[Two houses with an x]

For the existence of Eulerian trail it is necessary that no more than two vertices have an odd degree.

[9, 99 or 999?]

Is 90 / (your_number - 9) = 1?

[the only one]

Letters A, E, F, K, T, X, and Y require you to raise your pen once. Letter H requires you to raise it twice.

[the only one]

H is the only letter that requires you to raise your pen more then once to write it.

[Who am I?]

It is not possible to deduce about Bill and Susan the way problem is stated.

Some extra condition, for example about number of true statements, is required.

[Which die would you choose?]

No matter which dice I choose, if my opponent chooses properly, he has the same edge on me. My chance to win would be 12/36=1/3.

[Square inscribed in a triangle]

The area of the big triangle (S) is a sum of area of the small traingle (T) atop of the square and the area of trapezoid (let's call it M):

S = M + T.
Triangle T is similar to S, so:
T/S = (a/21)^2.
(where a denotes side of the square).
Area of trapezoid M is:
M = (21+a)*a/2.
Heron's formula tor triangle S gives:
S = 84.
This leads to equation:
84 = (21+a)*a/2 + 84*(a/21)^2
that has one positive solution:
a = 168/29.

[A society that prizes girls]

In a society in which people only want

boys, every family continues to have children while they have a boy. If they have a boy, they are enthusiastic and keep making babies. If they have a girl, they stop having sex in despair. What is the proportion of girls to boys in this society?

[Coins to Diamonds]

Six coins in 3x2 grid, diamonds are spaces between coins:

[Black over White]

Unfortunately not. I just wrote a recurrent equation (to count possible arrangements of marbles) and solved it using computer.

[Matryoshka]

CF|AG|BEH|D

[Penny's triple play]

Can you prove, that required arrangement is not possible with odd number of coins?

[Penny's triple play]

Is it required for each coin to touch fully table surface with one of its faces?

[Black over White]

1/5

[Matryoshka]

I've found 24 solutions:

CE|G|AFH|BD
CE|G|AFH|D|B
CE|G|AF|H|BD
CE|G|AF|H|D|B
CE|G|A|FH|BD
CE|G|A|FH|D|B
CE|G|A|F|H|BD
CE|G|A|F|H|D|B
CG|E|AFH|BD
CG|E|AFH|D|B
CG|E|AF|H|BD
CG|E|AF|H|D|B
CG|E|A|FH|BD
CG|E|A|FH|D|B
CG|E|A|F|H|BD
CG|E|A|F|H|D|B
G|C|E|AFH|BD
G|C|E|AFH|D|B
G|C|E|AF|H|BD
G|C|E|AF|H|D|B
G|C|E|A|FH|BD
G|C|E|A|FH|D|B
G|C|E|A|F|H|BD
G|C|E|A|F|H|D|B

Only the last one (found by k-man) has no nested dolls.

[1966]

The operation decreases the sum of all numbers by even number, but (1+2+...+1966) is odd.

[Rubber Stamping Dots]

there are 3 paths for all square roll..G N R, which way you pass?

G or R. Path N gives only 39 dots max.

[Rubber Stamping Dots]

42

[A Side Bet]

The dice is rolled 5 times. Mr G will be behind only if #1 never shows, so probability of Mr P winning the side bet is

    p = (5/6)^5

Therefore his average payoff is

    +5*p -5*(1-p) = 10p - 5 = 10*(5/6)^5 - 5 = -0.98

Casino wins on average

    +1*(5/6) -4*(1/6) = 1/6

with each roll, so expected casino's payoff after 5 rolls is

    5/6 = +0.83

The average payoff for Mr G is

    5 - 10*(5/6)^5 - 5/6 = +0.15

[Can you find a closed form equation for the sequence?]

Is a(n) = ⌊n/3⌋ (where ⌊x⌋ stands for

floor function) closed enough?

[Chess puzzle]

There is a legal solution, although not realistic in my opinion.

When one has eliminated the impossible, then whatever remains, however improbable, must be the truth.

This is what Sherlock Holmes used to say, who was a great master in retrograde chess.

You can learn more about it from a great book with retrograde puzzles:

The Chess Mysteries of Sherlock Holmes by Raymond M. Smullyan.

Highly recommended.

PS Well done, k-man

[The Last Set]

Well done, k-man.

[The Last Set]

The description of game Set in Wikipedia contains the following statement:

If 26 Sets are drawn from a collection of 81 cards, the remaining 3 cards form a Set too.

Can you provide a nice and smooth proof for it?

[Deal or No Deal]

1. We form all possible sequences of 40 cases (39 empty and 1 with the treasure).

2. Then we remove all sequences, where treasure is in the middle (1 < case# < 40). In other words we retain only sequences with treasure in case #1 or case #40.
3. Finally we ask what kind of sequences are more frequent among retained: those with treasure in case #1 or those with treasure in case #40.

It should be obvious that both types of sequences are equally frequent, since there is an obvious 1-1 correspondence between those two sets.

(k-man nicely proved the same by counting cardinality of both sets and showing their equality.)

[unanswered]

TSLF, is it true that fractional values are permitted?

Haven't you just proved, that there is no integer solution?