I need a tag team. I got this sucker within range, someone go for the knock-down

I got 3 possible pairs for the two words. I don't particularly relish the word matching part, so please help yourself to that part of the puzzle.

[1231415674, 1233454674]

[1231454674, 1233415674]

[1232415674, 1232454674]

Nice going! But how did you settle on the last digits of each pair?

The pattern of a word can be represented

by a sequence of numbers indicating

single and repeated letters. For

example, the pattern of BOYCOTT is

1234255 since the second letter is the

same as the fifth; the last two letters

are the same and the other letters are

all different. The patterns of two ten-

letter words each containing only seven

different letters, are added, treating

the pattern representations as ten-digit

numbers. The result is 2464870348.

What are the words?

The addition is done with carries, as usual.

Patterns always start with 1 and are created from left to right. The pattern digit for a letter that is not duplicated to its left is always 1 more than the largest digit in the pattern so far. If a letter has a duplicate to its left, then the pattern digit for the duplicate is used. So, for example, abracadabra would have pattern 12314151231.

This puzzle really has two parts. The first part is to come up with all possible pairs of patterns. The second part is trying to find pairs of words which fit those pairs of patterns. I consider the second part a bit of a drag and I don't suppose anyone would actually be able to do this without writing a program and using a large word list. Penney has a pair of words as an answer to this puzzle, but after programming and using a large word list, I discovered that they are by no means unique. So, if you can just short list the possible pairs of patterns, please post those as an answer. But some of you, like me, might actually go the full monty and write a program for the second part. I hope you have as much fun as I did in working this puzzle.

136/45= 3.022222222 (Is it >7? , No: Is it >5?, No: Is it >3?, No: Is it 3?, No: Is it 2?)

138/45=3.06666667(Is it >7?, is it >5?, is it >3?...)

An automatic card shuffler always

rearranges the cards the same way if the

setting is the same. A set of twelve

cards bearing the letters

     GOLDENBINARY

[/code]


is put through the machine, and the

cards, in the order in which they

emerge, are put through again.  They

come out

[code]

     ANINOYGREDBL

What message did the cards convey after

the first operation, assuming the

setting is left unchanged?

"setting" means the permutation to be applied to the cards.

If Perm is the permutation which the shuffler applies to the cards, then ANINOYGREDBL=Perm(Perm(GOLDENBINARY)) and you are asked to find Perm(GOLDENBINARY) which is a simple English sentence with spaces removed.

I am still working on this but I believe the "T" in MGACNFMCOTM is not correct. As I am filling in the matrix of what I do know, I get a conflict with the "T" in Mathematics and the "T" in MGACNFMCOTM. It may not matter but what should be the correct letter?

When I did the puzzle, I stumbled on the same T. It turns out to be correct. There is another explanation for the apparent conflict!

The Answer is ALAUZRNQNLI

No, sorry. But I see that you used the equivalents in my example. The example is only there to show you how the system works. You do not know the table of equivalents for the actual problem. Also, the answer is an actual 11-letter common English word.

The letters of the alphabet plus "space"

are to be replaced by the 27

permutations of the numbers 1, 2, 3,

repetitions allowed. These numerical

equivalents are to be written in a

column below each letter and taken off

horizontally in groups of three,

proceeding from the end of one row to

the beginning of the next. As a final

step these groups of three are to be

reconverted into letters using the same

equivalents. By this method MATHEMATICS

becomes MGACNFMCOTM and another eleven-

letter word becomes NFASOICAOTO. What

is the second word?

SUPERPRISMATIC EXAMPLE: Suppose we use

the following equivalents:

   A    = 3,3,3

   B    = 1,3,1

   C    = 2,2,3

   D    = 1,1,1

   E    = 2,2,2

   F    = 1,2,1

   G    = 2,3,2

   H    = 2,1,1

   I    = 1,3,2

   J    = 1,2,2

   K    = 3,1,3

   L    = 2,3,3

   M    = 2,3,1

   N    = 3,2,3

   O    = 3,3,2

   P    = 3,1,1

   Q    = 3,2,1

   R    = 1,3,3

   S    = 2,1,3

   T    = 3,3,1

   U    = 1,2,3

   V    = 3,2,2

   W    = 3,1,2

   X    = 1,1,2

   Y    = 1,1,3

   Z    = 2,1,2

"space" = 2,2,1

[/code]


The phrase "Bad Juju" becomes:

[code]

     BAD JUJU

     13121111

     33122222

     13112323

Pulling numbers off by rows then converting to letters, we get:

     B (1,3,1)

     H (2,1,1)

     Y (1,1,3)

     W (3,1,2)

     E (2,2,2)

     S (2,1,3)

     X (1,1,2)

     N (3,2,3)

[/code]

and so, BHYWESXN is the cypher.

A certain quotation containing 26

letters is written below a normal

alphabet and a third line is generated

by writing below each letter in the

second line the letter which is below

it in the first line. If, for

example, A in the first line has a B

below it in the second line, any A in

the second line will have a B below it

in the third line. Thus: "Give me

liberty or give me death" will lead

to the sequence:

    LBDMYMTBIMVMTRVLBDMYMEMGMI

[/code]


Another 26-letter quotation treated

in the same way yields the third line:

[code]

    YIYVNEASLROEEEEUAIYVAUAUAL

What does it say? SUPERPRISMATIC EXPANDING THE EXAMPLE:

    ABCDEFGHIJKLMNOPQRSTUVWXYZ

    GIVEMELIBERTYORGIVEMEDEATH

    LBDMYMTBIMVMTRVLBDMYMEMGMI

[/code]


SUPERPRISMATIC'S REQUEST:

If you solve this, please describe

how you did it.  My method is not

very slick, so I'd like to see a

better one -- perhaps one using a

result from Group Theory, e.g.

N electric light bulbs numbered 1 to

N are arranged in a circle and are so

wired that pressing the switch to

turn on (off) any light also turns on

(off) the adjacent lights. At present

all lights are off and it is desired

to turn on only the first light. What

sequence of switches should be

activated? (Assume that N is not a

multiple of 3)

SUPERPRISMATIC CLARIFICATION: Pressing

the switch for light K will change the

state of lights K-1, K, and K+1 (mod N).

So, if those lights were (ON,ON,OFF)

then pressing that switch will change

it to (OFF,OFF,ON).

SUPERPRISMATIC'S QUESTION: Are there

any on-off patterns of lights which

are NOT achievable from the all off

condition using some sequence of

switch pushing?

SUPERPRISMATIC'S OTHER QUESTIONS:

Is it always possible to get from

any particular on-off pattern to

any other with some sequence of

switch pushing? Is this essentially

different than my previous question?

im pretty sure i got it

learnsomewisdomfromconfuciusandangourlateryouredumbagain

learn some wisdom from confucius and angour later youre dumb again

goes as

72518346

7youredu

m2ewisdo

an5dango

lea1rnso

mbag8ain

mfrom3co

nfuciu4s

urlater6

and for the record im dumb again

Good work! You got it! I suppose I made a typo because it should have been "an hour later" instead of "an gour later". But, you have my admiration for this nice solution! Nice!

Just for grins, I went through all 3,628,800 possible assignments of digits to letters. Only one solution popped out. I realize that I could have gone through a much smaller number given the constraints that could be surmised, put the program only took a few minutes to write and under a second to run.

Then wouldn't each column have 7 letters, i.e. shouldn't the extracted solution look more like:

xxxxxxx xxxxxxx xxxxxxx xxxxxxx

xxxxxxx xxxxxxx xxxxxxx xxxxxxx

or is the spacing just a decoy?

Spacing isn't a decoy. It's just used to make it easy to read the text. Other than that, spaces aren't used. Look at my "To be..." example above and you can see that spaces don't matter.

Oh, I see what you mean now! In the "To be..." example, the letters just happened to come out in groups of 5. I guess that is confusing. But, just ignore spacing.

OK, that clears up the reassembly -- I was overwriting the numbers with letters. SO, is the puzzle to be worked out then two 6x6 squares based on the spacing you indicate?

Not the way I read the puzzle. I suppose it's an 8 by 8 square because 56 letters + 8 numbers would just fit.

Perhaps the "To be.." example will be enlightening:

6 S T I O N

O 3 B E T H

R N 2 O T T

H E Q 5 U E

T O B E 1 O

A T I S T 4

[/code]

So the beginning of the quote (the first 5 letters)

goes from right-to-left in the row containing the 1.

The next five letters go in the cow containing 2, etc.

Then, you pull out first (from top to bottom) from

the column containing the 1.  I hope this clears it up.

The numbers 1 to N are written in

scrambled order in the diagonal cells

of an N by N square (from upper left

to lower right). A message is written

horizontally in this square, starting

in the row in which the 1 appears,

then 2, etc. It is then taken out

vertically, taken in columns in the

order indicated by the numbers. Using

this method with N=6 and the sequence

6,3,2,5,1,4, the quotation "To be or

not to be, that is the question"

becomes

O T T U T   T B Q B I   S N E O T

N H T E O   I E O E S   O R H T A

[/code]


Read the following:

[code]

U W D G O   C A Y N E   B F F R E

S N N A U   E D D H S   I C R O E

A A R U L   U O O O N   O S M A L

M M N U R   I A R M I   T

A repairman fixing a typewritter

accidentally rearranged the electric

wiring so that each key printed a

letter other than the one shown on

the key (actually the letters were

wired in pairs so that if, for

example, on striking A a Z were

printed, then on striking Z an A

would be printed). A list of

9-letter words was being typed on

this typewriter and ten words were

written before the error was

discovered. These ten words

appeared as:

A J E Y F Z E W A

A N M K C G C W A

C Q Q A H C E W A

C X H C S X E X W

G W E C Z I E G A

G W Z A X K F K G

H A W A Z Q C X A

I D E K O O A K Z

O F Z W D T C W D

Z A O A Z A X I A

[/code]

What were the original ten words?

Just in case we get a visit from

a resident of Krypton (usually

called Superman) who is evil, it

behooves us to stock up on

kryptonite so as to have some

way to check his super powers.

I have found a chemical reaction

which produces kryptonite.

Unfortunately, I am unable to

balance this particular chemical

equation. Please help me by

balancing this equation.

You can Google "balancing

chemical equations" and get some

pretty good explanations.

Please help so that we can be

prepared! Here is the

unbalanced equation:

Te₁₆Ni₁₈O₈ + Te₃PtY + Ni₂₀Pt₁₄Kr₃ + O₄Y₁₃Kr₃ + Te₈Kr₅ + Ni₄O₄ --> Kr₆Y₁₇Pt₉O₈Ni₁₆Te₁₆

An eccentric billionaire knows that three people, A, B and C, who would dearly like to kill each other, are all perfect shots with a Colt 45 revolver at 25 paces. He arranges a three-way duel (a tri-el) whose rules are as follows.

(a) The contestants, each supplied with a large number of revolvers, are positioned at the apices of an equilateral triangle of side 25 paces. Each of A's revolvers is loaded with 2 live rounds and 4 blanks at random. Effectively, if he aims and fires at anybody, his chances of killing him with one shot are one third. In a likewise manner, B's revolvers are loaded with 3 live rounds and 3 blanks, while C's revolvers are loaded with 6 live rounds. Effectively, A's chances of hitting a target at which he is aiming are one third, B's one half, while C's are certainty.

(b) A is allowed to take 1 shot, after which he discards his revolver and picks another. Then B, if he is still alive, is permitted one shot, after which he discards his revolver and picks another. Then C, if alive, has his chance. The tri-el continues cyclically until there is just one survivor who is given one billion dollars (and immunitly from prosecution.)

Question 1 (easy).

Suppose C is dead and A and B are still alive. What are A's chances, if it is his shot, of eventually winning?

Question 2 (a bit more difficult.)

Is A's best strategy on his first shot to aim at C?

A should waste his first shot. Then B will shoot at C because C is the greater threat of A and C. If B shoots a blank, then C will kill B (again, because B is the greater threat of A and B). Then A will have the first at C. But if B shoots a live round, C will be dead and A will have first crack at B.

is the second to last letter capital i I or lowercase L l? I assume its uppercase since all the other letters are but just wanna make sure before i try

Upper case i.

The Fabulous Gazeeka Box converts

letters into their binary equivalents

(A=00001, B=00010, etc.), scrambles

these 5-bit groups -- always in the

same way -- and outputs the results

as reconstituted letters. A certain

word fed into the F.G.B. comes out

NYVRTBKPIJ. Can you unscramble this?

The letters of a certain phrase are

numbered in alphabetical order. Then

each letter is advanced in the normal

alphabetical sequence an amount equal

to its number. (A follows Z.) By

means of this scheme "Be a sport"

is converted into DHBZUSXB, since the

sequence of numbers is 2 3 1 7 5 4 6 8.

How about UHFIZBBJFOJ?

Sorry, I didn't do the table formatting right, I'll have to work on that.

Using my noggin' and a spreadsheet, I figured it out. Here's the generic sequence:

S0 = c

S1 = x

S2 = x^2

S3 = (x^2 + x + 1)/3

S4 = ( (x^2 + x + 1)/3 + x^2 + x )/3

= ( (4x^2 + 4x + 1)/3 )/3

If we rewrite this as

S4 = ( Aix^2 + Bix^2 + C ) / D

then watch the pattern of A,B,C,D, we get a table like this:

S Ai Bi Ci Di

0 1

1 1

2 1

3 1 1 1 3

4 4 4 1 9

5 16 7 4 27

6 37 28 16 81

7 121 85 37 243

Note that each Ai = A[i-1] + 3A[i-2] + 9A[i-3], same for Bi and Ci.

Now compute Bi/Ai, Ci/Ai, and Di/Ai:

S Ai Bi Ci Di b/a c/a d/a

0 1

1 1

2 1

3 1 1 1 3 1 1 3

4 4 4 1 9 1 0.25 2.25

5 16 7 4 27 0.438 0.250 1.688

6 37 28 16 81 0.757 0.432 2.189

7 121 85 37 243 0.702 0.306 2.008

where the ratios converge pretty quickly to this:

2/3 1/3 2

With T = target,

T = (x^2 + 2x/3 + 1/3)/2

and rearranging:

3y^2 + 2y - (6*T-1) = 0

For T = 321, this yields y = (25,-25-2/3).

Well, actually nothing ever actually converges until way, way out there at infinity. So, you are actually relying on a guess that this thing converges to (1+2*x+3*x^2)/6. If it differed by a small amount from this, the solutions may have been 25.00001, -25.6666678 say.

Just to add a little spice, may be a part 3 for complex numbers would be in order!

By that remark, I guess you're intrigued by this little problem. I never thought about it, but I guess if the convergent value were something like 321+87i, there may be some fruitful puzzling there! What next? Quaternions?

The first few terms of the series are:

1

x

x²

(x² + x + 1) / 3

(4x² + 4x + 1) / 9

(16x² + 7x + 4) / 27

It is trivial to observe that the third term (x²) must be > 321 and the second term (x) must be less than 321

Also, the 4th and 5th terms must be smaller than 321 and the 6th term must be greater than 321 in order to have a rolling average around 321

Otherwise the average will be either < 321 or > 321

Then,

x² > 321

x > 17

(x² + x + 1) / 3 < 321

x < 32

(4x² + 4x + 1) / 9 < 321

(2x+1)2 < 2889

x < 26

(16x2 + 7x + 4) / 27 > 321

(4x+2)² - 9x > 27*321

(4x+2)² - 9x > 8667

(4x+2)² - 9x > (93.something)²

So,

4x + 2 > 94

x > 23

But, x > 23 may not be exact as the square on the other side was > 93 and there is a term of -9x

So, if x = 24, we have

94² - 9*24 = 8620 which is less than 8667

So x > 24

Now we have 24 < x < 26

Since x is an integer, x must be 25

Nice job! I like your analysis. It's easy to follow. I wish I had thought of this way to do it.