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I like putting on a high amount of rotational symmetry and then using the "circle" brush to make one big off-centre circle.

The "circles" brush is quite fun even without using symmetry.

Not sure how easy it'd be, but one thing I'd like is the option to have a circular canvas.

Right now, depending on the type of symmetry you use, it's possible to draw something in the corner and have the corresponding drawings be offscreen.

Onako, could you share your proof that it doesn't work for p>1?

I have a feeling that a disproof for p=k could be tweaked to become a proof for p=1/k.

Nice page! I love this kind of puzzle, where all the clues refer to each other and you need to search for a foothold to get started.

One thing that annoyed me was that you can't turn a cell blank after you've entered something into it.

It'd be nice to get rid of numbers that you realise could be wrong.

Paradoxical statements don't bother the King.

If the prisoner gives a true statement, then the King is forced to have him eaten by lions.

If the prisoner gives a false statement, then the King is forced to have him trampled by buffalo.

If the prisoner gives a statement that is neither true nor false, then the King has no obligations and can kill the prisoner however he chooses.

The thing I like about the solution is that it shows that you'll definitely need that many breaks.

Calculating how many breaks you need if you do all horizontal breaks and then all vertical will get you one answer, but then there's the doubt that if you break it in a cleverer way you might be able to reduce the number.

Thinking of it the way the solution does shows that this number of breaks is indeed the best possible.

I've got another idea.

Haha, I love your thinking.

Proof by induction works in two stages.

First you prove the base case. In this case that's h=1, and you've done that.

Next you assume it's true for h=n, and use that to prove that it's true for h=n+1.

You've managed to get that 11^(n+1) - 4^(n+1) = 7*11^n + 4(11^h - 4^h).

The first term on the right is clearly divisible by 7, and by the inductive assumption, the second term also is. Therefore the whole right hand side is divisible by 7 and you're done. [True for h=n] implies [True for h=n+1].

I think the best way to do this is to build up from one horse.

With a one-horse race, there's only one way it can finish: [1].

With a two-horse race, the finish can either be [1,1] or [1,2].

With three horses, one of three things happens:

All horses tie.

The first two horses tie.

One horse gets an outright finish.

So the finishes for these would respectively look like [1,1,1]; [1,1,?]; [1,?,?].

To fill in the question marks, we can refer back to our results for the smaller races.

We know that there is one way to finish a one-horse race, and two ways to finish a two-horse race.

Therefore, all the ways a 3-horse race can finish are:

[1,1,1]

[1,1,3]

[1,2,2]

[1,2,3]

We can do a similar thing with 4-horse races.

First we observe that there are four patterns of firsts possible:

[1,1,1,1]

[1,1,1,?]

[1,1,?,?]

[1,?,?,?]

Then we refer back to every smaller race to fill in the blanks:

[1,1,1,1]

[1,1,1,4]

[1,1,3,3]

[1,1,3,4]

[1,2,2,2]

[1,2,2,4]

[1,2,3,3]

[1,2,3,4]

You should now be able to see a clear pattern in the number of results possible for n horses.

Forgive me if I've misinterpreted, but I think you want a function like f(x) = 1 - e^(1-x) would work.

When the ratio is 1 (ie the rocks are the same size), then f(x) = 0.

As the ratio rises, so does f(x). For example, f(2) ≈ 0.632 and f(3) ≈ 0.865.

As the ratio gets very large, f(x) tends towards 1.

A more general version of this function would be 1 - e^(a(1-x)).

"a" is some number that you could change according to your needs.

f(1) will always be 0 and f(x) will always tend to 1 as x gets large, but changing a will change the speed at which it does so.

For example, if your big rocks tend to be about 1000 times bigger than the small ones, you might want to make a fairly small so that you don't get a lot of index values in the region of 0.999.

The numeric keypad only has 9 numbers on it, so the amount of combinations to try is only 150,994,944.

I've just saved you a few hundred million years!

It doesn't really apply to the example you gave, but you can sometimes "scale" the numbers you want to average, find the average of the scaled numbers, and then reverse the scale to get the right answer.

eg.

Find the average of 55, 54 and 59.

You can scale each of these numbers by taking 50 away. The average of 5, 4 and 9 is 18/3=6, so the answer is 6+50 = 56.

You can also scale by multiplying and dividing.

To find the average of 16, 24 and 36, it could be easier to find the average of 4, 6 and 9, then multiply the answer by 4.

There's a fairly easy test to see which of two fractions is bigger.

Multiply the top of each fraction by the bottom of the other one.

The fraction that contributed its top to the smaller result is the smaller fraction.

For example, which is smaller, 3/14 or 2/9?

To work this out, work out which is the smaller of 3x9 and 14x2.

3x9 = 27.

14x2 = 28.

27 is smaller, and the 3/14 contibuted its top to that result, so 3/14 is smaller than 2/9.

I think it's important to point out that, in isolation, the expected rolls to get a "33" string is more than the expected rolls to get a "34" string.

However, when trying for them together, either string is equally likely to show up first.

Those two things aren't a contradiction.

Here's an intermediate step:

I found that it was easier to write numbers in terms of their prime factors.

When I wrote my code to write the first million terms, I used vectors.

The first term is [1,0,0,0,0,0,0,0,0,0].

Then it goes to [0,1,1,0,0,0,0,0,0,0]

Then [0,1,2,0,1,0,0,0,0,0], and so on.

The nth element of the vector means that the number contains that many powers of the nth prime.

You can tell by looking that no number in the sequence will ever have a prime factor higher than 29, which is why we can have length-10 vectors.

A number in the sequence will be significant if and only if all elements of the vector except the first are equal to 0. [In the code, I used A(1) == sum(A)]

To answer your other questions, you should always jump back to the top of the list. So if a number is divisible by 91, you should always multiply it by 17/91.

And the method was designed by John Conway.

I probably should have given my source in the first place, so here it is.

(The closest I am to solving that is figuring out a likely order of magnitude though)

**mathsyperson**- Replies: 4

I came across this method of generating prime numbers recently.

It can be used to list every prime there is, in order, with nothing else.

It's too slow to have useful applications, but I thought it was really interesting, and my mind boggles at the fact that it works at all.

Here is a list of fractions that makes it work:

17/91

78/85

19/51

23/38

29/33

77/29

95/23

77/19

1/17

11/13

13/11

15/2

1/7

55/1

Using this list, you generate a sequence in the following way:

The first element of the sequence is 2.

To get from one element of the sequence to the next, you start at the top of the list of fractions, and try multiplying by each one in turn. The next member of the sequence is the first integer you get as a result.

So, all of the fractions from 17/91 down to 13/11 will not produce integers when multiplied by 2.

However, the next one produces 15, and so that is the next element in the sequence.

The sequence starts like this:

2, 15, 875, 725, 1925, 2275, 425, ...

The interesting part is that every so often, an element of the sequence will be a power of 2.

If you ignore all elements of the sequence other than powers of 2 (and also ignore the first 2), you get this:

2^2, 2^3, 2^5, 2^7, 2^11, ...

That is, the powers of two that are produced are precisely the list of prime numbers.

Like I said, this works far too slowly to have any actual use.

Generating the first thousand terms of this sequence will just get you 2, 3, 5, 7.

Generating the first million will list the primes up until 89.

I still think it's neat though.

19CK is the short way of saying:

And you can write 19C(k+1) in a similar way.

Therefore, the fraction in the question is:

When you also expand out all the factorials (eg. k! = 1x2x3x4x5x...xk), then a lot of things can cancel out and leave you with something relatively simple.

The final answer is:

Maybe you're a computer.

A x A is a set made from pairings of dogs. Any pairing of dogs must be included in the set, and order is important.

For example, if the Dog Pound held the dogs Fido, Rex and Spot, then A x A would be:

{ (Fido, Fido), (Fido, Rex), (Fido, Spot), (Rex, Fido), (Rex, Rex), (Rex, Spot), (Spot, Fido), (Spot, Rex), (Spot, Spot) }

One relation on A could be that for any x, y in A, x~y if their names are the same length.

You can define any relation on A in terms of a subset of A x A.

In this case, it would be:

{ (Fido, Fido), (Fido, Spot), (Rex, Rex), (Spot, Fido), (Spot, Spot) }

For the first one, I'm not sure such a sequence exists.

The second one is easy enough though.

One example would be Xn = (-1)^n, Yn = 1.

A steady state is stable if d²N/dt² is negative for that value of N.

You can see why this is, because it means that for a value of N just less than a steady state, dN/dt would be positive (and so N would be increasing towards the steady state), and for an N just more than a steady state, dN/dt would be negative and so N would be decreasing towards the steady state.

Similarly, a steady state is definitely unstable if d²N/dt² is positive for that N.

If the second derivative is 0, then that test doesn't conclude anything.

Bit messy, but I think this one works:

a * b = { (a-1)² + b, if a > b

{ b(b-1) + a, otherwise

Here's how to start "counting" using this system.

1*1 = 1

2*1 = 2

1*2 = 3

2*2 = 4

3*1 = 5

3*2 = 6

1*3 = 7

2*3 = 8

3*3 = 9

4*1 = 10

...

Without doing any number crunching, here's why I think the 33 sequence should take a bit longer:

If you've already hit with the first throw, then the question is simple.

Hitting with the second throw means you win. Missing with it means you lose. The third throw doesn't matter either way.

Therefore, the probability is a nice round 40%.