MAT Prep Thread 2020

FWIW, I've not seen anyone post a solution to the "binary question" that I'd expect to lose marks if it was expressed reasonably clearly.

For the one proving 200 is not in S, I wrote the following:

As 4 | 200, the last function applied must be g, for if it is f instead, 200 would be odd, a contradiction.
Hence 200=g(50).
Note that 50 is not in S, as 50 is not divisible by 4 (which it would be if the last function applied is g) nor is it congruent to 1 or 3 (mod 4) (which it would be if the last function applied is f). Hence, 50 is not in S.
Hence, 200 is not in S, for if there exists some combination g^n1f^m1...(1)=200, then g^(n1-1)f^m1...(1)=50, a contradiction.



im not sure if this will lose marks :/

For the one proving 200 is not in S, I wrote the following:

As 4 | 200, the last function applied must be g, for if it is f instead, 200 would be odd, a contradiction.
Hence 200=g(50).
Note that 50 is not in S, as 50 is not divisible by 4 (which it would be if the last function applied is g) nor is it congruent to 1 or 3 (mod 4) (which it would be if the last function applied is f). Hence, 50 is not in S.
Hence, 200 is not in S, for if there exists some combination g^n1f^m1...(1)=200, then g^(n1-1)f^m1...(1)=50, a contradiction.

What do you guys think the offers mean will be this time? Low 70s?

I wouldn't expect this to lose marks. If I'm being picky, I think your last line actually adds confusion rather than clarity (although that may be partly an artifcant of you transcribing it to plain text), but I think you explained everything fine without it (in particular you took care to explain the various cases mod 4), so I wouldn't worry.

Note to everyone: I don't want to write individual commentaries on *every* solution to this, so just assume it will be fine - as I say, I didn't see any that I thought deserved to lose marks.

For ppl doing maths and comp sci what marks do u lot think u got?? I probs got around mid 60s so no idea if I'll get the interview

The marking for Q6 is going to be interesting. I've had a quick go through it, and couldn't get their given answer, solely because (in 3 attempts), I made at least one silly mistake in doing the sums. (Then wrote a few lines of code to both verify their answer and print out all the correct partial results so I could see where I messed up).

In terms of work, I think actually calculating g(7,5) has to be more than the rest of the Q6 put together. But at the same time, it's only (quite a lot of) fiddly arithmetic that a bright and careful year 9 could do. (And conversely, I got it wrong, and I'd feel it was an unfair reflection on my ability to lose many marks on this).

Edit: unless I'm mistaken there's a closed form solution: wonder if anyone argued that $g(n, k) = \binom{n+k-1}{k-1}$

same here... i think it was quite unreasonable for them to have asked about binaries. if they wanted that to be our method i think some more clarification as to how they work was needed as i’m sure most people will never have encountered them before

Let P(n) be the above claim. For n=1, no f nor g could be applied, hence unique. Hence P(1) is true.

For n=3, 3=f(1). Note that there cannot be any other combination, as applying g would make the number be at least 4, which is larger than 3. If another f is applied, the number would exceed 3. Hence P(3) is true.

Assume P(m) is true, i.e. for m in S there must exist a unique combination of f and g. This isn't really right for strong induction. You either wanted to say P(m) is true for all m in S with m < N, or define P(n) to be the statement "for all m in S with m <= N, there's a unique representation in terms of f,g" (or similar).

Suppose n = the least element in S larger than n = m. We split the following into two cases.

Case 1: m is divisible by 4.
Then m=g(n_1), for some n_1 in S that is less than or equal to n. By induction assumption, we have m=g(unique combination of f and g), so m has a unique combination. I don't really like seeing "for some n_1" here - you have a specific n_1 in mind (or should do), so it would be better to state it. It's also important because you need n_1 to be unique.

Case 2: m congruent to 1 or 3 (mod 4).
Then m=f(n_2) for some n_2 less than or equal to n, so by induction assumption m=f(unique combination of f and g), so m is expressed uniquely. Similar comment to case 1.

Note that by previous parts we proved that m cannot be congruent to 2 (mod 4)

Hence P(m) is true.
Hence the claim is true.

I know my presentation is a bit "eh" for this one. Thanks for reading my (unnecessarily) long solution

yeah i stated the last thing in the middle of the solution, and just used it. to get g(7,5) would I penalized for that?

I'm not sure what you mean by "the last thing in the middle of the solution".

Anyone else get question 1 wrong :/

i meant i wrote the last part of the of the previous comment, g(n,k) = is essentially the number of non integral solutions = (n+k-1)C(k-1), in the middle of the solution while doing part the 3/4 of the questions

You might lose marks if you didn't justify the formula. Bit hard to say, but TBH I'd have been uncomfortable just putting down a formula without justification.