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Hi everyone!

 My name’s John Papantoniou and I’m offering group and individual mathematics tutoring for the HSC, IB and anyone younger looking to do extension work or prepare for maths competitions.

 To start with some background information - I graduated from Sydney Grammar back in...

I'd say if you were really keen on getting involved with olympiad maths, the three things that might help would be to look up a book called "The Mathematics Toolchest" that the Australian Maths Trust sells, any of the past Australian Maths Olympiad papers you could find, and any past junior...

(I have to respond in two parts - my first message was too long) Sorry I didn't respond to this for so long - I only just noticed your message - but as to your question, the resources that would best help you get better at problem solving really depend on your age and experience. For example...

If such a polynomial existed, then as P has integer coefficients, a-b|P(a)-P(b) -> a-b|b-c, and similarly b-c|P(b)-P(c), so b-c|c-a and then c-a|P(c)-P(a), so a-b|b-c|c-a|a-b..., and then since the modulus of each thing is strictly increasing at each point (Since none of a-b, b-c, c-a can be 0...

So if you can prove that it's true for any particular prime factor you're done, so we'll show that if a prime p is a factor of k, then the exponent it has in the prime factorisation of k is congruent to 0 mod 3. So note that each column has the same power of p dividing it and then let...

Just doing it recursively, if Tn is what you want T1=45 by observation, then T(n+1) = Tn + (Sum of digits of n+1 digit numbers) = Tn + 9Tn (Training n digits of n+1 digit numbers) + 45*10^(n-1) (Leading digits cycle and occur 10^n-1 times each) So Tn = (45*n)*10^(n-1) if I didn't mess up...

This seems untrue, e.g. what about k=8 with 2 2 2 2 2 2 2 2 2 Also just in general, if any grid worked with k a square, you could multiply every entry by a non-square, preserve the property you were talking about and make the new k non-square...

It just factors to (x+y)(y+z)(x+z) = 0, so x=-y, y=-z or x=-z are the solution sets.

If I understand the question correctly, and it's asking if there are any numbers for which every multiple has a 0,1,2...,9 in it somewhere, then I think the answer is no. We can do this constructively and actually show that every positive integer has infinitely many multiples which contain only...

For n=1 it's 1, and we claim for all n>=2 it's (x+b1)(x+b2)...(x+bn)/b1.b2....bn, which follows easily by induction. I.e. Let the sum of n terms be Sn, then S2 = 1 + x/b1 = (x+b1)/b1 which is what we want, Then if it follows up to some n=k, for n=k+1, we have S(k+1) = Sk +...

The number 5 can occur in the hundreds, tens or units digit of an integer in [1,1000], so we do each case: It's clear only the numbers in [500,599] have the 5 as the hundreds digit, so there are 100 instances of 5 as the hundreds digit. Every number in the sets [50,59], [150,159], ...

All they're really doing is finding a nice way to encode the restraint of the problem in steps: So first given you have 8 people, and the order in which you put them in a line matters, it's clear that there are 8! arrangements possible, since you have 8 people to put first, 7 to put second and...

This is actually a pretty neat consequence of the fact that FDEB is a harmonic quadrilateral I.e. Since DE/DF = CE/CD = CB/CD = BE/BF, FDEB is a harmonic quadrilateral - then consider taking a perspective from G of FDEB onto the line FE, F and E stay where they are, D goes to X and B goes to...

Let the integers in the ratio 1:2:3 be a,2a and 3a and let the 4th integer be b, then 11/6a + 1/b = 19/20. Note b>1 as 1/b < 19/20, so 1/b <= 1/2, so if a>=5, then 11/6a + 1/b <= 11/30 + 1/2 < 19/20. Hence it suffices to check a=2,3,4. Luckily a=2 gives 1/b = 19/20 - 11/12 = 1/30, Hence...

If BDX collinear, then by power of a point, XD.XB = XA^2 = XC^2, so XDA and XDC are similar to XAB and XAC respectively (Since you have side ratios and the included angle and stuff...). Then AD/AB = XA/XB = XC/XB = DC/BC, so AD/AB = DC/BC, so AB.CD = AD.BC. I'm not entirely sure what you're...