bonanova

Two sets have the same size (cardinality) if their elements can be placed in a 1-1 correspondence. (A bijection exists.) For finite sets this is a simple concept. The sets {a b c d e f g} and {days of the week} are the same size; they both have a 1-1 correspondence with the first 7 counting numbers. For infinite sets things get dicey. The counting numbers, the even counting numbers (seemingly half as many) and the rational numbers (seemingly infinitely more numerous) have equal cardinality. Bijections among these sets can be easily described in words (or with a simple picture.) Can you verbally describe a bijection of the counting numbers with all of their subsets (their superset power set)? Or, remembering that neither set is finite, verbally prove no such bijection exists?

Now that I have a solution I can say, what a pretty problem!

Entire set. The numbers are consecutive, and they're placed on the grid. Same for 'Magic squares, sort of' puzzle

The figure shows nineteen circles arranged hexagonally in fifteen rows of length 3, 4 or 5. Place consecutive numbers in the circles so that . At least one row in each of the three directions contains exactly one pair of numbers (e.g. 17 and 34) that differ by a factor of 2. Exactly one of the six outer vertices contains a prime. (1 is not prime.) Exactly one two of the six inner vertices (connected to the center) contains a prime. The fifteen row sums are identical.

I'm always 1 to 6. I'm always 15 to 20. I'm always 5. I'm never 21 unless I'm flying.

The figure shows two squares that share two of their circles, and seven rows of length 3 or 4. Place consecutive numbers in the circles so that One of the squares contains three pairs of numbers (e.g. 13 and 26) that differ by a factor of 2. The other square has no such pairs. The seven row sums are identical.

Still open.

I'm trying to think of a rule for this.

Open book

What is the most efficient way to time a 15-second interval using a 7-second and an 11-second hourglass?

The intent of the blue outlines is to show the size and shape of the quadomino. The OP originally referred question 4 to two images, which was an error. Question 4 should now be clear.

There are 12 ways to form a pentomino from five squares. Can you arrange them to cover a 3x20 rectangle? Why or why not?

Thirty-two dominoes can always cover a standard 8x8 chessboard. But can 31 dominoes always cover a chessboard if we remove two of its squares? Evey problem solver knows it can never happen if the removed squares have the same color. 1. But what if the removed squares have opposite colors? If you try to cover a 7x7 chessboard with dominoes, one square will be left over. 2. Which one? 3. Can this board be tiled with dominoes? Why or why not? 4. Can this board be tiled with quadominoes? Why or why not?

Alice and Bob run a standard 26.2-mile marathon race. Bob's pace is a constant 8 minutes/mile, start to finish. Alice's pace is such that for every 1-mile segment of the race, she takes 8:01 minutes. That is, between mile markers 10 and 11, Alice consumes 8:01 minutes. And between miles 6.374 and 7.374, Alice consumes 8:01 minutes. Is Bob the sure winner of the marathon?

[spoiler='Demonstration for irrational case']Amazingly, everything is on the Web. This guy used Mathematica to simulate the problem. For 185 degrees, it takes 4 cuts. For 1 radian (an irrational number) it takes 84 cuts. Check it out.

[spoiler='How about']Clair says, quite wrongly, "The moon is made of cheese."