Math riddles

viagrandad
viagrandad

Give me what you got.

Riddle #1 (easy-medium):
There is a line of people wearing black and white hats. Each person doesnt know which color their hat is and they don't know which color the people behind them wear. One at a time, every person shouts "black" or "white" from the last person in the line (the one that can see all hats but not his own) until the first one.
Their job is to shout the color of the hat that they are wearing except the last one, which doesn't have to say the correct color.
How do they make that possible?

Riddle #2 (hard):
There are 3 rooms.
Room 1: 2 people
Room 2: 100 drawers numbered 1-100, each drawer has inside it a number between 1 and 100, no duplicates.
Room 3: Empty
Person 1 goes out of room 1 and into room 2. He can look inside the drawers all he want. After he made his calculations he can switch the numbers between 2 drawers (1 switch) and then he is escorted out to the 3rd room.
Person 2 goes into room 2 and is given a random number between 1 and 100. His job is to find that number within 50 drawer openings.
When they are on room 1 they both know what is about to take place but they don't know which drawer has which number and they don't know what random number is going to be given to person 2.
How do they make this possible?

Burnblaze
Burnblaze

An eleven year old blind orphan girl weighing 73 pounds is shot from a cannon at a muzzle velocity of 89 feet per second and an angle of 19 degrees from horizontal. If she strikes a solid brick wall located 4 yards away, how long does she stick to the wall and how big is the stain left behind when she drops off? Use SI units in your answer.

Evilember
Evilember

Here's a hard one. Continue this sequence:

1, 11, 21, 1112, 3112, 211213, 312213, 212223, ...

Spazyfool
Spazyfool

@Evilember
12113213?

Sir_Gallonhead
Sir_Gallonhead

@Evilember
411312 but i already knew this

King_Martha
King_Martha

@Sir_Gallonhead
I mean, 114213

Evil_kitten
Evil_kitten

@viagrandad
#1 Seen this one 2000 times before, can't be bothered.
#2 Consider this strategy: The next drawer you open is the one with the number you've found inside the current drawer.
Opening drawers using this strategy will result in cycles forming, as in, it's inevitable that at one point, you'll return to the drawer you started with. This also means that if you start with the drawer that has the number you have to find on it, as long as you manage to get through the whole cycle, you'll find your cycle. Of course, you only have 50 moves, so if the cycle is longer than 50, you're out. This, however, is where the first man who came before you helps you out. Since there is only 100 drawers, there can't be more than 1 cycle with more than 50 drawers, and any cycle can be broken in half with a single swap (Swap last and middle drawer). So basically, the first man ensures all cycles are <50, second man walks the cycle starting with his number.

viagrandad
viagrandad

@Evil_kitten
you'll find your number**** god damn it.

iluvmen
iluvmen

@Evilember

Actually it's not hard at all. The next number in the sequence registers the number of particular digits in the previous one.
1 - one
11 - 1 count of the digit "1"
21 - 2 counts of the digit "1"
1112 - 1 count of "1" and 1 count of "2"
3112 - 3 counts of "1" and 1 count of "2"
211213 - 2 of 1, 1 of 2, 1 of 3
312213 - 3 of 1, 2 of 2, 1 of 3
212223 - 2 of 1, 2 of 2, 2 of 3
114213 - 1 of 1, 4 of 2, 1 of 3

Deadlyinx
Deadlyinx

Well, since we're out of riddles now, I have a pretty good one.
Can at any given time a point on earth's surface be found where the corresponding point on the opposite side has the same temperature? (For ease of finding the opposite side point, you can assume earth is a sphere)

Dreamworx
Dreamworx

Come on, I don't want this thread to die.

Nude_Bikergirl
Nude_Bikergirl

@Deadlyinx
Yes, walk around the equator and you will find a point because intermediate value theorem.

VisualMaster
VisualMaster

@Nude_Bikergirl
Killin' it.

How about this one (fairly easy though):

Your name is Zac Danger, you are the superdetective. Your intuition is always right. Currently, you are solving a case of a clown murderer. You have four suspects, You've already interrogated all of them:
The first suspect, Richard Willy, testified:
It's the third suspect, John west, who is the clown.
The second suspect, Honker Rednose, testified: I am not the clown.
The third suspect, John west, testified: I am not the clown.
The fourth and final suspect, Josef Mengele, testified: The first suspect, Richard Willy, is clearly a jew so it's obvious that he is the murderous clown.

Your infallible intuition tells you that only one of them tells the truth.

Who is the clown?

CodeBuns
CodeBuns

3

Ignoramus
Ignoramus

@CodeBuns
Wrong.

Supergrass
Supergrass

Sorry meant 2

SomethingNew
SomethingNew

@Evil_kitten
Correct

askme
askme

@VisualMaster

a clown murderer or a murderous clown?

Inmate
Inmate

@VisualMaster

since my intuition is always right according to your post, i'll say that the fourth suspect is the murderer.

lostmypassword
lostmypassword

@viagrandad
Oh boy, someone let OP watch the TED "Can you solve the _____ riddle?" YouTube channel again.

Gigastrength
Gigastrength

@VisualMaster
Honker Rednose.
Since only one of them tells the true, we just check by logic which case have only one true option:
If Richard is telling the true, so does Honker Rednose, we have:
TTFF(T for True and F for false).
If Josef Mengele is telling the true, then Richard Willy is telling a lie, which means John west is telling the true:
FFTT
Now, in the case Honker Rednose tells the true, we can either assume John West is telling the true, or he is lying and Richard Willy is telling the true, thus:
FTTF or TTFF
In the case John West is telling the true, than Richard Willy is lying and so does Honker Rednose and nothing can be assumed from Josef Mengele. If Mengele tells the true, we have: FFTT
If Mengele is lying, we have: FFTF
Thus, the only case we have only one true is when John West is telling the true, thus, Honker Rednose is the clown.

Emberburn
Emberburn

This is a good one. Very simple. Kind of hard, but doable.

massdebater
massdebater

@Gigastrength
Correct

Nude_Bikergirl
Nude_Bikergirl

@Gigastrength
There's a faster way to solve @VisualMaster
:
The truth-teller must be one of Honker or John, since they can't both be the clown (i.e. both be lying).
Since there's only one truth-teller, the other one is lying and hence is the clown.
Also, Richard is lying, so John is not the clown. Therefore Honk is the clown.

Harmless_Venom
Harmless_Venom

@Emberburn
Is arcsin cheating? Because if it's not, this is a cakewalk.

Lord_Tryzalot
Lord_Tryzalot

@Emberburn
An utterly inelegant but workable method:

1. Draw the obvious isoceles triangle with three sides r,r,L known
2. Get the angle subtended using the cosine formula
3. Use angle*pir*^2 to get area of the sector
4. Calculate area of the r-r-L triangle using sine formula (or Heron's semiperimeter formula, or whatever floats your boat)
5. Then the required area is (3) - (4).

Soft_member
Soft_member

@Nude_Bikergirl
The most elegant solution.

Evil_kitten
Evil_kitten

@Lord_Tryzalot
You're pulling the angle out of the cosine (arc cos), this usually isn't what you're supposed to do in those riddles.

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