 9. I think if you included that smaller numerator then you wouldn't get a lower bound of 15456 but something smaller and hence you will not be able to conclude that it's greater than 10000.

Oh I see, but say if I worked out the actual "number" of 10 digit multiples of 7, it wouldn't really affect anything, just a smaller lower bound heh

[TrueTears]

1.)
Clue:

the stricly increasing sequence 2,4,6,8...998 (with 1 and 1000 on the ends) can be represented as CNCNCNC...NCN

where C means chosen, N means not chosen.

Oh yeah, I was trying to find out an encoding method but didn't get it. Thanks for that I got it now.

2 choices for each number.

Consider the sequence

That is encoded by

Now say you have slots for numbers from 1 to .

Since and are fixed at both ends the number of increasing sequences is

[kamil9876]

question about multiples of 7: I think it assumes that say, 9 is really 000 000 000 9. I think if you included that smaller numerator then you wouldn't get a lower bound of 15456 but something smaller and hence you will not be able to conclude that it's greater than 10000.

Oh I see, but say if I worked out the actual "number" of 10 digit multiples of 7, it wouldn't really affect anything, just a smaller lower bound heh

yeah but if it is too small, we cannot say at least 10000