1. Originally Posted by blahblahblah
cool stuff redbacks..

u got any problems u willing to share?

u r welcome =)
I uncovered some old 'comp maths' books we were given at uni so I have hundreds of puzzles for you to enjoy

a) Just after three o'clock the school fire alarm rings for exactly 10 minutes. If the angle between the minute hand and hour hands was three times bigger when the alarm started than when it finished, when exactly did the bell start ringing. (draw a diagram )

b) Prove that if p and q are odd numbers then p^2 - q^2 is divisible by 8.

2. Originally Posted by Redbacks
I uncovered some old 'comp maths' books we were given at uni so I have hundreds of puzzles for you to enjoy

a) Just after three o'clock the school fire alarm rings for exactly 10 minutes. If the angle between the minute hand and hour hands was three times bigger when the alarm started than when it finished, when exactly did the bell start ringing. (draw a diagram )

b) Prove that if p and q are odd numbers then p^2 - q^2 is divisible by 8.
A. 3pm. At 3pm, you have a 90 degree angle. at 3:10pm, you have a 30 degree angle.

B. I've done this problem before, so it's unfair . Hint: Factor!

3. in (a) you have to take into account that the hour hand moves also.

4. b), on a piece of little paper.

Let q = 2a-1, b= 2b-1, a,b in N*
p^2-q^2 = 4(a-b)(a+b-1)

Case a-b is even: a-b = 2n, n in N*, hence p^2-q^2 = 8n(a+b-1) | 8 as required
a-b odd: a+b is odd (=(a-b)+2b, odd+even = odd), hence a+b-1 is even ie a+b-1=2n etc.as above

QED

Here's one from an exam today that gave me the ****s

Find an element x (or prove there does not exist one) of S5 such that when a is conjugated by x it gives b, where a = (15)(243) and b = (135)(24)

And no I don't know the answer.

5. Originally Posted by silentstriker
A. 3pm. At 3pm, you have a 90 degree angle. at 3:10pm, you have a 30 degree angle.
The hour hand moves too in that ten minutes so it's not exactly 3pm. The hour hand moves 1/12th of the speed of the minute hand.

Letting M be the number of minutes that have passed since 3pm, H be the position of the hour hand and x be the number of minutes past 3 the clock originally was, we can derive three equations:

H=15+M/12
M=x+10
H-M=(H-x)/3

so
x=M-10,
x=3M-2(15+M/12)=17M/6-30

17M/6-30=M-10, M+20=17M/6, 6M+120=17M, 11M=120, M=10.9090..

But we actually want x, which is M-10. So unless I've yet again ballsed up the algebra at some point, the alarm went off at exactly 0.90r minutes past 3.

6. Ah, right. Of course .

7. The answer is slightly out, H will move also during the 10 mins but appears to be given equivalent in the 3rd equation

8. Here is how I attacked the problem + the uppercut method which give the same results

9. Love your hand writing. Very mathematician-esque.

10. ∫e^(x^2) dx

Came across this in a homework and have no idea how to do it, any help?

11. You can differentiate that, right? Integrating it is just doing the opposite. Divide by the factors you'd multiply by when differentiating, keep constant what you'd keep constant when differentiating.

12. Well yeah I can differentiate it and I can integrate lots of other ones like that, but I can't do that specific one.

13. Maybe I'm missing something... why not?

14. Lack of something like x or 2x before the e term...

Try if you can get an answer for it..

15. The coefficient of e is 1.

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