This thread contains two explanations:

By user typographicalerror:

Every prime is a tree. Every number is a forest whose trees correspond to the primes its prime factorization. The tree for 2 is a single vertex. The tree for any other prime is a root vertex connected to the forest representing your prime’s place in the list of all primes. (e.g., 7 is the 4th prime, so you create a root vertex and connect it to the representation for 4.)

By user root45:

Each prime number has it’s own tree and we simply put these trees together to represent multiplication. So, once we know what the trees look like for 3 and 7 we can put them next to each other to make 21.

To find out what the tree looks like for a prime, we build it recursively using the number of primes less than or equal to our prime. To start off with, 2 is just a dot (*).

2: *

To make the tree for 3 we note that there are 2 primes less than or equal to 3, so we take the tree for 2 and connect it to another * like so:

3:

*
|
*

To make 4 we first prime factor 4 as 2*2 and then put the trees next to each other.

4: * *

To make 7 we note that there are 4 primes less than or equal to 7, so we take the forest for 4 and connect it to a root.

7:

*   *
 \ /
  *

Now we can make the tree for 21 = 3*7.

21:

* *   *
|  \ /
*   *

So basically you can recursively define each number using prime factorization and the number of primes less than or equal to a prime. Note that the function that gives this number is called π(x) (that is, ‘pi of x’).

Each column contains 3 squares, so each of the 7 columns must contain 2 squares of the same color. There are 3 ways to position 2 black squares in a column, and 3 ways to position 2 red squares, for 6 possibilities overall. Since there are 7 columns in the grid, some 2 of them must bear the same pattern, and thus there are 2 different columns that each contain 2 squares with the same color and position. These form the corners of a rectangle.

(Jeff Hooper, “Pólya’s Paragon,” Crux Mathematicorum 33:3 [March 2007], 146-148.)

Suppose that the task is possible, and that we’ve done it. Due to their shape, each of our 25 tiles contains either 1 or 3 black chessboard squares, always an odd number. So, taken together, the tiles contain an odd number of black squares. But the original board contained 100/2 = 50 black squares. That’s a contradiction, so the task is not possible.

(From Prasolov’s Problems in Plane and Solid Geometry.)

The prime factors of 32,118 are 2, 3, 53, and 101. There are only six ways to decompose it into three factors that are different from 1:

6 × 53 × 101
3 × 101 × 106
3 × 53 × 202
2 × 101 × 159
2 × 53 × 303
2 × 3 × 5,353

Of these possibilities, only the first includes two factors in the range from 4 (the minimum number of children) to 99 (the captain’s maximum age). So the captain has 6 children, he’s 53 years old, and his boat is 101 feet long.

From George Polya, The Stanford Mathematics Problem Book, 1974.

(a) Not knot, (b) Knot, (c) Knot, (d) Not knot.

Three.

11/09/2023 UPDATE: There’s another solution: Possibly he has no horses, sheep, or pigs but two llamas (or any other sort of animal). That works too. (Thanks, Gareth and Bob.)

Further information about the construction of such crosswords can be found at https://www.leesallows.com/files/Reflexicons%20NEW(4c).pdf

(Thanks, Lee!)

Yes! The formula for the volume of a sphere with radius r is . Let a, b and c be the radii of the base, torso, and head, respectively, which are integers greater than 0. The problem then amounts to Frosty solving the equation

By canceling out the factor of from both sides, we’re left with

As stated in the problem, a ≥ b ≥ c. The biggest that a can be is 5, so let’s try that first. Subtracting 5³ = 125 from both sides gives us

The biggest that b can be is 4, so let’s try that next. Subtracting 4³ = 64 from both sides gives us

Frosty is in luck, since 3³ = 27. Thus we have that 6³ = 5³ + 4³ + 3³. Now Frosty can build his new snowman friend to his specifications. In fact, (5, 4, 3) is the only combination of numbers that will work.

From NSA’s Puzzle Periodical series.

This solution is by Mandy Rodgers and Josh Trejo. C accuses B of lying, so they can’t both be knights or both knaves. Assume first that B is a knight and C is a knave. That would mean that A really did tell the visitor that there was one knight among the three islanders. But if A’s statement were true then both A and B would be knights, which is a contradiction; and if A’s statement were false then B really would be the only knight, which again is a contradiction. So B can’t be a knight. That means that B is a knave, which makes C a knight.

Note that we don’t know whether A is a knight or a knave. It’s possible that he’s a knight and said that the trio includes two knights, which would be true (in this case A/B/C are knight/knave/knight). And it’s possible that he’s a knave and said that all three were knights or all knaves, both of which statements we know are false. These possibilities are both self-consistent, so we can’t say anything about A’s identity, only that B lies and C tells the truth.

Obligatory John Finnemore sketch: