Solutions to the Problems

1. a) Assuming the break points are chosen uniformly over the length of the match, the pieces will form a triangle 1/4 of the time, b) The answer is changed if we break the match in stages - now the probability is 2 ∫₀^1/2 x/(1-x) dx ≈ 0.386
2. Consider √ 2 ^{√ 2}. If this number is rational we are done and dusted, if it is not then raising it to the power √ 2 gives a rational number: √ 2 ^{√ 2 √ 2} = 2. (Beware, some browsers have trouble displaying these nested superscripts correctly.)
   
   
3. In terms of the arrangement of the ants, a collision between two ants is equivalent to the ants just passing through one another. The ants are effectively indistinguishable for the purposes of the problem and can be treated as independent. Therefore, regardless of the number of ants, all of them will fall off within one hour. There are lots of arrangements that can attain this one hour time.
   
   
4. Both numbers are the Golden ratio, (1+√ 5 )/2
   
   
5. Using the fact that the total number of ways of pairing off the class is:
   
   tot(N) = (N-1)*(N-3)*(N-5)...*1
   
   Part 1) The fraction of same sex only combination is:
   
   p(same sex only|N) = 0 if N odd
   p(same sex only|N) = tot(N/2)²/tot(N) if N even
   
   Part 2) The fraction of mixed only combinations is:
   
   p(mixed only|N) = N!/tot(N)
   
   
   
   
6. The top end accelerates quickly towards the bottom end, but the bottom remains (approximately) fixed until the slinky has contracted. To convince yourself, consider running the same experiment using a spring with (unequal) masses at each end. Then generalise to a chain of masses and springs.
   
   
7. A surprisingly small 2 π metres (i.e. about 6metres)
   
   
8. The solution is the Fibonacci sequence: f(N) = f(N-1) + f(N-2), where f(0)=1 and f(1)=1
   
   
9. An approximate solution is informative: Imagine the related problem where the mile of track was hinged at the middle and clamped at the two ends, how high must the (hinge at) the mid-point rise when you add a foot of extra track:
   
   h = ( 2640.5² - 2640² )^1/2 feet = 51 feet
   
   Which was 51 times my initial ball-park guess and quite close to the answer for a circle, which is 45 feet.
   
   
10. Using the following strategy, the duck can escape even if the fox is four times faster: First, the duck swims to a point where it can circle round the centre of pond at a rate which is infinitesimally faster than the fox. Second, the duck circles until it is on the opposite side of the pond to the fox (this will take some time...). Then, thirdly, the duck dashes to the pond's edge. The duck will escape so long as its speed is greater than the 1/(1+π) times the fox's speed.
    
    
11. Prime numbers are odd and so we can let p=2k+1 where k=1,2,3,4 ... So, p²-1 = 4 k(k+1). As either k or k+1 are even, this means p²-1 must be divisible by 8. Now we need to show that it is also divisible by 3. First we note that a prime number greater than 3 is always one away from a number which is a multiple of 3. Thus, p ± 1 = 3 n (where ± means either/or but not both and n=1,2,3,4 ...). Therefore, p²-1 = 3(3 n ± 2), which indicates that 3 is a factor. This completes the proof.
    
    
12. The square numbers (1,4,9,16...) remain heads up at the end of the process.
    
    
13. For a full discussion of the problem and a solution please read this note.
    
    
14. For a full discussion of the problem and a solution, please read this note.

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