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Your answer is correct. The 'one other digit' I used was simply 9. My proof is similar to yours, but uses Euler's Totient Theorem. Suppose N is stubborn. Then N=2^a5^bM, where M is coprime to 2 and 5. But 2^b5^aN=10^{a+b}M is also stubborn, hence all multiples of M must contain all nine...

I'll be attempting the 2012 paper, but not officially. That is, I will still be doing it under exam conditions, but no official mark.

Please explain how you acquired that answer.

My apologies, everybody. I am not thinking clearly at the moment. The initial value T_0 = T.

Best: Extension 2 Mathematics. Worst: Extension 1 Mathematics and English.

Sorry, I made a typo. The first term should read T_{n+1}.

$Determine all pairs of rational numbers $ (x,y) $ such that $ x^3+y^3=x^2+y^2

\\ $Define the sequence $ T_{n+1} = \frac{T_n}{1+nT_n}. \\\\ $Find the value of T_{2012}.$

\\ $A number $ N $ is said to be 'stubborn' if when multiplied by some positive integer $ k $, the end product always contains the digits $ 0,1,2,...,9 $ in any permutation, allowing repetition.$ \\\\ $Prove, or disprove, that the number $ N=526315789473684210 $ is 'stubborn'. If so, do there...

\\ $The Fibbonaci Sequence $ f_1, f_2,..., f_n $ is defined by $ f_1 = f_2 = 1 $ and $ f_n = f_{n-1} + f_{n-2}, n \geq 3. \\\\ $Let $ Q = \begin{bmatrix} 1 & 1 \\ 1 & 0 \end{bmatrix}.$ Prove that $ Q^n = \begin{bmatrix} f_{n+1} & f_n \\ f_n & f_{n-1} \end{bmatrix} $ and hence prove that $ f_{3n}...

Not sure at the moment. But I'm doing STEP.

Have fun during your gap year then. Chances are, I won't see you at University. But regardless, do what you like. 'Pre-study' can also be fun, especially with more beautiful topics.

I very rarely see 'Probability Proofs'. I am not sure what you're talking about here. Trigonometric Proofs can be found in the Cambridge Mathematics book.

Read through and attempted some questions from the paper. Some questions are wrong and your 'warning' at the front, , is very silly and demonstrates a lack of understanding about what Mathematics is all about. Your paper is not 'hard' in the same way as the HSC Examination is. However, it is...

Make the denominator of the entire term become (a+1)(b+1)(c+1).

Of course I will be! I cannot imagine myself doing anything else. What about you?

I think I might know someone who may know someone who did the test. I'll ask them to scan a copy of it. I'm really interested to see.

The solution falls out fairly quickly if you use the following identities, but I'm just doing this over the top of my head at the moment. I'll look for a quicker method when I have the time ^^ \\ \frac{a+b+c}{3} \geq \sqrt[3]{abc} = 1 \\\\ a^2+b^2+c^2 \geq ab+bc+ac \\\\ (a+b+c)^2 =...

Personally, I would use an inductive proof.

Could I please see the examination paper that students did today?