Homework 1 was due on Thursday 23rd January 2014. I graded Q4,5 and the grader graded Q1,2,3. All questions are marked out of 6.
Question 1. The most common error was checking that the values of $a$ you obtain actually worked. If you didn't do this, it is likely that you showed that if $a$ works then $a = \pm 1$; but you still need to show that if $a=\pm 1$ then $a$ works!
Question 2. This problem was novel in that it was very wordy, and solutions were wordy too. This is absolutely fine, and in fact is preferable than abrupt answers with details omitted, but even if you give a wordy answer it must be precise! A lot of people had the correct idea of taking different numbers of coins from each machine, but to justify that this method would work they gave examples of what would happen if you picked a particular machine. This does not constitute a proof unless you do it for all 20 machines! But there's no need to do it for all 20 machines; the best answers would have generalised to the scenario where there were any number of machines.
Question 3. A lot of people used the words 'worst case' without really explaining what it meant. A bit more precision was perhaps needed here. The idea is that in order to have at most one sock of each colour, the number of socks you have must have at most the number of colours there are (and this scenario is possible just by picking one of each colour). Then if you have one sock of each colour already, picking another will match one of the socks you have already picked.
Question 4. Much like with Question 1, by far the most common error was neglecting to prove that the solutions you obtain actually work. Most people correctly identified that if $\sqrt{x-1} = x-3$ then $x=2$ or $x=5$. The trap here is that $x=2$ doesn't work, since $$\sqrt{2-1} = 1 \ne -1 = 2-3$$ (Remember: the square root symbol always refers to the positive square root!) This is due to a phenomenon called injectivity, which you'll learn about later: informally speaking, a function is injective if you can undo it. Squaring isn't injective, since for instance $(-2)^2 = 2^2 = 4$ (so, given the number $4$, we can't know which of the numbers $2$ or $-2$ it came from), and that's what gives rise to these extraneous solutions. Injectivity is a concept that will come up later in the course.
Question 5. The most common error here was people asserting without justification that if $p^2$ is divisible by $3$ then $p$ is divisible by $3$. The vast majority of people gave a very long-winded argument involving whether $p$ and $q$ are even or odd... this typically gave rise to correct answers, but the proofs weren't very nice.