# I'm bored. Give me a challenging maths problem to work on. watch

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1. (Original post by multiratiunculae)
hyperbolic arctan?

I said A-levelish
That is A-level. FP2
2. (Original post by multiratiunculae)
hyperbolic arctan?

I said A-levelish
I'm doing it now at A-level.....Further maths
3. (Original post by 13 1 20 8 42)
If so, note that the derivative of x^2 + 6x is 2x + 6 = 2(x + 3)
I was thinking of using u-substitution: why is the derivative important?
4. X and Y go shopping. X buys 1 apple and Y buys 2 apples. How many apples all together?
5. (Original post by multiratiunculae)
What can I deduce about f(sin(x))? If the original function was f(x)=1 for example, then f(sin(x)) has no effect. Obviously, the original function wasn't so, but what can be deduced?
The info gives away that the first part can be done by IBP. When the first part is done, the second deduction is just an application using the same logic.
6. (Original post by multiratiunculae)
I was thinking of using u-substitution: why is the derivative important?
Well this is quite a standard integrating technique; you note that by the chain rule differentiating something of the form (7 - 6x - x^2)^n will give you -2n(x +3)(7 - 6x - x^2)^(n-1); all you need to do is find the right power and the right constant
7. (Original post by 13 1 20 8 42)
Idk what is challenging for you

Here's one that's probably too easy, prove that n^3 - n is divisible by 6 for n a natural number
You can write it as n(n^2-1) which is the same as (n-1)n(n+1).
(n-1)(n)(n+1) is 3 consecutive numbers multiplied together.

Therefore, one of your numbers must be divisible by 2 and one must be divisible by 3.
As we know any multiple of 2 multiplied by any multiple of 3 equals a multiple of 6; because 2x3=6
You then have a third number to multiply, and whatever that number is your solution to the equation must still be divisible by 6, since a multiple of 6 times any number is still a multiple of 6.

So you will have (multiple of 2)(multiple of 3)(multiple of 1).
Which is the same as (multiple of 6)(multiple of 1).
Which is the same as (multiple of 6).

With this method you're only really proving that 3 consecutive numbers multiply into a multiple of 6.

I hope there's a more mathematical way to do it
8. (Original post by multiratiunculae)
A-levelish.
here you go:
http://www.damtp.cam.ac.uk/user/stcs...99paperIII.pdf
9. what is 9 + 10
10. (Original post by MathsAstronomy12)
X and Y go shopping. X buys 1 apple and Y buys 2 apples. How many apples all together?
that moment when you realise the answer is too obvious lol

3?
11. (Original post by serah.exe)
9+10
Edit: o **** Ripper Phoenix said this before me.
I could have sworn you posted that originally and I responded to the "9+10" not "0/0" wtf
12. (Original post by serah.exe)
Ur a funny ol' scallywag
13. (Original post by Jamdroid)
You can write it as n(n^2-1) which is the same as (n-1)n(n+1).
(n-1)(n)(n+1) is 3 consecutive numbers multiplied together.

Therefore, one of your numbers must be divisible by 2 and one must be divisible by 3.
As we know any multiple of 2 multiplied by any multiple of 3 equals a multiple of 6; because 2x3=6
You then have a third number to multiply, and whatever that number is your solution to the equation must still be divisible by 6, since a multiple of 6 times any number is still a multiple of 6.

So you will have (multiple of 2)(multiple of 3)(multiple of 1).
Which is the same as (multiple of 6)(multiple of 1).
Which is the same as (multiple of 6).

With this method you're only really proving that 3 consecutive numbers multiply into a multiple of 6.

I hope there's a more mathematical way to do it
Yeah that's the way. I suppose asserting that the number is divisible by 2 and 3 isn't completely rigorous but it is obvious and in terms of this question seems a bit unnecessary to prove, although it can be done quite easily also.
14. (Original post by Awesome Genius)
Do they even do Maths at Norwich?
In the University of East Anglia? Yes, yes they do.
15. (Original post by serah.exe)
#lifegoalz
On a more serious note did u actually manage to solve

(Original post by serah.exe)
9+10
Edit: o **** Ripper Phoenix said this before me.
16. how do you do 1f?
17. What about 9b, I keep on getting 1 as k but the answer is 1/2 but how?
18. (Original post by coconut64)
how do you do 1f?
Use F=ma
The key word is maximum, so you can use values 3ms-2 (acceleration) and 12000N

Let X equal the number of people
Using f=ma
12000N-(300g+75g[x])=(300+75x)3

And find X by making it the subject.

(12000-300g-75gx)/3 = 300+75x
(12000-300g-75gx)/3 - 300 - 75x = 0
(12000-300g)/3 - (75gx)/3 - 300 - 75x = 0

Multiply by 3, and switch sides around.

225x + 75gx = 12000 - 300g - 900
X(225+75g) = " "
X = (12000 - 300g - 900) / (225+75g)

And when rounding, round downwards to the nearest whole number.
So I got 8 people, I hope that is correct!
19. (Original post by Jamdroid)
Use F=ma
The key word is maximum, so you can use values 3ms-2 (acceleration) and 12000N

Let X equal the number of people
Using f=ma
12000N-(300g+75g[x])=(300+75x)3

And find X by making it the subject.

(12000-300g-75gx)/3 = 300+75x
(12000-300g-75gx)/3 - 300 - 75x = 0
(12000-300g)/3 - (75gx)/3 - 300 - 75x = 0

Multiply by 3, and switch sides around.

225x + 75gx = 12000 - 300g - 900
X(225+75g) = " "
X = (12000 - 300g - 900) / (225+75g)

And when rounding, round downwards to the nearest whole number.
So I got 8 people, I hope that is correct!
I did something similar to yours and got 8.5 but the answer is 11 which doesn't ring any bells. also why didnt you round it to 9? Could you please do the other question I posted? Thanks
20. (Original post by coconut64)
I did something similar to yours and got 8.5 but the answer is 11 which doesn't ring any bells. also why didnt you round it to 9? Could you please do the other question I posted? Thanks
I hate lift questions haha

You have to round downwards because you are using the maximum values to work out X. So the number you get, such as 8.5, is the absolute maximum number of people you can have.

Since you can't have 8.5 people, the maximum, you can only have 8 people.
If you have 9 people the lift will break because it can't hold that force, it can only hold 8.5.

Assuming if 8.5 was the correct answer, of course.

+ the answer for your second question should be 1, I can't see why not.

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