Mathematics marathon

[SpacePanda15]

Ummm I have a question(not math) How old are you guys. I only understand some of this stuff. I'm 13. I don't want ot fell that I am very slow

 

[Akash Rupela]

aaaah

Spacepanda15, i m 18

 

[MTGjumper]

This has been annoying me:

Using the Chinese Remainder Theorem, prove that there are a million consecutive integers, each of which is divisible by at least a million different prime numbers.

Some kind of hint would be nice, but feel free to include your solution

 

[y235]

I have a problem:
Let A be a group in the plane that its area is bigger than 1. prove that you can move it such that, so it will contain 2 integer points (integer point - point (a,b) and a,b are integers)

 

[Erdos]

This has been annoying me:

Using the Chinese Remainder Theorem, prove that there are a million consecutive integers, each of which is divisible by at least a million different prime numbers.

Some kind of hint would be nice, but feel free to include your solution

Denote the set of consecutive integers as X, so |X| = 10^6., and n is the first of the consecutive integers. Now suppose n is divisible by at least a million different prime numbers. Therefore, consider the group Z_n. Using CRT, what is this isomorphic to? How does Z_n+1 relate? Then use induction for formalities, although since |X| is a finite number, you don't really need to.

 

[macky]

This has been annoying me:

Using the Chinese Remainder Theorem, prove that there are a million consecutive integers, each of which is divisible by at least a million different prime numbers.

Some kind of hint would be nice, but feel free to include your solution

Hint: Use more than a million primes.

A solution:

Spoiler

It's no harder to show that there are \( N \) consecutive integers, each divisible by at least \( M \) different primes. Choose \( NM \) primes indexed as \( p_{ij}, 1 \le i \le N, 1 \le j \le M \). We will take the string \( a+1, \ldots, a+N \) so that \( a+i \) is divisible by \( p_{ij}, 1 \le j \le M \). For this it suffices to take \( a \equiv -i \pmod{p_{ij}} \) using CRT.

I have a problem:
Let A be a group in the plane that its area is bigger than 1. prove that you can move it such that, so it will contain 2 integer points (integer point - point (a,b) and a,b are integers)

Reworded: Given a region in the plane whose area [whatever that means] is greater than 1, prove that some translation contains 2 integer points.

A solution:

Spoiler

Reduce mod \Z \times \Z so that the region lies in [0, 1) \times [0, 1). Since it has area greater than 1, some two points have the same reduction, and these can be translated to integer points.

 

[hic0057]

Got given this on a math comp and I have no idea how to work it out.

The first digit of a six-digit number is 1. This digit is now moved from the first digit position to the end, so it become the last digit. The new six-digit number is now 3 times larger than the original number. What are the last three digits of the original number.

Can you please explain how to get the answer so that a fairly smart 15 years old could understand. Thanks

 

[y235]

Got given this on a math comp and I have no idea how to work it out.

The first digit of a six-digit number is 1. This digit is now moved from the first digit position to the end, so it become the last digit. The new six-digit number is now 3 times larger than the original number. What are the last three digits of the original number.

Can you please explain how to get the answer so that a fairly smart 15 years old could understand. Thanks

A clue: try to think how the numbers are related to each other.

Spoiler

the new number is 3 times bigger than the original one, which means that the last digit of the original one must be 7 (because 7 is the only digit that if you multiply you get 10X+1). the second digit from the end must be 5, because 5 is the only digit that if you multiply it by 3 and add 2 (because 7*3=21) you get 7.
in the same way the third digit from the end is 8.

Sorry for my bad english, Hope you'll understand.

 

[hic0057]

Spoiler

the last digit of the original one must be 7 (because 7 is the only digit that if you multiply you get 10X+1)

Spoiler

Thanks for that, really helpful

 

[chicken9290]

wow i see problems like this every wednesday at math team.

 

[hic0057]

Got another one that my maths teacher couldn't help me with.

The diagram shows the net of a cube. On each face there is an integer: 1, w, 2011, x, y and z.

[w]
[x][y][2011][z]
[1]



IF each of the numbers w, x, y and z equals the average of the numbers written on the four faces of the cube adjacent to it, find the value of x.

Could you please explain how to work this out so that a fairly smart 15 years old could understand.

 

[blah]

Got another one that my maths teacher couldn't help me with.

The diagram shows the net of a cube. On each face there is an integer: 1, w, 2011, x, y and z.

[w]
[x][y][2011][z]
[1]



IF each of the numbers w, x, y and z equals the average of the numbers written on the four faces of the cube adjacent to it, find the value of x.

Could you please explain how to work this out so that a fairly smart 15 years old could understand.

Spoiler

Opposite faces must have the same value. If you can't figure out why, then you haven't given this problem the thought it deserves, which really isn't much. So w = 1, x = 2011, and y = z = 2012/2 = 1006.

Your math teacher probably didn't even try >_>

 

[hic0057]

Spoiler

Opposite faces must have the same value. If you can't figure out why, then you haven't given this problem the thought it deserves, which really isn't much. So w = 1, x = 2011, and y = z = 2012/2 = 1006.

Your math teacher probably didn't even try >_>

Umm, The answer is only 3 digits.

 

[blah]

Umm, The answer is only 3 digits.

And what does that tell you?

 

[qqwref]

And what does that tell you?

That 1 = 2011?

 

[hic0057]

That you are wrong.

This question was on an Australian National Mathematics Competition.

Quote from the text paper
"Requires a whole number answer between o and 999"

 

[y235]

That you are wrong.

This question was on an Australian National Mathematics Competition.

Quote from the text paper
"Requires a whole number answer between o and 999"

Blah's solution seemes correct to me.


EDIT:

The question is worded kinda poorly. It says that w,x,y,z are equal to the average of their adjacent sides but doesn't say 1 and 2011 are.

forgot about that

 

[hic0057]

Would someone be able to explain how to get an answer between 0 and 999?

 

[aronpm]

The question is worded kinda poorly. It says that w,x,y,z are equal to the average of their adjacent sides but doesn't say 1 and 2011 are.

 

[hic0057]

The question is worded kinda poorly. It says that w,x,y,z are equal to the average of their adjacent sides but doesn't say 1 and 2011 are.

Just wondering, did you do the test.