Math Is Fun Forum

Can we narrow down the range of areas for a given diagonal?

Hi, you can use Pythagoras' theorem to find the length of the diagonal in terms of the length and breadth. Answering your question requires manipulating the two formulas A = l*b and D = sqrt(l^2 + b^2)

Edit: I may have jumped to conclusions. I am having a bit of trouble using the algebra to get anything in terms of just D

Hi, you are absolutely correct!

By the way, p(x=0) is two orders of magnitude smaller than p(x=1). So their sum is precisely 1.068115234375 * 10^-4. Perhaps this was the source of your confusion.
This probability is about one in 9362. I would call that a small probability (:

Hi,

I found a way to fit 17 lights. There is enough space available for 18, but I would be astonished if it was usable.
I'm not really sure how to describe the (my) solution without an image. Basically:
There is a light at each corner.
Two of the corner lights opposite each other have 1.5m gaps on either side, and then there are three lights in a row 1m apart on each side (including the other corner).
There is a light as close as possible to both corner lights with a gap on either side. It looks like these lights are sqrt(2)/2 metres up/down and sqrt(2)/2 metres across from the corner lights, but that is just a guess based on physical measurement.
There is a light as far across as possible from the two corner section lights, i.e. 1m across and 1m up from one of the other corners.
Finally, there is room for two more lights as close as possible to each corner section light (those lights have a calculable distance that is about 2.95m)

Hello,

I agree with bobbym. This is how you answer the problem:

The probability of any child having a green pod is 1/4 = 0.25. To have nine children with no green pods, this has to occur nine times. So you raise it to the ninth power.

For the probability of one green pod, no green pods has to occur eight times and a green pod has to occur once. Therefore we raise 1/4 to the eighth power and multiply it by 3/4. Furthermore, since there are nine different ways this can occur (the green pod could be possessed by any one of the nine children), we must also multiply this by 9.

When we add these together, we get 7/65536

I have just learned that there is a result in mathematics where 1 + 1 + 1 + 1 + 1 + ... = -1/2

Perhaps that is the expectation, and the banker profits after all xD

Hi Agnishom,

I have heard of a solution where dollar value is replaced by utility, or risk-aversion, and so the higher rewards that are much less likely become exponentially less valuable and the expectation is a bit over $4. Can you please explain what is involved in this resolution?

LIHJ,

If you are interested in primes (I do not think their frequency in the whole numbers in general is directly relevant to this problem), I have just learned of three useful approximations that have to do with them.

1. A good approximation for the number of primes less than or equal to x is:

2. A good approximation for the nth prime number is:

3. A good approximation for the probability of a whole number x being prime is:

Hello! I look forward to seeing the results of further hexagons, as I won't be attempting to go much further myself. It is an interesting question what the proportion of primes converges to. Note that the proportion of numbers with a particular lowest factor seems to stay the same, but more factors keep appearing; therefore, the proportion for new factors must be subtracted from the proportion of primes. So in order for the proportion of primes to converge instead of becoming very small, the proportion of new factors must become increasingly small.

I believe we have a reasonable answer to the original question, however. There are so many primes because all of the numbers must satisfy 6x + 1 and may not have any composite factors (besides themselves) or factors that do not satisfy 6x + 1. And there are probably so many primes among whole numbers that meet these conditions.

Just for the fun of it, I collected some stats.

Of the first 200 nested hexagons (not counting H=1):
71 are prime (35.5%)
56 have lowest prime factor of 7 (28%)
22 have lowest of 13 (11%)
11 have lowest of 19 (5.5%)
8 have lowest of 31 (4%)
7 have lowest of 37 (3.5%)
4 have lowest of 43 (2%)
4 have lowest of 61 (2%)
All others have less than 4

If we check for depths other than 200, we find that the proportions for any given factor are almost identical (as if they may be cyclic). But at greater depths, the proportion of prime numbers becomes diluted because of the greater number of factors to be represented. For example, in the first 50 nested hexagons, there are 26 primes (52%) because the factors above 31 do not appear and the factors 31 and below maintain their original proportions very closely.

Oh, I see what has happened. My formula includes 1 in the sequence, whereas yours starts with 7.

It seems that all the composite numbers in the sequence only have prime factors. Furthermore, those prime factors, and the terms of the sequence, must be of the form 6x + 1 where x is a positive integer. If we could list the numbers that have only prime factors of the form 6x + 1, and list the prime numbers of the form 6x + 1, we might find a similar ratio of prime numbers and the question would be answered.

But that creates a whole new question of why nesting hexagons makes numbers with no composite factors.

It's brilliant! I love conceptual problems like this.
This is how I derived my answers