Rugby scoring

Formosa

What (if anything) is the lowest score above 4 that can't be achieved in rugby (ignoring that game only lasts 80 mins so impossible to score 23,477 etc etc)...

Fremen

You can make all other numbers.

To see this, note that it's possible to make 10,9 or 8: try and coversion gives you 10. Three conversions give you 9, and a penalty and a try with no conversions give you 8.

Now pick any number greater than 10. Keep subtracting 3s from the number (i.e. scoring penalties), and eventually you will be left with 10, 9 or 8. Now use the methods above to score the remaining points.

MathsManiac

Or the same rationale works with 7, 6 and 5.

LeixlipRed

Similar problem, show that it's possible to divide a square into k square pieces for all k>=6.

hivizman

LeixlipRed wrote: »

Similar problem, show that it's possible to divide a square into k square pieces for all k>=6.

Nice one.

It's also possible for k=4, which is, I think, the key to the puzzle (and the reason why this is a similar problem to the rugby scores).

As a corollary, I suppose one should show that it's not possible for k=2, 3, and 5.

AwayWithFaries

Funny. Only the other day we were explaining to an American group what scores are possible in rugby. Took a while for them to grasp that anything over 4 was possible.

LeixlipRed

hivizman wrote: »

Nice one.

It's also possible for k=4, which is, I think, the key to the puzzle (and the reason why this is a similar problem to the rugby scores).

As a corollary, I suppose one should show that it's not possible for k=2, 3, and 5.

Works for 4 yeh, key is that any square can be divided into 4 squares meaning you get 3 extra for each division.

hivizman

LeixlipRed wrote: »

Works for 4 yeh, key is that any square can be divided into 4 squares meaning you get 3 extra for each division.

Yes, that's how I got my proof. I worked out the division for k=6 and noted that the basic approach worked for all even values of k, in particular k=8.

The division for k=9 was obvious but not the division for k=7. But then I noticed that dividing the larger square in the k=6 division into four squares each 1/4 of the size of the larger square gave the k=9 division, and realised that I could do the same process to one of the squares in the k=4 division to give four smaller squares, plus the remaining three bigger squares, hence solving the problem for k=7. Once I'd found solutions for k=6, 7 and 8, and noted that dividing one square in a solution for k into four smaller squares gave a solution for k+3, then this meant that there was a solution for all k>=6.

I have given a general formula in the spoiler that produces a division for all k>=6. I think, though, that there are at least two distinct ways of dividing a square into k square pieces for all k>=9.

The general solution for even values of k divides a 1 x 1 square into one larger square (k-2)/k x (k-2)/k and k-1 smaller squares each 2/k x 2/k. For k=4, this gives a division into four squares each 1/2 x 1/2. For k=2, the formula "works" in the sense that it gives one "larger" square 0 x 0 and one "smaller" square 1x1, but this doesn't count as a true division into two squares.

There is a similar general solution for odd values of k, based on taking the solution for k-3 and dividing the larger square into four, hence adding three squares. Here, there will be k-4 squares each 2/(k-3) x 2/(k-3) and 4 squares each [(k-5)/2(k-3)] x [(k-5)/2(k-3)]. For example, for k=7, there will be three squares each 1/2 x 1/2 and four squares each 1/4 x 1/4. If k=5 is substituted into the formula, it gives the degenerate solution of one square 1 x 1 and 4 squares each 0 x 0.

LeixlipRed

You would have got 1/10 right at least on last week's olympiad trial test anyway :P Got one more right and you'd have won

Fremen

Eh? The winner only scored 2/10? Was it a really tough test or are this year's leaving cert students just a bunch of duffers? :P

LeixlipRed

I only seen about 75% of the papers and that was just in NUIM. A couple got one question right. Awful, awful standard. The first question was two equations which you had to solve for positive integer solutions. Every single paper had something with square roots, etc,. Kind of worrying considering they do months of classes and they should be exposed to that stuff in school too.

moycullen14

if you limit it to tries and conversions, whats the highest score you CANNOT get (and prove it)

equivariant

moycullen14 wrote: »

if you limit it to tries and conversions, whats the highest score you CANNOT get (and prove it)

23