Fun questions :)

[kamil9876]

then how about everything except ?

0=0+2*0
1=1+2*0
2=0+2*1

and then x=x+2*0 for x>3

[pi]

A lot of these look interesting

http://www.physicsforums.com/showpost.php?p=3433157&postcount=272

[rife168]

Here is the 2012 University of Melbourne School Mathematics Competition (Senior division) paper for those interested.

[Nagisa]

ok guys, im don't quite remember was while ago. but here, consider the 3 digit number, 370. the sum of the cubes of its digits equal itself, 3^3 + 7^3 + 0^3 = 370. if my memory is correct, there is only 4 such numbers (that have 3 digits). can anyone else think of them. also, number cant begin with 0(wouldnt be 3 digits).

So...

3^3 + 7^3 + 0^3 = 370 ... works

what else??

[paulsterio]

I'm gonna have a go at the UoM Maths Comp questions, I haven't done one in a while and I was never really good with the Senior Papers, but I've been able to do a few of the first ones, so I'll see what happens. Let's see if we can have a discussion + collaboration on the last few questions (which are really, really, really hard).

Question 1)

Essentially we just need to find the overlapping area, which is the triangle. Find the base using Pythagoreas' Theorem.



We can find the height using tan, let the angle between a and the diagonal be







Which can be simplified further, I think, but I'll leave that here for now, I'll be back to do more.

[kamil9876]

I'm gonna have a go at the UoM Maths Comp questions, I haven't done one in a while and I was never really good with the Senior Papers, but I've been able to do a few of the first ones, so I'll see what happens. Let's see if we can have a discussion + collaboration on the last few questions (which are really, really, really hard).

I remember posting two solutions to two of them here: Re: Did anyone do the Melbourne Uni Maths Competition today?

There are also some discussions of the other ones there.

[kamil9876]

ok guys, im don't quite remember was while ago. but here, consider the 3 digit number, 370. the sum of the cubes of its digits equal itself, 3^3 + 7^3 + 0^3 = 370. if my memory is correct, there is only 4 such numbers (that have 3 digits). can anyone else think of them. also, number cant begin with 0(wouldnt be 3 digits).

So...

3^3 + 7^3 + 0^3 = 370 ... works

what else??

Try to use this as an excuse to learn some programming if you havn't. This exercise inspired me to do some programming which I havn't done in ages. Results I got are:

153
370
371
407

[Nagisa]

haha nice man, u got em all. i just have em in my head .

Was it a long code? and do you think there'd be a way of doing it without a computer testing everything in 1 second.??

[b^3]

Using Matlab it doesn't seem too long a code (although forgive me for using matlab....)

Code: [Select]

i=0;
for n=100:999
    a=num2str(n);
    a1=str2num(a(1));
    a2=str2num(a(2));
    a3=str2num(a(3));

    num=a1^3+a2^3+a3^3;
    if num==n;
        i=i+1;
        b(i)=n
    end
end
Result

Code: [Select]

b =

   153   370   371   407
Not sure if there is a proper way of doing it though.

[Nagisa]

yeah, would be pretty cool though. to show it on paper.

[Planck's constant]

Is the answer to q7,

1 + 0.5*n*(n-1)

?


It is my no means a proof, but I started playing the game and my intuitive plan was to try and put as much distance as possible between the first and second coins and then iterate the process to the right.

eg, starting with 6 coins

oooooo

First move the 5th coin, then the 4th, then the 3rd then the 2nd

o----ooooo

Now repeat the process with the 5 rightmost coins

o----o---oooo
o----o---o--ooo
o----o---o--o-oo  endgame


The sequnce of moves if we start with 7 coins,

ooooooo
o-----oooooo
by iteration (the 6 coin problem was 'solved' above)
o-----o----o---o--o-oo  endgame

The arithmetic sequence thus formed gives the answer.

Is there a smarter way to play the game ?

[b^3]

Ok I got bored so I tried taking that programming one up a step. I.e. For 4 digits taking the sum of fourth power digits and so on.
3 Digits, 3rd power - 153           370          371        407
4 Digits, 4th power - 1634        8208        9474
5 Digits, 5th power -  54748     92727     93084
6 Digits, 6th power - 548834

[Nagisa]

wow i really got you going didn't i. This is really cool, i wish i could program and experiment around with this stuff.

[saradom136]

45. A job needs three men to work for two weeks (10 working days).
Andy works for all 10 days.
Bert works for the first week and Clive works for the second week.
Dave works for 6 days, but then is too sick to work.
Eddy takes his place for 3 days, then Fred does the last day.
When the job is finished they are all paid the same amount. At first they could not work out how much each man should have, but then Fred says: "If I give my wages to Andy, and Eddy gives 100 to both Dave and me, then the wages will be correct". How much was paid for the whole job, and how much does each man get? 
3
Total days worked= 3x2x5 = 30         
Fred's work as a fraction of total work = 1/30
 $100/x = 1/30, X= 100x30 = 3000= total amount payed
$3000/6 = uncorrected payment per man= $500
corrected pay= fraction of total days worked x total payment

total pay is $3000
Andy 10/30 x 3000= $1000
Bert 5/30 x 3000= $500
Clive 5/30 x 3000= $500
Dave 6/30 x 3000= $600
Eddy 3/30 x 3000= $300
Fred 1/30 x 3000= $100