6, and there are four different combinations of the last two digits in n that get it.
Is it dishonorable or insulting in BAUT culture to give away my turn to anyone who likes to take it? I am still working on the airplane circumnavigation problem.
I don't understand the question?
I understand the question!
no wait...I don't.
"I" is the only English word that could possibly fit ("a" doesn't quite work, neither does the contraction "o" ).
Does the ? need to be only a single letter?
Not sure what you mean. The number of letters in each word correspond to the digits of e, insofar as your sentence is a finite representation of the first few digits of e.
So, the ? should be a single letter long. So, I guess tdvance was correct as far as the number of letters is concerned. But ?=e is probably better, except that that leads to a paradox since you clearly remember the mnemonic for e, you can determine quite a few digits and that would hardly constitute a disaster. So what's the rub?
Unfortunately success has only 7 letters! I was only a kid when I devised this mnemonic, so the illogic escaped me at the time. But it has served me well over the years.... But ?=e is probably better, except that that leads to a paradox since you clearly remember the mnemonic for e, you can determine quite a few digits and that would hardly constitute a disaster. So what's the rub?
Over to you, mrobvious.
Take a chessboard (8 by 8 grid) and remove one set of opposite corners (A1 and H8, for example, for those that know chess notation). There are 62 squares left. Take a single knight and place it anywhere on the board.
(For those that don't know how knights move, see Diagram 2 here:
Your job is to describe a sequence of moves such that the knight visits every square exactly once (the square you start on counts as visited immediately), or explain why this is not possible.
It should be impossible, since you removed two black squares, leaving 30 black and 32 white squares. When a knight moves, the squares it lands on are alternately black and white. So after visiting 30 black squares and 31 white squares, the knight will run out of black squares.
I assume that is the correct answer.
There are exactly 2 proofs regarding number theory that I remember from my high school days. I hesitate to ask for either one as I'm sure they will be easily available using Google. Still, they are the only mathematical questions I can think of at this moment, so here's one of them:
Prove that sqrt(2) is an irrational number.
Let a/b = the rational expression of sqrt(2). Further, assume that a and b have gcd=1, or else reduce them so that their gcd =1.
a=sqrt(2) * b
a*a = 2*b*b (1)
Thus, a must be even (if a is odd, then the '2' on the right can't divide a).
let a = 2*n
Substituting in equation 1,
Dividing both sides by 2,
By the same reasoning as previously applied, b must be even. However, now both a and b are even, contradicting the initial condition that a and b must have gcd=1. Therefore, such a and b do not exist, therefore by definition sqrt(2) must be irrational.
For whatever reason, a king has decreed that he wishes to increase the female population in his kingdom. Without advanced medicine, he orders all families to stop having children once they have a son. His reasoning is that even though the chances of male vs. female in a given birth is 50-50, after a while, there will be families with one girl, two girls, three girls, etc. whereas there will be no families with more than one boy (and of course, there may be families with no boys - the people are not forced to have children until they have a boy; they simply can't have more once they do have a boy).
Assuming this rule is extended to infinite time, that families have many children (e.g., no self-imposed abstinence), that natural resources are not a concern (this is a math problem, after all), what is the asymptotic ratio of females to males?
Last edited by mr obvious; 2007-Sep-25 at 12:45 AM. Reason: typo, sorry
Nevermind. It seems to me that the asymptotic ratio would still be 1.
Correct, but an explanation would be nice
If the probability of a boy is 1/2, then 1/2 of all families will have no girls, 1/4 of all families will have one girl, 1/8 of all families will have two girls, and so on. The expected number of girls is:
This series converges to one, so each family has children until it has one boy, but it also has one girl on average.
It is simpler than that, though, if the probability of a boy is 1/2, then on average 1/2 of all children are boys, regardless of what rule families apply in when to stop.
Fair enough. One of you two gets to ask a question. Please go ahead.
DanishDynamite gave a correct answer, so probably he did everything correctly even though he did not explain. So I think it is his turn