Suppose you are playing with fractions 1,1/2,1/3,.... 1/100. At each step, you take two of the fractions x,y and replace them by a function f(x,y). You keep repeating this process until only one number remains. What are all the possible values that the remaining number can take if f(a,b)=ab+a+b? Can you solve the puzzle again if f(a,b)=ab/(a+b)?

- via Australian mathematical society puzzle corner 34

1. For the first one, write it as f(a,b) = (a+1)(b+1)-1, which gives the final answer as (1+1)*(1+1/2)*...*(1+1/100)-1 = 100.

2. For the second one, write it as 1/f(a,b) = 1/a + 1/b, giving the final answer as 1/(1+2+...+100) = 1/5050.