The worlds first riddle!

WEIGHTS

[size=7]Rap is correct, however, he could have arrived at the answer more directly had he not abandoned his initial strategy.

As an aside, it's somewhat fortunate that the strategy works, because on the surface, there is nothing that requires 1 to be a weight. The weights must sum to 40, and all integers between 1 and 40 must be uniquely represented by the various sums and differences, but 1 could be covered by 5 and 6 (6-5=1) for example.

To generalize rap's strategy:
Step 1: Start with 1.
Step 2: To get the next weight, double what you've already got and add 1. If the current total weight is X, then the next weight is 2X+1 because 2X+1 - X = X+1.
Step 3: If not finished, go to step 2.

1, 3, 9, 27

Note that these are powers of 3. For a given weight, to determine the arrangement of weights on the balance scale, follow this procedure:
Step 1: Write the desired weight in base 3 (digits are 0, 1, and 2)
Step 2: From right to left: If the current digit is a 0 or a 1, leave it unchanged. If the current digit is a 2, change it to a -1 and add 1 (adhering to base 3 addition rules) to the next digit.
Step 3: If not finished, go to step 2.
Step 4: If a ternary digit for a given weight (1, 3, 9, 27) is 0, it is not used in the weighing. If the digit is 1, place the corresponding weight on the left pan. If the digit is -1, place the corresponding weight on the right pan.

Example: 25 = 0221 base 3
The sequence of digit modifications is:
0 2 2 1 (leave the 1 alone)
0 2 -1 1 (change the 2 to a -1)
1 0 -1 1 (add 1 to the next 2 which results in a carry: 2+1 = 10)

27 goes on the left
9 is not used
3 goes on the right
1 goes on the left

(27+1) - 3 = 25

This works because it leaves the original base 3 number unchanged. When you convert a 2 to a -1, you are subtracting 3 times the place value. By adding one to the next digit, you are adding back 3 times that place value (which in base 3 is the next place value).[/size]

markr noted (et al and burped)

London Calling

6 or so people and 216 thousand looks said the trying one

sounds like 36pi in a pipe thru a sphere

Q?

If the answer is 36pi then what is the question?

solipsister:
[size=7]If a hole is drilled through the center of a solid sphere (to insert a pipe?), and the length of the hole is 6, what is the volume of the remaining portion of the sphere?

I'll take Math for $1000, Alex.[/size]

Math for 1,000?

he formed the equation m = e/c^2 ?

(al was reappraised generally)

bumpf

[size=7]What is Algebra, Einstein?[/size]

Rap

May I thank Slippy for holding the fort during my enforced absence, I think she adds a touch of class to the proceedings; speaking of which, I made bail so I could say…


Lzzie:

?'GIT'

Gosh, It's Tough
Good Imagination TryAgain
Good, I Trust
Genuinely In Touch
Generous In Thought
Gyroscopic Inertial Thruster
Great Inventive Thinkster
Gin In Tonic
Grinning I Think

Everyone a winner!



Rap:

Weights

A+B+C+D=40
One has to be 1?-let A=1
To get two the difference between A and the next weight (B) has to be 2
B-A=B-1=2
So B=3
Now A+B+C+D=4+C+D=40
Or C+D=36
So C & D have to be evenly distributed around 36
Call this difference X
C+X=D-X
C=D-2X
C+D=D-2X+D=36
2D-2X=36
D-X=18
C+X=18
Most reasonable X is 9
C=18-9=9
C=18+9=27
So the 4 weights are
1,3,9,27

I checked and this works, but I'm not sure if it is unique


Now, you would think there is no more to say about that; but…

Mark:

WEIGHTS

Rap is correct, however, he could have arrived at the answer more directly had he not abandoned his initial strategy.

As an aside, it's somewhat fortunate that the strategy works, because on the surface, there is nothing that requires 1 to be a weight. The weights must sum to 40, and all integers between 1 and 40 must be uniquely represented by the various sums and differences, but 1 could be covered by 5 and 6 (6-5=1) for example.

To generalize rap's strategy:
Step 1: Start with 1.
Step 2: To get the next weight, double what you've already got and add 1. If the current total weight is X, then the next weight is 2X+1 because 2X+1 - X = X+1.
Step 3: If not finished, go to step 2.

1, 3, 9, 27

Note that these are powers of 3. For a given weight, to determine the arrangement of weights on the balance scale, follow this procedure:
Step 1: Write the desired weight in base 3 (digits are 0, 1, and 2)
Step 2: From right to left: If the current digit is a 0 or a 1, leave it unchanged. If the current digit is a 2, change it to a -1 and add 1 (adhering to base 3 addition rules) to the next digit.
Step 3: If not finished, go to step 2.
Step 4: If a ternary digit for a given weight (1, 3, 9, 27) is 0, it is not used in the weighing. If the digit is 1, place the corresponding weight on the left pan. If the digit is -1, place the corresponding weight on the right pan.

Example: 25 = 0221 base 3
The sequence of digit modifications is:
0 2 2 1 (leave the 1 alone)
0 2 -1 1 (change the 2 to a -1)
1 0 -1 1 (add 1 to the next 2 which results in a carry: 2+1 = 10)

27 goes on the left
9 is not used
3 goes on the right
1 goes on the left

(27+1) - 3 = 25

This works because it leaves the original base 3 number unchanged. When you convert a 2 to a -1, you are subtracting 3 times the place value. By adding one to the next digit, you are adding back 3 times that place value (which in base 3 is the next place value).


Print it out folks, you will never see better!

Hi Try,,, i like being poor and wish to stay that way.

Having nothing but ~Thee ~One and only ~God~ forever,,, he ~GoD~ (i pray forever) can have my souls.

I only want ~they ~God and Yeggy as parents' and always' will. I can't think of a better combination than that ^,,, nor do i want one,,, i hope everyone understands,,, if not ~may they walk in all of my shoes and lives from The Beginning of my inception in Gods head until they understand (this i pray for Amen.


Try wrote: royal flushes I have uncomfortable beginners luck when it comes to playing cards,,, people end up hating me.

So how have you been? Well i hope Amen.

Jovay wrote, "…I have uncomfortable beginners luck when it comes to playing cards,,,"

What a wonderful turn of phrase; ?'uncomfortable'; who would have thought of putting that word in the sentence! I wish I had thought of it.

I am well as always; how about yourself?


Two players are each dealt a card face down. Each player may look at his own card. The highest card wins. (It's ok, Jovay is not playing).

Cards are valued as in poker with aces being low. The first player may either keep his card or switch with the second player.

The second player may keep his card, whether it be his original card or one that the first player gave him after switching, or trade it with the next card on the deck, which is also face down.

The loser pays the winner $1 and if both cards are equal then no money exchanges hands. For the sake of simplicity assume an infinite number of decks. Both players are infinitely logical.

At what point should the first player switch

At what point should the second player switch if the first player doesn't switch

What is the expected gain of the first player

Before Einstein

CARDS
[size=7]First player switches with A-6, doesn't switch with 7-K.
If first player doesn't switch, second player switches with A-8, doesn't switch with 9-K.
Expected gain for first player is 151/2197 = 0.06873 (almost 7 cents).

This is based on examining all 2197 (13^3) possible permutations of three cards. It assumes that decisions are based strictly on probability and doesn't take into account bluffing (which I doubt is beneficial).[/size]

How about Henri Poincaré?

markr identified my conjecture.

To be fair, rapper did name the guy that deduced a far more general theorum from a particular case: don't ya just love it when you do that.

Presumably the origin of the formula was the main reason Al didn't cop the Nobel gong for those nonpareil insights.

The story reads like a thriller to me.

bluffing is perfectly logical in cards because in a moment it transcends probability

Q?

Who said, "keep your friends close but your enemies closer"?

Sun Tzu in "The art of War"..

Mark:

CARDS

First player switches with A-6, doesn't switch with 7-K.
If first player doesn't switch, second player switches with A-8, doesn't switch with 9-K.
Expected gain for first player is 151/2197 = 0.06873 (almost 7 cents).

This is based on examining all 2197 (13^3) possible permutations of three cards. It assumes that decisions are based strictly on probability and doesn't take into account bluffing (which I doubt is beneficial).

I have the expected gain for the first player as: ~ 0.081930. But I accept yours may be closer!


Slippy asks, "Who said, "keep your friends close but your enemies closer"?"

That would be me!


Your job allows you to work any number of hours per week you desire.
Your take home pay is proportional to the number of hours worked (no overtime).

After subtracting time for sleeping and routine daily tasks you have 80 hours per week left for work and pleasure.

You wish to maximize your income multiplied by the amount of pleasure time you have to enjoy it.

How many hours per week should you work

Working for the Man

[size=7]Tp is pleasure time
Tw is work Time
I=R*Tw
M=I*Tp=R*Tw*Tp
Tp+Tw=80
Take Differentials of both
dM=0=R(Tp*dTw+Tw*dTp)
Tp*dTw=-Tw*dTp
dTw/dTp=-Tw/Tp
dTp+dTw=0
dTw=-dTp
dTw/dTp=-1
-Tw/Tp=-1
Tw=Tp
Since Tw+Tp=80 & Tw=Tp then Tw=40
So maximum is to work 40 hrs
Check Tw=40hrs, Tp=40 hrs M=R*40*40=1600R
Tw=41 hrs, Tp=39 hrs, M=R*39*41=1599R
It's a max Ans 40 hrs[/size]

Rap

Netymology. According to Wikipedia the line is by Michael Corleone in Godfather II and misattributed to Sun Tzu and/or Machiavelli.

CARDS
I see what I did wrong. I evaluated the first player's strategy when he has a 7 without taking into account the second player's strategy. The first player should switch with 7. That raises the expected gain from 151/2197 to 180/2197 = 0.0819299 (which is what you had).