Any complicated maths problems?

[nathan444]

Hey, I'm an A-level maths student and I was wondering if anybody has any pretty complicated equations that can be solved using just the rules of indicies and general algebra.

I ask because we're not really doing much in year 13 yet and I love a challenge when it comes to maths lol.

 

[micromass]

Here are some fun ones:

Easy:
- Find two perfect squares such that their difference is 63.

- Find a 5-digit number (which is not 00000) such that the cube of the first two digits is that 5 digit number. Thus find XYZUV such that $XY^3=XYZUB$

Moderate:
-Find 1000000 consecutive integers which are not prime.

- Take a regular n-gon whose vertices lie on the unit circle. Take one vertex and call it x. Draw lines between all the vertices and x. Prove that the product of the lengths of these lines is exactly n.
Generalize.

Hard:
Take the numbers $1,2,3,...,2^{1000}+100$. The average of these numbers is not an integer. How many ways are there to remove 3 consecutive numbers in this sequence such that the average is an integer?

Unsolved:
Take an arbitrary positive integer $x_0$. If this is even, then write $x_1=x_0/2$. If $x_0$ is odd, then write $x_1=3x_0+1$. If $x_1$ is even, then $x_2=x_1/2$. If $x_1$ is odd, then $x_2=3x_1+1$. Continue to define $x_n$.
Prove: For every arbitrary $x_0$ we pick, there exists an n such that $x_n=1$.

Good luck!

 

[nathan444]

Okay, I got the very first one with no mathematics at all lol, 12 and 9 were just the only logical answers that sprang to mind straight away.
However, with the second one can you give me a hint as to how I should start it, I've written down:
ABCDE=AB^3
Obviously that isn't an algebraically correct way of setting it out as that assumes A*B*C*D*E=A*(B^3).
what would you do from there, please don't tell me the answer, just give me a little nudge :P

 

[micromass]

Okay, I got the very first one with no mathematics at all lol, 12 and 9 were just the only logical answers that sprang to mind straight away.
However, with the second one can you give me a hint as to how I should start it, I've written down:
ABCDE=AB^3
Obviously that isn't an algebraically correct way of setting it out as that assumes A*B*C*D*E=A*(B^3).
what would you do from there, please don't tell me the answer, just give me a little nudge :P

OK, I'll write down the equation that you need to solve. You need to find $0<x<100$ such that

$$x^3-1000x<100$$

Do you see what I did?

 

[micromass]

Okay, I got the very first one with no mathematics at all lol, 12 and 9 were just the only logical answers that sprang to mind straight away.

Follow-up question: let a be an odd number. Is a necessarily the difference of two squares?
Follow-up question: what if the condition that a is odd is dropped?

 

[micromass]

Some tricky questions:

Easy:
Say you have all the integers between 2 and 100 written on cards. Take the card "2". Fold the card and fold every card divisible by 2. Then take "3". Fold the card and fold every card divisible by 3 (and 6 gets to get be folded twice now). Do the same with 4,5,6,...,100.
Find the end position of the card. Which ones are folded??

Moderate:
There are 1000 ants walking on a stick. They can only walk right or left. If they bump into another ant, then they change their direction. If they come to one of the ends of the stick, then the ant falls of the stick.
Assume that the stick is 1m long and that every ant walk 0.1 m/s. How long does it maximally take for the stick to have no ants on it??

 

[nathan444]

OK, I'll write down the equation that you need to solve. You need to find $0<x<100$ such that

$$x^3-1000x<100$$

Do you see what I did?

Where did the 1000x bit dome from?
Sorry, it's 2:30am, I'm really tired and my head isn't working properly lol.

 

[micromass]

Where did the 1000x bit dome from?
Sorry, it's 2:30am, I'm really tired and my head isn't working properly lol.

Oops, it's wrong. It has to be

$$0<x^3-1000x<1000$$

Well, try to solve it now. I won't be giving any more hints now

 

[Anonymous217]

Here's another one: Can you express any whole number as the sum of distinct powers of 2? If so, can you do it for distinct powers of 3? 4? n?

Besides the actual proof-writing, it's something anyone can play around with and see what happens. And depending on your background, you could be surprised what type of things come out of this.

And here's a fun one: Are there an infinite number of primes? Are there an infinite number of primes whose difference with another prime is 2?

 

[nathan444]

Oops, it's wrong. It has to be

$$0<x^3-1000x<1000$$

Well, try to solve it now. I won't be giving any more hints now

I understand the 1000x now.
I've got x=32

I rearranged it to 0<x(x2-1000)<1000
As x had to be a 2 digit number, I tried x=10 and x=99 but both were far out of the acceptable range.

I then used trial and error going up in multiples of 10 from 10->30 and got x=32 :)

That method seemed pretty messy though, is there a more mathematical way of working it out?

 

[nathan444]

Moderate:
There are 1000 ants walking on a stick. They can only walk right or left. If they bump into another ant, then they change their direction. If they come to one of the ends of the stick, then the ant falls of the stick.
Assume that the stick is 1m long and that every ant walk 0.1 m/s. How long does it maximally take for the stick to have no ants on it??

Okay, I had no clue with this one so I scaled it down to 10 ants on a 1cm long stick all moving at 10 cm/s

They were arranged like this:
>>>>><<<<<
and by my calculations it took 0.14 seconds for the stick to be completely empty.

Assuming it works to scale then the 1m long stick should take 14 seconds to empty.

 

[micromass]

Okay, I had no clue with this one so I scaled it down to 10 ants on a 1cm long stick all moving at 10 cm/s

They were arranged like this:
>>>>><<<<<
and by my calculations it took 0.14 seconds for the stick to be completely empty.

Assuming it works to scale then the 1m long stick should take 14 seconds to empty.

Close, but it only takes at most 0.1 seconds. I'm curious as to which calculations you did, though...

 

[micromass]

I understand the 1000x now.
I've got x=32

I rearranged it to 0<x(x2-1000)<1000
As x had to be a 2 digit number, I tried x=10 and x=99 but both were far out of the acceptable range.

I then used trial and error going up in multiples of 10 from 10->30 and got x=32 :)

That method seemed pretty messy though, is there a more mathematical way of working it out?

Well, you need to find x such that

$$0<x(x^2-1000)<1000$$

So let's solve $x(x^2-1000)=0$, this yields $x\sim 32$. So 32 should be the first number such that $0<x(x^2-1000)<1000$.

 

[nathan444]

Yeah I guess I messed the ant thing up lol.
How would you start that one off?

 

[micromass]

hint: bumping doesn't matter. You can do like the ants walk through each other.