DeGe

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For $10 or $11 cost of pizza, pay $12 (including tip).
Both Alice and Bob pay $6 each to make the $12. Charlie pays $1 each to Alice and Bob.
Everyone "pays" $1 per slice and the vendor is happy too!
 1

The floor is not biased. By logic and also mathematically:
Lets say the floor is biased and induces (on the final result) a factor of x for the biased side of the coin and a factor of y for the unbiased side
Also, for the coin toss, H H and T T produce a result while H T and T H are discared and a retoss is done.
Then,
0.52x + 0.49y = 0.5 (probability for heads on both coins)
0.51x + 0.48y = 0.5 (probability for tails on both coins)
Subtracting both equations,
x + y = 0
meaning that x = y
Therefore the floor induces no bias to one side or the other.

2 possible methods
Let a = 3 + r5
r5 means 5^{0.5}
Then a² = 14 + 6r5
Then a² = 6a  4
Then, for any n
a^{n }= 6a^{n1 } 4a^{n2}
Does this keep the 3 decimal places intact?
(3 + r5)^{n} = x + y(r5)
This can be easily computed for any n
X is a digit so no problems there for decimal places
now, Let y(r5) = S
Then, Ln S = Ln y + (0.5) Ln 5
Once y is known, unsing log and then antilog the first 3 decimal places for S can be known accurately and hence for a^{n}

Certain!
The average is <2 arms 1

0.5
the digit sum will be either odd or even with equal probability1/N
All numbers have equal probability 
First Name Years Profession Transport
Travis 3 Sec officer Subway
Roberto 30 Project mgr Car
Mike 5 Consultant Bus
Stephanie 15 Attorney Bullet train
Vickie 1 Specialist Bicycle

sq rt (N)?

Are cylindrical casts allowed?

Cut into 3 pieces of 1x4, 1x4, and 6x4
Bend the 6x4 so that it has a base of 4x4 and 2 sides of 1x4 each
Weld the 2 pieces of 1x4 on the remaining 2 sides of the base
You have a 4x4x1 = 16 cu ft volume box 
I think the probability of triangle being obtuse is near 1 for an infinite plane
Choose any 2 points shown with the blue line below;
Now, as long as the third point is within the two red lines, the triangle is acute otherwise obtuseFor an infinte plance, the probability of third point within these two lines is near zero.
 1

Edward was the murderer
Adam is the seer
If Bruce was murderer, then Clark should have been killed
Clark was blocked so he is not the murderer
Dick obviously wasn't as he was the one killed 
If the caretaker remembers 5 eggs which he might have swapped two of them,(5c2) or 10 possible combinations,how he could find the pair with 2 trials?
The solutions above seem to be for the first 5 eggs. I understood that it was any 5 eggs out of the 12. Try this then!

Nice Witzar. I had another approach which was to consider any expected dot rather than only 6...
Since the expected number is also added, the expected number of dots would be same for any number from 1 to 6.
Probability of each is 1/6
That means in 6 tosses all the numbers appear once (expected)... therefore, total dots = 1 + 2 + 3 + 4 + 5 + 6 = 21
This means that not counting the last dot, the expected numbers of dots for 1 to 6 are 20,19,18,17,16,15 respectively!

Between any 2 H, the probability of Ts being even or odd is equal.
The first H is expected to be seen in 2 tosses (probability 0.5 for T/H).
The expected number of Ts between Hs is:
0/2 + 1/4 + 2/8 + 3/16 + 4/32 + ... = 1And a final toss for the last H.
Since the probability of Ts being even or odd is 0.5, multiply these tosses by 2
You get (2 + 1 + 1)*2 = 8 tosses 
For area, had to check the web on how to calculate the radius of incircle
So, the first circle has radius 10/3
So the first circle covers 2/3 of the altitude length
Without going into details, I would say by symmetry that each circle covers the next 2/3rd
So the second circle would be of radius = 1/3(10  20/3) = 10/9
ratio of each consecutive circle radius = 1/3
So sum of areas = pi(r1² + r2² + ...)
= pi.(10/3)² / (1  1/9) = (25/4)pi 
Circumference is quite straight forward
Circumference of each circle is 2.pi.r
So for all the circles it will be 24pi
Because the length of altitude is 12 (with sides 13 & 5) 
If Bi > Bi+1, put the green ball in pile 1
If Bi+1 > Bi, put the green ball in pile 2
At the end, both piles will weigh the same
Effectively the weight of green ball = absolute (Bi  Bi+1)
By sorting the piles as above, we take care of the absolute value of this subtraction.
Since the weights are in a full cycle, the net (without absolute) values of green balls will be zero.
I don't understand this solution. What is Bi?
Bi is a blue ball at position i (i is from 1 to 10). For the tenth ball bi is blue ball 10 and Bi+1 is blue ball 1

Come to think of it...
There are 12c2 = 66 combinations possible.
In each weighing there are 3 outcomes possible > equal, left heavy, right heavy
That mean, in 3 weighings, there are 3^{4} combinations possible; therefore all the 66 combinations can be coded in 4 weighings.
Now, just need to work out what to weigh in each weighing to make the weighing results different for all 66 combinations :)

If Bi > Bi+1, put the green ball in pile 1
If Bi+1 > Bi, put the green ball in pile 2
At the end, both piles will weigh the same
Effectively the weight of green ball = absolute (Bi  Bi+1)
By sorting the piles as above, we take care of the absolute value of this subtraction.
Since the weights are in a full cycle, the net (without absolute) values of green balls will be zero.
But weight difference is always positive.
Yes, that is why we sort them into two piles depending on whether the next ball is heavier or lighter. And the two piles would weigh the same

If Bi > Bi+1, put the green ball in pile 1
If Bi+1 > Bi, put the green ball in pile 2
At the end, both piles will weigh the same
Effectively the weight of green ball = absolute (Bi  Bi+1)
By sorting the piles as above, we take care of the absolute value of this subtraction.
Since the weights are in a full cycle, the net (without absolute) values of green balls will be zero. 
Ah the editor did not let me change the formatting while editing.
In the previous post, I copy pasted from excel. While editing it showed things fine but the final result was all messed up as you can see in the previous post.
This should look better
I would recommend to use the digital scale.
In max 5 weighings, the swapped pair of eggs can be identified:Weigh 1 to 6 number eggs It should weigh 15
If it weighs other than 15, refer to weighings A below  max 2 more weighings needed
If weight > by Possible egg switches
1 6 & 7
2 6 & 8 5 & 7
3 6 &9 5 & 8 4 & 7
4 6 & 10 5 & 9 4 & 8 3 & 7
5 6 & 11 5 & 10 4 & 9 3 & 8 2 & 7
6 6 & 12 5 & 11 4 & 10 3 & 9 2 & 8
7 5 & 12 4 & 11 3 & 10 2 & 9 1 & 8
8 4 & 12 3 & 11 2 & 10 1 & 9
Taking worst case example where there are 5 possibilities
Weigh 6 10 4 8
If weight is +5, possibilities are 6 &11, 4 &9
If weight is 5, then possibilities are 5 &10, 3 &8
If there is no difference then eggs 2 &7 are swapped
Depending on the weight difference, whoose either 6&9 or 5&8 to weigh
Depending on the next weigh difference (+5 or 5), the swapped eggs are identified
Otherwise Weigh 2 4 6 8 10 12
It should weigh 42
If it weighs other than 42, refer to weighings "B" below  max 3 more weighings needed
If weight difference Possible egg switches
5 1 6 7 12
3 1 4 6 3 10 7 12 9
1 1 2 3 4 5 6 7 8 9 10 11 12
1 2 3 4 5 8 9 10 11
3 2 5 8 11
Next, Weigh alternate weights from the weight difference
Taking worst case example where there are 6 possibilties (1 weigh difference)  weigh 1 4 5 8
Possbilities Possible egg switches
+1 1 2 5 6
1 3 4 7 8
0 9 10 11 12
Now, If weight difference Weigh
+1 1 & 6
1 3 & 8
0 9 & 12
A positive or negative weight difference in this weighing identifies the swapped pair
For example if in weighing 1&6, the weight is 6, then the swapped eggs are 5 &6
If the weight is 8, then swapped eggs are 1 & 2
Otherwise Weigh 1 2 11 12
It should weigh 26
If it weighs other than 42, refer to weighings "C" below
Weight Difference Possible egg switches
4 1 5 2 6
2 1 3 2 4
2 10 12 9 11
4 8 12 7 11
0 4 6 3 5 8 10 7 9
For the two possible pairs weigh any one egg out of the 4. If the weight is ok, the other pair is swapped. If the weight is not ok, you have identified the swapped pair
Otherwise Weigh 3 4 10 It should weigh 17 If it weighs 17, then eggs 7 &9 are swapped If it weighs (2) 15, then eggs 8 & 10 are swapped Otherwise, it weighs 19 (+2) and the possibilites are 3&5 or 4&6 are swapped Weigh 3 If it weighs 5, then eggs 3 & 5 are swapped Otherwise eggs 4 & 6 are swapped
 1

I would recommend to use the digital scale.
In max 5 weighings, the swapped pair of eggs can be identified:If it weighs other than 15, refer to weighings A below  max 2 more weighings needed
If it weighs other than 42, refer to weighings "B" below  max 3 more weighings needed
Taking worst case example where there are 6 possibilties (1 weigh difference)
Possbilities Possible egg switches +1 1 2 5 6 1 3 4 7 8 0 9 10 11 12A positive or negative weight difference in this weighing identifies the swapped pair
For example if in weighing 1&6, the weight is 6, then the swapped eggs are 5 &6
If the weight is 8, then swapped eggs are 1 & 2
If it weighs other than 42, refer to weighings "C" below
For the two possible pairs weigh any one egg out of the 4. If the weight is ok, the other pair is swapped. If the weight is not ok, you have identified the swapped pair

Lets say they are in rooms A to E in that order.
All signals are sent after the previous signal from the adjoining room has been heard
First all of they signal their own positions one by one from A to E
Now after hearing the signal from E,
D signals position of C
C signals position of D
B signals position of A
Then C signals position of B
B signals position of D and D signals position of B
Finally, B signals position of C
Each of them knows the position of 4 persons and can deduce the position of the 5th.
Code for sending the signals can be any for example 1 = F, 2 = T....

If you are in France, irrespective of the year, this looks like August when almost everybody is on 3 or more weeks vacation!
How ?
in New Logic/Math Puzzles
Posted
Someone hit Jack's car while the car was parked and Jack was away. The ambulance came for the driver of the vehicle that hit Jack's car.