Jewish Donut
I am a jelly-filled donut that is fried in oil. I am a Jewish food that starts with the letter "s" and a favorite Hanukkah treat. What am I?
Hint:
Given Each Night Riddle
Hint:
Harvesting Eggs Riddle
Hint:
Rope Burn Riddle
You have two ropes. Each rope takes one hour to burn. These ropes are not identical, nor are they uniform; i.e. it does not necessarily take half an hour for half the rope to burn (if you have trouble visualizing this, imagine a rope of varying thickness across its length). With only these two ropes and a way to light them, how do you measure out 45 minutes?
Hint: You can light multiple ends and/or multiple ropes at the exact same time.
Light both ends of one rope, and only one end of the other rope. This will cause the first rope to burn out in 30 minutes. When the first rope burns out, there will be 30 minutes left on the second rope. So then, light the other end of the second rope, and the rest of it will burn out in 15 minutes. 30 + 15 = 45 minutes. Did you answer this riddle correctly?
YES NO
YES NO
Missing Dollar Riddle
Three guests check into a hotel room. The clerk says the bill is $30, so each guest pays $10. Later the clerk realizes the bill should only be $25. To rectify this, he gives the bellhop $5 to return to the guests. On the way to the room, the bellhop realizes that he cannot divide the money equally. As the guests didn't know the total of the revised bill, the bellhop decides to just give each guest $1 and keep $2 for himself. Each guest got $1 back: so now each guest only paid $9; bringing the total paid to $27. The bellhop has $2. And $27 + $2 = $29 so, if the guests originally handed over $30, what happened to the remaining $1?
Hint: Make a list of all of the people involved and how much money they ended up with/spent.
The $9 paid by each guest accounts for the $2 that went to the bellhop. So rather than adding $27 to the $2 kept by the bellhop, the $27 accounts for the bellhops money. The $27 plus the $3 kept by the guests does add up to $30. Did you answer this riddle correctly?
YES NO
YES NO
Prince Age Riddle
A princess is as old as the prince will be when the princess is twice the age that the prince was when the princess's age was half the sum of their present ages.
What are their ages?
What are their ages?
Hint:
Current Future Past
Princess x 2z (x+y)/2
Prince y x z
I then created three equations, since the difference in their age will always be the same.
d = the difference in ages
x y = d
2z x = d
x/2 + y/2 z = d
I then created a matrix and solved it using row reduction.
x y z
1 -1 0 d
-1 0 2 d
.5 .5 -1 d
It reduced to:
x y z
1 0 0 4d
0 1 0 3d
0 0 1 5d/2
This means that you can pick any difference you want (an even one presumably because you want integer ages).
Princess age: 4d
Prince age: 3d
Ages that work
Princess:
4
8
16
24
32
40
48
56
64
72
80
Prince:
3
6
12
18
24
30
36
42
48
54
60 Did you answer this riddle correctly?
YES NO
Princess x 2z (x+y)/2
Prince y x z
I then created three equations, since the difference in their age will always be the same.
d = the difference in ages
x y = d
2z x = d
x/2 + y/2 z = d
I then created a matrix and solved it using row reduction.
x y z
1 -1 0 d
-1 0 2 d
.5 .5 -1 d
It reduced to:
x y z
1 0 0 4d
0 1 0 3d
0 0 1 5d/2
This means that you can pick any difference you want (an even one presumably because you want integer ages).
Princess age: 4d
Prince age: 3d
Ages that work
Princess:
4
8
16
24
32
40
48
56
64
72
80
Prince:
3
6
12
18
24
30
36
42
48
54
60 Did you answer this riddle correctly?
YES NO
Skeleton Keys Riddle
Hint:
Monster Meals Riddle
Hint:
The Safest Room Riddle
Hint:
Three People In A Room
Three people enter a room and have a green or blue hat placed on their head. They cannot see their own hat, but can see the other hats.
The color of each hat is purely random. They could all be green, or blue, or any combination of green and blue.
They need to guess their own hat color by writing it on a piece of paper, or they can write 'pass'.
They cannot communicate with each other in any way once the game starts. But they can have a strategy meeting before the game.
If at least one of them guesses correctly they win $50,000 each, but if anyone guess incorrectly they all get nothing.
What is the best strategy?
The color of each hat is purely random. They could all be green, or blue, or any combination of green and blue.
They need to guess their own hat color by writing it on a piece of paper, or they can write 'pass'.
They cannot communicate with each other in any way once the game starts. But they can have a strategy meeting before the game.
If at least one of them guesses correctly they win $50,000 each, but if anyone guess incorrectly they all get nothing.
What is the best strategy?
Hint:
Simple strategy: Elect one person to be the guesser, the other two pass. The guesser chooses randomly 'green' or 'blue'. This gives them a 50% chance of winning.
Better strategy: If you see two blue or two green hats, then write down the opposite color, otherwise write down 'pass'.
It works like this ('-' means 'pass'):
Hats: GGG, Guess: BBB, Result: Lose
Hats: GGB, Guess: --B, Result: Win
Hats: GBG, Guess: -B-, Result: Win
Hats: GBB, Guess: G--, Result: Win
Hats: BGG, Guess: B--, Result: Win
Hats: BGB, Guess: -G-, Result: Win
Hats: BBG, Guess: --G, Result: Win
Hats: BBB, Guess: GGG, Result: Lose
Result: 75% chance of winning! Did you answer this riddle correctly?
YES NO
Better strategy: If you see two blue or two green hats, then write down the opposite color, otherwise write down 'pass'.
It works like this ('-' means 'pass'):
Hats: GGG, Guess: BBB, Result: Lose
Hats: GGB, Guess: --B, Result: Win
Hats: GBG, Guess: -B-, Result: Win
Hats: GBB, Guess: G--, Result: Win
Hats: BGG, Guess: B--, Result: Win
Hats: BGB, Guess: -G-, Result: Win
Hats: BBG, Guess: --G, Result: Win
Hats: BBB, Guess: GGG, Result: Lose
Result: 75% chance of winning! Did you answer this riddle correctly?
YES NO
Unwilling To Kiss
First think of the person who lives in disguise,
Who deals in secrets and tells naught but lies.
Next, tell me whats always the last thing to mend,
The middle of middle and end of the end?
And finally give me the sound often heard
During the search for a hard-to-find word.
Now string them together, and answer me this,
Which creature would you be unwilling to kiss?
Who deals in secrets and tells naught but lies.
Next, tell me whats always the last thing to mend,
The middle of middle and end of the end?
And finally give me the sound often heard
During the search for a hard-to-find word.
Now string them together, and answer me this,
Which creature would you be unwilling to kiss?
Hint:
Five Potatoes Riddle
A mother has six children and five potatoes. How can she feed each an equal amount of potatoes? Do not use fractions.
Hint:
Tell Us What You See
Have a look at the pic and tell us what it is. It definitely is something btw and once you know it's super obvious!
Still can't see it? Look harder!
Still can't see it? Look harder!
Hint: Stare at the white contrast.
3 Gallon Jug And 5 Gallon Jug
You have a 3-gallon and a 5-gallon jug that you can fill from a fountain of water.
The problem is to fill one of the jugs with exactly 4 gallons of water. How do you do it?
You've got to defuse a bomb by placing exactly 4 gallons (15 L) of water on a sensor. The problem is, you only have a 5 gallon (18.9 L) jug and a 3 gallons (11 L) jug on hand! This classic riddle, made famous in Die Hard 3.
The problem is to fill one of the jugs with exactly 4 gallons of water. How do you do it?
You've got to defuse a bomb by placing exactly 4 gallons (15 L) of water on a sensor. The problem is, you only have a 5 gallon (18.9 L) jug and a 3 gallons (11 L) jug on hand! This classic riddle, made famous in Die Hard 3.
Hint:
Fill the 5-jug up completely. There will be, of course, 5 gallons in the 5-jug. You must fill all the gallons up to the top, otherwise you don't actually know how much you have.
Use the water from the 5-jug to fill up the 3-jug. You're left with 3 gallons in the 3-jug and 2 gallons in the 5-jug.
Pour out the 3-gallon jug. You're left with nothing in the 3-jug and 2 gallons in the 5-jug.
Transfer the water from the 5-jug to the three jug. You're left with 2 gallons in the 3-jug. And nothing in the 5-jug.
Fill up the 5-jug completely. You now have 2 gallons in the 3-jug and 5 in the 5-jug. This means that there is 1 gallon (3.8 L) of space left in the 3-jug.
Use the water from the 5-jug to fill up the 3-jug. Fill up the last gallon of space in the 3-jug with the water from the 5-jug. This leaves you with 3 gallons in the 3-jug, and 4 gallons in the 5-jug.
Fill the 3-jug completely with water. You now have 3 gallons (11.4 L) of water.
Transfer this water into the 5-jug. You now have nothing in the 3-jug, and 3 gallons (11.4 L) in the 5-jug.
Re-fill the 3-jug with water. You now have 3 gallons (11.4 L) in the 3-jug and 3 gallons in the 5-jug.
Fill the 5-jug with water from your 3-jug. You now have 1 gallon (3.8 L) in the 3-jug and 5 gallons (18.9 L) in the 5-jug. This is because, in the last step, you only had 2 gallons (7.6 L) of space left over, so you could only pour 2 gallons.
Pour out the 5-jug and refill it with your 1 gallon. You now have nothing in the 3-jug and 1 gallon in the 5-jug
Fill up the 3-jug. You now have 3 gallons (11.4 L) in the 3-jug and 1 in the 5-jug.
Transfer the 3 gallons (11.4 L) of water into the 5-jug to end up with 4 gallons (15.1 L). Simply pour over your three gallons into the 5-jug, which only had 1 gallon (3.8 L) in it previously. 1+3=4, and a successfully defused bomb. Did you answer this riddle correctly?
YES NO
Use the water from the 5-jug to fill up the 3-jug. You're left with 3 gallons in the 3-jug and 2 gallons in the 5-jug.
Pour out the 3-gallon jug. You're left with nothing in the 3-jug and 2 gallons in the 5-jug.
Transfer the water from the 5-jug to the three jug. You're left with 2 gallons in the 3-jug. And nothing in the 5-jug.
Fill up the 5-jug completely. You now have 2 gallons in the 3-jug and 5 in the 5-jug. This means that there is 1 gallon (3.8 L) of space left in the 3-jug.
Use the water from the 5-jug to fill up the 3-jug. Fill up the last gallon of space in the 3-jug with the water from the 5-jug. This leaves you with 3 gallons in the 3-jug, and 4 gallons in the 5-jug.
Fill the 3-jug completely with water. You now have 3 gallons (11.4 L) of water.
Transfer this water into the 5-jug. You now have nothing in the 3-jug, and 3 gallons (11.4 L) in the 5-jug.
Re-fill the 3-jug with water. You now have 3 gallons (11.4 L) in the 3-jug and 3 gallons in the 5-jug.
Fill the 5-jug with water from your 3-jug. You now have 1 gallon (3.8 L) in the 3-jug and 5 gallons (18.9 L) in the 5-jug. This is because, in the last step, you only had 2 gallons (7.6 L) of space left over, so you could only pour 2 gallons.
Pour out the 5-jug and refill it with your 1 gallon. You now have nothing in the 3-jug and 1 gallon in the 5-jug
Fill up the 3-jug. You now have 3 gallons (11.4 L) in the 3-jug and 1 in the 5-jug.
Transfer the 3 gallons (11.4 L) of water into the 5-jug to end up with 4 gallons (15.1 L). Simply pour over your three gallons into the 5-jug, which only had 1 gallon (3.8 L) in it previously. 1+3=4, and a successfully defused bomb. Did you answer this riddle correctly?
YES NO
12 Islanders Teeter Totter Riddle
There is an island with 12 islanders. All of the islanders individually weigh exactly the same amount, except for one, who either weighs more or less than the other 11.
You must use a see-saw to figure out whose weight is different, and you may only use the see-saw 3 times. There are no scales or other weighing device on the island.
How can you find out which islander is the one that has a different weight?
You must use a see-saw to figure out whose weight is different, and you may only use the see-saw 3 times. There are no scales or other weighing device on the island.
How can you find out which islander is the one that has a different weight?
Hint:
Six on one side - six on the other = one side is heavier.
Take the heavier six men, divide them into three and three (random).
Three on one side - three on the other = one side will one heavier.
Divide that three men from the heavier side side, have one on one side - one on the other.
Two results can determine which of the last three men weight is a different weight than each other.
With the last group of three men, have two men go head-to-head. The see-saw will either weight different: one weights more than the other man meaning the heavier man is the "12th man" or the see-saw will balance between the two men because they are the same weight. That means the third man standing on the sidelines by default weights more than the last two men weighted. Thus making that man on the sidelines the "12th man" that weights more than other 11.
Heavier wins 6v6; winner gets divided. Heavier wins 3v3; winner gets divided. Heavier wins 1v1 (12th man) or Equal 1v1 = third man weight more, he's the 12th man.
You could find the same results changing the process and picking from the lighter group three times. You’re only trying to find the difference in weight. Not the exact weight (more or less) of that "12th man."
Lightest 6v6; Lightest 3v3; Lightest 1v1 or Equal 1v1 = third man weight less. Did you answer this riddle correctly?
YES NO
Take the heavier six men, divide them into three and three (random).
Three on one side - three on the other = one side will one heavier.
Divide that three men from the heavier side side, have one on one side - one on the other.
Two results can determine which of the last three men weight is a different weight than each other.
With the last group of three men, have two men go head-to-head. The see-saw will either weight different: one weights more than the other man meaning the heavier man is the "12th man" or the see-saw will balance between the two men because they are the same weight. That means the third man standing on the sidelines by default weights more than the last two men weighted. Thus making that man on the sidelines the "12th man" that weights more than other 11.
Heavier wins 6v6; winner gets divided. Heavier wins 3v3; winner gets divided. Heavier wins 1v1 (12th man) or Equal 1v1 = third man weight more, he's the 12th man.
You could find the same results changing the process and picking from the lighter group three times. You’re only trying to find the difference in weight. Not the exact weight (more or less) of that "12th man."
Lightest 6v6; Lightest 3v3; Lightest 1v1 or Equal 1v1 = third man weight less. Did you answer this riddle correctly?
YES NO
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