Math Is Fun Forum

Are you sure you're supposed to use induction?  It may be possible, but don't see a good way of solving the problem with it.  I would go for a more direct approach.

First, let's simplify the expression.  The LHS is equal to

for all x other than x=1.  Since we're only interested in x>3, we can use this simplification.

Let's assume that this expression is a perfect square.  That means we can factor it as follows:

Solve for a, b, c:

The function you described is simply a root.

There are also systems already in place for describing large numbers.  Look up things like Knuth's up arrow notation (http://en.wikipedia.org/wiki/Knuth%27s_up_arrow_notation) and Conway chained arrow notation (http://en.wikipedia.org/wiki/Conway_chained_arrow_notation).

Is that longer or shorter than i months?

Trial and error works really well here since we can narrow down possible answers to a workable amount.

First, we know that

is an integer, so x must be rational.  Let a/b = x, GCD(a, b) = 1, then we have 88/x = 88b/a, which again is an integer.  a must be a factor of 88, so a can be 1, 2, 4, 8, 11, 22, 44, 88.  We can also see that 3 < x = a/b < 4.  That let's us rule out 1, 2, 4, and 8 as possibilities for a.

We have now narrowed down our possible answers to these fractions:

11/3
22/7
44/13
88/23
88/25
88/27
88/29

You can then do trial and error on these 7 numbers.

You're not using their definition.  Whenever you see numbers in parentheses you need to mentally replace them by the set.  For example, the first question asks if

.  To answer the question you need to replace (a,b) with {{a},{a,b}}.  You can see that the answer is actually false now (remember like Jane said that

).

Firefly

How do you solve these just by knowing the positions of the vowels?

That is an interesting problem mathsy.  Unfortunately I'm not familiar with giving rigorous proofs of something like this, but I can see it visually.

First thing is to show that you can cover any 2^n X 2^n board with a corner square missing.  n = 0 is a trivial case, so let's look at n = 1.  We have a 2 X 2 board, and since all 4 squares are corners we can assume WLOG that the bottom-right square is missing, so our board looks like

X X
X _

It's quite obvious how we place our triomino to cover the remaining squares on the board.  So now we know how to do a 2 X 2 board.  Let's see what happens when we turn that into a 4 X 4 board, again only considering cases where the bottom-left square is removed:

X X X X
X X X X
X X X X
X X X _

Let's place our first triomino the same way that we did in the 2 X 2 case (H represents squares covered by a triomino):

X X X X
X X X X
X X H H
X X H _

Let's pretend that the board is divided into four 2 X 2 squares.  The bottom-right square is either covered by a triomino or has an empty square.  That leaves us with three 2 X 2 squares to cover.  How can we do that?  By placing another triomino such that each of the 3 tiles that it covers are corner squares to one of the 2 X 2 squares:

X X X X
X H H X
X H H H
X X H _

Now those other three 2 X 2 squares are identical to our original 2 X 2 board problem, which we already know how to solve.  Thus, we now know how to solve a 4 X 4 board.

How do we do an 8 X 8 board?  We use the same trick as before.  We mentally divide the board into four 4 X 4 boards.  We can solve the bottom-right board with the method we just used, which will leave us with three other completely uncovered 4 X 4 boards.  If we place our next triomino such that each tile covers a corner of one of the remaining 4 X 4 boards they'll each become idential to the original 4 X 4 board problem, which we already know how to solve.  Covering those gives us a completely filled 8 X 8 board.  The process can continue indefinitely to any 2^n X 2^n board.

Now, mathsy's actual question was to show that it could be done with any square missing.  This is actually not much of an issue thanks to how triominoes work.  Let's say that instead of the lower left corner tile being removed, we actually removed the piece immediately to the left of that piece.  Here's what that would look like on a 4 X 4 board:

X X X X
X X X X
X X X X
X X _ X

To solve this new problem we can simply rotate our first triomino such that the corner tile is in the top-right instead of top-left, so it would look like this:

X X X X
X X X X
X X H H
X X _ H

Notice that once again we are left with a completed 2 X 2 board in the bottom right, so the rest of our 4 X 4 solution can remain unchanged.  The same thing goes for if we move the empty square one tile up:

X X X X
X X X X
X X _ H
X X H H

Any other tile that is removed from the 4 X 4 board is just a reflection or rotation of what we've just shown, so now we can solve a 4 X 4 board with any square missing.  And if we can solve any 4 X 4 board with a missing square we can generalize that to the 8 X 8 case, since we need solve only one 4 X 4 board and then we add a triomino that turns the other three parts of the board into the "corner missing" case.  And if we can do any 8 X 8 board we can do any 16 X 16 board, etc.

I'd say there's a good chance that it doesn't have 4 O's in the name, so I'll go with E.

You know that your numbers will have 3 digits, that all 3 digits must be either 1,3,5,7, or 9, and that all 3 digits must be unique.

My suggestion would be to first find the cubes of all five of the possible digits.  They are 1^3 = 1, 3^3 = 27, 5^3 = 125, 7^3 = 343, and 9^3 = 729.  Now just try systematically adding cubes together to see if you get a number with all 3 of the cubes you used to form it.

For example, I would start with 9.  Since 729 + 343 > 1000 you know that no numbers with both 9 and 7 will work.  However, 729 + 125 = 854, and you can plainly see that adding neither 1 nor 27 will give you a number that fits the parameters, so you can rule out all 3 digit numbers with a 9.

Then just keep working your way down until you've exhausted all possible numbers.  There are surprisingly few that you even need to consider.

Would you mind reposting your argument?  I don't want to sift through nearly 40 pages of the 0.999... post.

Although I think what we're really getting at is something fundamental to mathematics.  It appears that George either disagrees with one or more axioms related to set theory, limits, or whatever other fields of math are involved in this discussion, or believes that one or more of the theorems arising from those axioms is wrong.  Would you mind explaining which it is George?