[Balancer's Chess]

Whoops, I see my error.

B-------
--------
--K-----
--------
----R---
-----Q--
------N-
-------P

[Balancer's Chess]

--------
-N------
--K-----
---P----
----R---
-----Q--
------B-
--------

[Rotating Tires]

a) 30,450 miles can be driven.

b) Rotate the tires after each mile:
Spare -> Right Rear -> Right Front -> Left Front -> Left Rear -> Spare

Explanation:

For each 5 miles driven using this rotation, each tire will

lose (2/29000)+(2/21000) of its tread. So, suppose we drive

5*N miles like this. Then, each tire will lose

2*N*((1/29000)+(1/21000)) of its tread. We wish to

determine N so that the latter expression gets to be 1,

representing all of each tire's tread. That happens when

N=6090, which is when 5*N (the total miles driven) is

30,450.

[four of a kind]

40

[Something From Nothing]

0!

[The most productive year?]

There are two years which have the maximum number of three such dates:

2017 with dates 1/17/17, 7/31/17, 9/13/17 and 2019 with dates 1/19/19, 7/17/19, 11/29/19.

If, however, we drop the prime and larger than 12 requirements for the year modulo 100, the year
2020 contains the most such dates, 2040 comes in close behind. The 13 such dates in 2020 are:
1/20/20, 2/10/20, 4/5/20, 4/30/20, 5/4/20, 5/24/20, 6/20/20, 8/15/20, 10/2/20, 10/12/20,
10/22/20, 11/20/20, 12/10/20.

[4 points, 2 lengths]

Do you know of any proof that this is a complete set?

[clever Timmy]

use machines which give Timmy a mechanical advantage such as a block and tackle, a lever, a ramp, etc. Timmy could use one of these.

LOL

No, the second option is rather realistic, logical, and actually stated in the OP

A lever is stated in the OP?

[clever Timmy]

use machines which give Timmy a mechanical advantage such as a block and tackle, a lever, a ramp, etc. Timmy could use one of these.

[Calendar Math]

Sunday

[Area of the overlap]

Just like Bonanova, I spent most of my career using algebra

and computers to do my job. There were a handful of times
when I had to optimize some function, which I did using a
little differential calculus. So, when I saw this problem,
I decided to try to rub off a little rust from my calculus
and derive the definite integral which solves the problem.
It took longer than I expected. Anyway, here are the
particulars:

I refer to the Captain's first diagram which delineates
the area he calls X. The leftmost point of X is the leftmost
point in common with the two circles centered at (1,1) (whose
equation is (x-1)²+(y-1)²=1) and (1,0) (whose equation is
(x-1)²+y²=1). Solving these simultaneously gives the leftmost
point of X as (1-sqrt(3)/2,1/2). Because of symmetry, I only
needed to integrate to get the left half of X, then double it.
So, I only needed to integrate from x=1-sqrt(3)/2 to x=1/2.
Now for an x in this range, the top curve is from the circle
centered at (1,0). Using its equation, I get that
y=sqrt(2x-x²). Similarly, the bottom curve's equation gives
y=1-sqrt(2x-x²). So, the function to integrate is the
difference of the two which is 2sqrt(2x-x²)-1. As I must
double this to get the full area of X, the thing to integrate
is 4sqrt(2x-x²)-2 from x=1-sqrt(3)/2 to x=1/2.
This is where my real troubles began. In my student
days, I could integrate away like nobody's business. I found,
to my horror, that all that ability had atrophied! After a
few unsuccessful hours, I gave up and decided to install a
math manipulation program on my machine and, thereby, cheat.
I looked around for a good open source program and found
wxMaxima which was intuitively easy to use (I didn't need
to resort to reading documentation!). Within minutes, I had
the integration done.

I really enjoyed the Captain's derivation. It never
occurred to me to try doing it that way. Nice going there,
Captain!

[Area of the overlap]

(pi/3)+1-sqrt(3) which is the definite integral of 4*sqrt(2*x-x^2)-2 from 1-sqrt(3)/2 to 1/2.

[Blind man's jigsaw puzzle]

A matrix showing the expected number of draws necessary to complete an

n by m (n,m<11) blind jigsaw using 1,000,000 iterations for each (n,m):

001.00 002.00 003.67 006.00 009.00 012.66 017.01 022.00 027.67 033.99
002.00 004.50 008.00 012.51 018.00 024.49 031.99 040.54 050.02 060.50
003.67 008.00 013.66 020.66 028.93 038.54 049.46 061.72 075.33 090.19
006.00 012.51 020.66 030.39 041.78 054.76 069.38 085.62 103.53 123.08
009.00 018.00 028.93 041.78 056.49 073.14 091.75 112.26 134.75 159.05
012.66 024.49 038.54 054.76 073.14 093.72 116.62 141.59 168.83 198.33
017.01 031.99 049.46 069.38 091.75 116.62 143.85 173.69 206.02 240.83
022.00 040.54 061.72 085.62 112.26 141.59 173.69 208.56 246.02 286.52
027.67 050.02 075.33 103.53 134.75 168.83 206.02 246.02 289.33 335.68
033.99 060.50 090.19 123.08 159.05 198.33 240.83 286.52 335.68 388.04

[All numbers showing]

this was an expected value problem with the expected number of rolls needed to get all six numbers equal to the sum of the expected number of rolls needed to get each new number. The average number of rolls to roll a new number being the inverse of the probability to roll a new number. So 14.7?

Please explain further. I can't parse this in any sensible way.

[All numbers showing]

I can get an answer, but no simple explanation. But that should do for everyday gambling needs.

It's relatively easy to calculate in a spreadsheet.For m=13, the winning chance is approximately 0.51. For m=12, it's about 0.44.The probability to see just one and the same number after n rolls is 1/6

^n-1.The probability P(k,n) to get exactly k different numbers (1 < k <= 6) after n rolls is P(k,n-1)*k/6 + P(k-1,n-1)*(7-k)/6.Make a 6-column table for k=1 through 6. Populate the first row with 1,0,0,0,0,0. The second row with the formulas above. Then drag the formulas down.I can't think at the moment of a simple formula instead of all that.

Ah, but there is one. No correct answers yet.

Since I got my answer using Prime's method, I'll now consult Strunk and White to see what else the OP could possibly mean. Possibly a weighted average m? Anyway, I must look in the basement for that copy of Strunk and White which I know is down there somewhere.

[Six integer equations, Eight digits]

Please check the equations for typos. I wrote a program to look for solutions and it found none.

I looked for bugs to no avail. It is certainly is easier to check the equations for typos than to find

an easily overlooked bug.

[1 = 2 revisited]

Start with the trig identity

cos(2x)=cos²(x)-sin²(x)
re-arrange the equation to
cos(2x)+sin²(x)=cos²(x)
take square roots
sqrt(cos(2x)+sin²(x))=cos(x)
divide by sqrt(2) to get
sqrt(cos(2x)+sin²(x))/sqrt(2)=cos(x)/sqrt(2)
add (3/2) to both sides
(3/2)+sqrt(cos(2x)+sin²(x))/sqrt(2)=(3/2)+cos(x)/sqrt(2)
evaluate at x=(3*pi/4) to get
2=1.
Voila!

[1 = 2 revisited]

Let T be the square root of 2. Consider the infinite continued

exponential, X=T^(T^(T^(T^(T^..... What is X? Well, we clearly
have X=T^X. X must be 2 because 2=sqrt(2)^2. Also, X must be 4
because 4=sqrt(2)^4. Ergo, 2=4. Dividing both sides by 2 gets
1=2. QED

[All numbers showing]

Lucky 13

[1 = 2 revisited]

We will prove a more general result using a simple induction argument.

We prove: Any set of numbers are all equal.

Proof:

(Base Case) -- If we have a set consisting of one number, it is clear

that all the numbers in this set are equal.

(Inductive Step) -- We assume that any set of N numbers are equal.

Now, we will show that any set of N+1 numbers are all equal. Let us

take one number out of the set of N+1. We are now left with a set

of N numbers and, by our induction hypothesis, all N of these are

equal. So, we put the number we took out (call this number X)

back in the set to again make it a set of N+1 numbers. We now

take a different number out of this set (call this number Y).

Now, we once again have a set of N numbers which, by the induction

hypothesis, have all equal numbers. The only way this can happen

is if X=Y because each of X and Y are equal to the other numbers in

the set. So, any set of N+1 numbers must all be equal.

(Therefore) By induction, any set of numbers must contain numbers

which are all equal.

In particular, the set {1,2} must contain numbers which are all

equal. Thus 1=2.

Counter example to this proof =)

We use the same set- {1, 2}. Following the reasoning above, then

X = 1

Y = 2

but it doesn't follow that X = Y. I guess N=2 is the weakest link of this induction chain.

it's the ONLY weak link....Give a guy a little break here!

[1 = 2 revisited]

We will prove a more general result using a simple induction argument.

We prove: Any set of numbers are all equal.

Proof:
(Base Case) -- If we have a set consisting of one number, it is clear
that all the numbers in this set are equal.
(Inductive Step) -- We assume that any set of N numbers are equal.
Now, we will show that any set of N+1 numbers are all equal. Let us
take one number out of the set of N+1. We are now left with a set
of N numbers and, by our induction hypothesis, all N of these are
equal. So, we put the number we took out (call this number X)
back in the set to again make it a set of N+1 numbers. We now
take a different number out of this set (call this number Y).
Now, we once again have a set of N numbers which, by the induction
hypothesis, have all equal numbers. The only way this can happen
is if X=Y because each of X and Y are equal to the other numbers in
the set. So, any set of N+1 numbers must all be equal.
(Therefore) By induction, any set of numbers must contain numbers
which are all equal.

In particular, the set {17.3,1,-6,2} must contain numbers which are all
equal. Thus 1=2.

[Three-letter Tic Tac Toe]

I downloaded a list of 1012 Scrabble words

from the interwebs. I used them for my
word list. The best 3X3 array I made was:

BOM
RAP
GET

which contains the 85 3-letter words:

BOB BOA BOP BRO BRA BAM BAR BAP BAG BAT
OBA ORB ORA ORE OAR OAT OPE OPT MOB MOM
MOR MOA MOP MAR MAP MAG MAE MAT ROB ROM
RAM RAP RAG RAT REP REG RET ABO ABA AMA
AMP ARB ARE APO APE APT AGA AGE ATE POM
POP PAM PAR PAP PAT PER PEA PEP PEG PET
GAB GAM GAR GAP GAG GAE GAT GET ERA ERG
ERE EAR EAT ETA TAB TAO TAM TAR TAP TAG
TAE TAT TEA TEG TET

[Find the missing combination]

Merry Christmas, wolfgang!

TGNUH

[Crossword Puzzle Maze]

Thank you for taking the time to try my puzzle. I'm afraid your answer is incorrect though. If you follow the link I gave, there's a more detailed description of how to move through the puzzle. What might be off though is you're counting the space you already landed on. This is incorrect. I hope the link helps.

OOps! I made a typo. It should have been

RLDRLUDURLDRLDLULDURRLRLRUDUDLR

The 6th letter should have been a U, not an L.

[Crossword Puzzle Maze]

RLDRLLDURLDRLDLULDURRLRLRUDUDLR

where, of course,

R=move 3 to the right;

D=move 3 down;

L=move 2 to the left;

U=move 2 up.

This is a very nice puzzle.

I applaud your creativity. Thanks,

and welcome to the Den!