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I answered this question: http://answers.yahoo.com/question/index?qid=201307...

Then I thought up this related question:

Consider a natural number N. Then what is the largest natural number x so that it is impossible to have N numbers without containing a pair whose sum is divisible by x?

Characteristic properties of the graph of a function of the type

f(x) = a·x²+b·x+c

where a≠0, is given by the information about whether each of a, b, c and d is positive, negative or equal to zero. Here d denotes the discriminant d = b²-4·a·c.

Each of the four values has three possible cases, positive, negative or zero, except for the constant a which never equals zero. So we have at most 2·3³ = 54 types of parabolas. But for instance:

a,c<0=b<d is not possible

So the number of possible cases is less than 54. On the other hand any choice of a, b and c will provide a parabola, so we have at least 2·3² = 18 types. How many types are there and why?

Suppose we let k monkeys type on k typewriters at random typing any character next with a probability of q. When a monkey dies it is immediately replaced by another monkey. Each typewriter is pressed 10^7 times a year. We let this experiment run for 10^10 years (a little less than the entire life of Earth). What the monkeys do not know is, that they are actually set up for the specific task of randomly producing the following quote from Shakespeares 'As You Like It', Act V, Scene 1:

The fool doth think he is wise, but the wise man knows himself to be a fool.

This quote is 76 characters long (counting punctuation and spaces). The capital 'T' at the beginning is the only capital 'T' throughout the phrase. The question is:

what is the longest substring (starting from the beginning of the phrase) that has an expected time of at most 10^10 years to occur.

Feel free to consider simpler cases and provide answers to these - it might aim towards a solution of the full sized problem. This question was inspired by the following:

http://answers.yahoo.com/question/index;_ylt=Aljm....

Consider the number x = 31250. It is divisible by any number in the following list containing 10 numbers: {1, 2, 5, 25, 50, 125, 250, 1250, 3125, 31250} which are all substrings of the decimal representation of x. The number 31250 has 14 divisors and you can form 15 substrings of it. Let the 'self-divisibility' of x be denoted by the product:

selfdiv(x) = (10/14)·(10/15) = 10/21 ≈ 47,619%

This far I have found the numbers 1, 15, 125, 1250 and 31250 as the numbers of greatest 'self-divisibility' when considering a fixed number of digits (they have 1, 2, 3, 4 and 5 digits repspectively). They are 100%, 75%, 66.7%, 64% and 47.6% 'self-divisible' respectively.

Surprisingly, for each fixed number of digits a unique number of maximal 'self-divisibility' came out in these five cases. I do not know whether this is the case for numbers having 6 digits or more.

Who can determine the number(s) that is most 'self-divisibile' among numbers of 6 and 7 digits. Can anyone go beyond 7 digits?

Let an 'unpredictable walk' denote a path of unit steps from (0,0), where for each step a random direction is chosen uniformly from 2π radians.

Let P(n,d) denote the probability that we end up within a distance of d from (0,0) after n 'unpredictable' steps. Also let Q(n,r) denote the probability that all steps of an 'unpredictable walk' of length n lies within a circle of radius r centered at the origin. Clearly P(n,n) = Q(n,n) = 1.

Can you determine P(n,d) and Q(n,r) for any d or r between 0 and n?

Suppose we have a 3x3 unit grid. Define a connection as a path starting from the lower left to the upper right traversing a part of the grid without intersecting itself. The longest connection to my knowledge consists 14 unit steps.

Let L define the number of times moving left on a given connection. Correspondingly R for right moves, U for upwards and D for downwards. Then the following considerations may be useful:

I believe that

1. R = L+3 and 0 ≤ L ≤ 3

2. U = D+3 and 0 ≤ D ≤ 3

About a boy and a girl who both are at least seven years old we are told:

The boy is just as old as the girl will be when the boy turns twice as old as the girl was at the time when the boy had half the age of the sum of their current ages. Every age mentioned is an integer. How old are they.

I am looking for an elegant answer to the question as I already know the answer...

Characterize the quadruples (a,b,x,y) of integers 0<a<x≤y<b that satisfy the equation:

1/a+1/b = 1/x+1/y

The case a=2 has been completely covered in my previous question:

http://answers.yahoo.com/question/index;_ylt=AhZyh...

But note how the number of solutions increases for a=3 as we have:

(3,b,4,y):

1/3+1/6 = 1/4+1/4

1/3+1/12 = 1/4+1/6

1/3+1/24 = 1/4+1/8

1/3+1/36 = 1/4+1/9

1/3+1/60 = 1/4+1/10

1/3+1/132 = 1/4+1/11

(3,b,5,y):

1/3+1/15 = 1/5+1/5

1/3+1/30 = 1/5+1/6

1/3+1/105 = 1/5+1/7

Is there any clever pattern to the solutions other than the fine restrictions given in the answer to my previous question? How can the method be extended to larger a's and how many solutions will there be for a given a>3?

When adding 1/2 and another unit fraction it might be possible to obtain the same sum by adding two other unit fractions:

1/2+1/12=1/3+1/4

and

1/2+1/30=1/3+1/5

Is it possible to generate all solutions, ie. triples (n,x,y) of positive integers satisfying the equation 1/2+1/n = 1/x+1/y?

Consider the number in the interval [0,1) given by decimal expansion x = ∑d(i)∙10^(-i) where the sum is over i>0. We avoid d(i) to end in an infinite tail of 9's. Is it possible to list all rationals in [0,1) not ending in an infinite tail of 5's so that the i'th rational on the list has its i'th decimal not equal to 5?

If this is not possible, is it then possible to list all rationals with finitely many 5's in the decimal expansion so that the i'th rational on the list has its i'th digit not equal to 5?

The question is based upon the following question asked by Michael:

http://answers.yahoo.com/question/index;_ylt=ApWDF...

Let x and y be vectors in m-dimensional real space and x~y denotes that x and y have the same coordinates (not necessarily in the same order). Let xⁿ denote the sum of n'th powers of the coordinates xⁿ = (x1)ⁿ+...+(xm)ⁿ where x1,...,xm denotes the m coordinates of vector x. Prove that x~y if xⁿ=yⁿ for all n=1,...,m.

If you ever cared to measure the distance around a circle plotted on your screen following the edges of the pixels used to draw it you would have noticed something shocking:

It is 4D, i.e. four times the diameter of the circle drawn.

Even when you increase the resolution (theoretically even infinitely much) this holds good. And it implies π=4 or does it?

Yesterday I did quite some work typing in an edit of my answer to a question. Since I know that one can't rely on the internet to work every time I copied the edited text so I still have it.

But although I am still able to edit my answer to another question I keep getting an error message when I try to edit my contribution to this particular question.

What is there to do? And could it have something to do with the fact that I used quite some time typing in the answer so maybe the website with the edit form somehow timed out?

Since I already did the work I would really appreciate to be able to post it!

Let a real number, z, be given. Does there always exist an infinite sequence, {bₓ}, of natural numbers such that some multiple of 1/bₓ is closer than 1/(bₓ)² to z.

Put otherwise: is there a infinite sequnce of naturals, bₓ, such that |z-a/bₓ|<1/(bₓ)² for some integer, a.

Suppose z is irrational and |z-a/b| < 1/b^n for some rational a/b. By this I define the rational approximation a/b of the irrational z to be "level n good" when n is the maximal integer satisfying the inequality.

For any irrational it is always possible to find an approximation that is "level 1 good" whatever b may be. But does there always exist approximations that are good at higher levels than 1?

One of my students removed a tiny USB-plug from my computer today. I didn't allow him to plug it in in the first place and I don't know what it was. Only that it was so little that it looked more like an adapter than a USB-pen.

Is it possible and if so how do I check whether my student just rapidly installed a backdoor on my PC to spy on me?