[Binary lock]

Spoiler

This is correct, Pickett. No "extra" moves are required (obviously we cannot do better, since there are 2ⁿ-1 combinations to test).
Comparing this problem to Towers of Hanoi looks like a good insight to me, since both problems are naturally defined with exactly the same recurrence.
I've played a game today where I had to brute-force such lock. I was just testing subsequent binary numbers (thinking of levers as bits), hence I made more moves then optimal. Even after I gave this problem some thoughts, I'd still be afraid to use "optimal procedure", since I feel that the gain in speed is not worth the risk of making a mistake and starting over.

 

[Binary lock]

There are n binary levers: each lever can be in position 0 or position 1.
Exactly one out of 2ⁿ possible combinations of levers opens the lock.
The lock opens immediately as soon as each lever is in proper position.
Changing position of one lever is called a move.
Suppose all levers are initially in position 0.
What is the minimal number of moves that guarantees opening the lock?
In other words: how many moves are required to test each position of levers (the worst case scenario)?

Can you also describe the optimal procedure of moving the levers?

[Number of discrete "curves" in two dimensions]

Fibonacci(2n+1)-1

[LCD Watch]

For example you can encode MM using first digit except bottom segment (6^2=64>=60),

you just need to agree on some order or the above 6 display elements, so that they represent MM in binary.
(Each element of display is a binary digit "on"=1, "off"=0.)
Same way you can use second digit to encode seconds.
The remaining 4 elements of display (two bottom segments plus two dots) can be used to encode HH (4^2=16>=12).
I don't see any particular ordering that feels natural.

[LCD Watch]

To display HH:MM:SS in 12-hour format you need ⌈log

₂(12*60*60)⌉ = 16 bits of information, but your display has only 15 elements. Therefore the task is impossible if each element can only be "on" or "off". You need to cheat a little bit, for example add blinking or something like that.

 

Edit:

I miscounted the elements:) The display has 16 not 15 of them, so you

can use it without cheating.

[The Risky Choice of Going to Work]

Let:

p = prob. of getting killed
L = value of your life
S = value of the package

Expected value of going to work:
EVW = p*(-L) + (1-p)(500 + 0.2*S)
Expected value of staying home:
EVH = S
You are indifferent to going to work if and only if above EV's are in equilibrium:
EVW = EVH
p*(-L) + (1-p)(500 + 0.2*S) = S
-p*L + 500 + 0.2*S - p*500 - p*0.2*S - S = 0
p = (500 - 0.8*S)/(500 + L + 0.2*S)

Note however, that if (500 - 0.8*S) < 0, then also p<0, but probability can't be negative,
therefore you can't be indifferent to going to work when S > 625 (in such case staying home is always better for every possible value of p).

[Covering perforated hexagon with triminoes]

Is there a more elegant reasoning to show why that is?

I was analyzing D

_n+1-D_n instead of dividing D_n into 6 triangles, but I don't think that it was more elegant

[Covering perforated hexagon with triminoes]

Well done, gavinksong.

At first counting colors of D

_n looks hard for big n. The trick is to realize that all that matters is n mod 3.

[Covering perforated hexagon with triminoes]

Paint unit hexagons with three colors so that no two hexagons sharing side have the same color (like

this)
and observe then no matter how you place a trimino, it will cover one unit hexagon of each color.

[six dice = four numbers]

There are two ways to get exactly four numbers: aaabcd and aabbcd.

First you can get in 6_choose_3*6*5*4*3 ways, and
second you can get in 6_choose_4*3*6*5*4*3 ways,
making total of (6_choose_3+6_choose_4*3)*6*5*4*3 = (20+15*3)*6*5*4*3 = 23400 ways.
We should not accept the bet if 23400/6^6 > 1/2.
It's close, but we are slight underdog, so we reject.

[Covering perforated hexagon with triminoes]

Recall a classic: When a chessboard with two squares removed can be covered with 2x1 dominoes? The same approach works here.

[Diameters]

Given definition of diameter of region directly implies, that any disk of radius 1 (diameter 2) and center in region R will contain the whole R, provided diameter of R <=1. Am I missing something?

[Set Theory 101]

The set of all subsets of a set S is called a power set (powerset) of S.

According to Cantor's theorem, cardinality of the power set of any set S
is strictly greater then cardinality of S.

PS The Schröder–Bernstein theorem was called Cantor–Bernstein theorem at my set theory classes.

[Three 8's to make 24 (my modified version)]

Here is what I found.

F stands for factorial, S for square root.

I didn't use negation, since it doesn't add anything interesting.

Some parentheses are redundant, feel free to ignore them.

(8+(8+8))
F(S((8+(S(8)*S(8)))))
F((8/S(S((8+8)))))
F((8/F(S(S((8+8))))))
F(S((8*S(S((8+8))))))
F(S((8*F(S(S((8+8)))))))
F((8-S((8+8))))
F(S((F(S((8+8)))-8)))
F(S((8+S((8*8)))))

[Covering perforated hexagon with triminoes]

This puzzle is inspired by posted by bonanova.

Again we work on a hexagonal tiling of a plane, and the question is about
possibility of covering some shape with triminoes.
Trimino is a "triangle" formed by three unit hexagons sharing common vertex.

The shape to cover is defined as follows:
Let's pick a unit hexagon and call it H₁.
Now we recursively define H_n+1 as a sum of H_n and all unit hexagons adjacent to H_n.
So basically H_n is a "hexagon" with side of length n (unit hexagons).

Let D_n be H_n with one unit hexagon at it's center removed.
So, can you cover D₂₀₁₅ with triminoes?

[Covering chessboards with dominoes]

The answer is "no".

Consider those two squares in the center, that - when removed - disconnect the board into two pieces.
Those two squares can be covered either by one or by two dominoes.
In both cases this disconnects the board into two pieces, that cannot be covered by dominoes,
since they have different number of black and white squares.
To see this, first inspect the case with one domino (just count black squares and white squares of the upper piece).
The difference equals 2 in this case.
Now it is easy to realize that replacing the single domino with two dominoes can fix this difference only by 1 (not enough).

[Covering chessboards with dominoes]

Suppose the bottom-left corner is black (and so are the other corners).

There is one more black square than white,
therefore the remaining square should be black.
The question remains: could it be any black square?
The answer is "yes".
To see this you can draw a path for the lame King (the one that cannot move diagonally),
that visits each square exactly once (there is no requirement for path to be closed this time as in solution to #1).
Removing one black square breaks the path into two segments of even length (only one segment in case when terminal square of the path is removed),
and obviously we can easily cover each segment with dominoes.

[Covering chessboards with dominoes]

Consider a lame King (chess piece) that cannot move diagonally.

It's easy to draw a closed path for him to inspect all his kingdom:
a closed path that visits each square exactly once (and returns to starting point).
Obviously such path can be covered by dominoes starting from any square,
as well as any continuous segment of this path of even length.
Removing two squares of different color we break the path into two such segments,
so the answer is "yes".

[Covering chessboards with dominoes]

The answer is "no".

Consider a rook that can only move on black squares.
Let's call "reachable" the squares that can be reached by this rook
(in finite number of moves) from the bottom-left corner.
Only black squares in rows: 1 (bottom), 3, 5, 7 and 9 are reachable.
This accounts to 25 (5*5) reachable squares total.
Now observe that a quadomino, now matter how placed on the chessboard,
always covers an even (0 or 2) number of reachable squares.
But 25 is odd.

[Alice and Bob's marathon]

Let s(t) be the distance Alice covers in time 't'.

We assume s(t) is a continuous and differentiable function of 't' - a reasonable assumption: Since Alice can't disappear and appear in a different spot in zero time (implying infinite speed), and can't instantaneously change her speed to a completely different value (implying infinite acceleration).

This is exactly what meant by the last sentence of my answer:

The question remains: is it possible to run like this?

I'd also add one more condition: s(0)=0, since the runners start moving when the the race starts.

Same restrictions should also apply to Bob,

so another question is: how can he run with a constant positive speed?

Obviously he can't without braking his speed function at 0.

[Alice and Bob's marathon]

Alice could run fast for 0.2 mile then slow for the rest of the first mile,

and repeat this pattern for the whole marathon.
This way her pace is constant on every 1-mile segment.
Now, after 26 miles of the race she is 26 seconds behind Bob,
and she is going to run fast on remaining 0.2 mile segment of the race.
If she runs fast enough the fast segments, she wins.
The question remains: is it possible to run like this?

[Paradise anyone?]

@witzar: what was your approach?

Just brute force.

[Paradise anyone?]

There are five solutions:

a=[0, 1, 1, 2, 2, 3] b=[2, 4, 5, 6, 7, 9]
a=[0, 1, 2, 3, 4, 5] b=[2, 3, 4, 5, 6, 7]
a=[0, 2, 3, 4, 5, 7] b=[2, 3, 3, 4, 4, 5]
a=[1, 2, 2, 3, 3, 4] b=[1, 3, 4, 5, 6, 8]
a=[1, 2, 3, 4, 5, 6] b=[1, 2, 3, 4, 5, 6]

[Comparing numbers]

How about saying "smaller" with probability (1/2)+(1/2)^n (where n is the number Victor sees)?

[Removing pawns - the game]

If at least one of m or n is odd, player 1 wins. First he removes middle line, then do actions symmetrical to player 2's actions with the removed line.

This is incorrect.

Consider 3*3 for example. After player 1 removes middle line, player 2 removes some line perpendicular to line just removed, leaving player 1 with 2*2.

 

if m and n both even, player 2 wins. Every move player 1 make, player 2 make symmetrical move with center point.

This is correct.