7 Riddles That Will Test Your Brain Power - Episode #6
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Date: 2023-11-15
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Comments and reviews: 29
MalinP.
A faster way to figure out the water barrel is to add all the barrels = 119 and then add the numbers 1+1+9=11. 11 is not dividable with 3 -> 119 is not dividable with 3. If you take off 2 (11-2=9) it is dividable with 3, which means the water barrel is a number dividable with 3 +2 (witch is only 20, 20-2=18. If you take away 20 from 119 the total of the milk barrels will be dividable with 3.
This makes it faster and easier to do quickly without writing it down.
If you add all separate numbers in a big number until yo have a number you can easily divide by 3 in your head you know if the big number is dividable with 3 or not. You can also split it up into sections to make it easier (and take away all 3s, 6s and 9s - since those obviously are dividable by 3. For example, is 264718305619437 dividable with 3?
1: take away 6, 3, 6, 9 and 3.
2: 2+4=6 (take away, 7 (save, 1+8=9 (take away, 5+1 (take away, 4+7=11 (save.
3: 7+11=18 (dividable with 3) -> 264718305619437 is dividable with 3.
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A faster way to figure out the water barrel is to add all the barrels = 119 and then add the numbers 1+1+9=11. 11 is not dividable with 3 -> 119 is not dividable with 3. If you take off 2 (11-2=9) it is dividable with 3, which means the water barrel is a number dividable with 3 +2 (witch is only 20, 20-2=18. If you take away 20 from 119 the total of the milk barrels will be dividable with 3.
This makes it faster and easier to do quickly without writing it down.
If you add all separate numbers in a big number until yo have a number you can easily divide by 3 in your head you know if the big number is dividable with 3 or not. You can also split it up into sections to make it easier (and take away all 3s, 6s and 9s - since those obviously are dividable by 3. For example, is 264718305619437 dividable with 3?
1: take away 6, 3, 6, 9 and 3.
2: 2+4=6 (take away, 7 (save, 1+8=9 (take away, 5+1 (take away, 4+7=11 (save.
3: 7+11=18 (dividable with 3) -> 264718305619437 is dividable with 3.
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JP
Riddle 5 is flawed. Assumes constant speed. But people accelerate. So maybe Bob has higher top end speed but slower acceleration. You don-t know. Maybe Bob threw race 1 so his bookie could everyone to bet on Alice. Maybe they started on the edge of a river so 10m back is in the middle of the water and she can-t swim so she-ll never make it to the finish line. Maybe Bob got too many participation trophies as a child and getting beat by a woman in a foot race made him question his life choices and his lack of effort. And maybe this was the new start Bob needed to have the confidence to get ahead in life. And he-ll try harder both in the race and in his life.
You don-t know Bob. He-s a new man now.
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Riddle 5 is flawed. Assumes constant speed. But people accelerate. So maybe Bob has higher top end speed but slower acceleration. You don-t know. Maybe Bob threw race 1 so his bookie could everyone to bet on Alice. Maybe they started on the edge of a river so 10m back is in the middle of the water and she can-t swim so she-ll never make it to the finish line. Maybe Bob got too many participation trophies as a child and getting beat by a woman in a foot race made him question his life choices and his lack of effort. And maybe this was the new start Bob needed to have the confidence to get ahead in life. And he-ll try harder both in the race and in his life.
You don-t know Bob. He-s a new man now.
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Venkat
When you put something into a spinning wheel it tries to get to the centre. So there are no fields. No magnetic or electric. Structures are the positional dependency of spin. Similarly there are no imagery numbers Associated spin quantum of numbers. Why do numbers spin because of the base or symbols. The duality of nature is just a vector positive or negative relative. Light takes two planes because of the negative drag effect. Curvature of space is a relative drag momentum of light. Drag effect is always one third because of next to half. For sound it is about 70 meters per second.
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When you put something into a spinning wheel it tries to get to the centre. So there are no fields. No magnetic or electric. Structures are the positional dependency of spin. Similarly there are no imagery numbers Associated spin quantum of numbers. Why do numbers spin because of the base or symbols. The duality of nature is just a vector positive or negative relative. Light takes two planes because of the negative drag effect. Curvature of space is a relative drag momentum of light. Drag effect is always one third because of next to half. For sound it is about 70 meters per second.
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Bonus riddle 1:
I solved it without actually touching the pen. There-s a tape there, X is just two Vs mirrored together from top to bottom. You don-t need to use the pen. Use both the paper and the tape to halve the top half of IX, which is basically IV. Paste the tape on IV. Then turn the book 180 degrees. VI equals six. Wasn-t that hard. Yet Ammar said to draw an S at IX, which is basically lifting the pen, which is not allowed for not lifting the pen out of the paper to be a condition.
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Bonus riddle 1:
I solved it without actually touching the pen. There-s a tape there, X is just two Vs mirrored together from top to bottom. You don-t need to use the pen. Use both the paper and the tape to halve the top half of IX, which is basically IV. Paste the tape on IV. Then turn the book 180 degrees. VI equals six. Wasn-t that hard. Yet Ammar said to draw an S at IX, which is basically lifting the pen, which is not allowed for not lifting the pen out of the paper to be a condition.
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George
I'm lucky if I can get 1 or 2 solutions per video. The most important thing to remember is, to - keep it simple-
Very good. I recently started watching higher math tutorials, way out of my league. Again, if you listen diligently, you can pick up some information. That's not say the abstract & arcane. Even the brightest folks would need several years of training.
Great videos by the way. I doubt many kids graduating today would understand.
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I'm lucky if I can get 1 or 2 solutions per video. The most important thing to remember is, to - keep it simple-
Very good. I recently started watching higher math tutorials, way out of my league. Again, if you listen diligently, you can pick up some information. That's not say the abstract & arcane. Even the brightest folks would need several years of training.
Great videos by the way. I doubt many kids graduating today would understand.
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Pusspaws
Your 1st riddle/puzzle is incomplete and confusing in the requirements. Your 2nd riddle breaks the rule of having odd coins in 3 glasses. Even if you have 4 or whatever even number in the bottom or even top inserted glass it breaks the rule you gave us to the riddle/puzzle where the odd number of coins in the glasses is the main exercise. Even though glass B is inside glass A, there are still an even number of coins in glass A.
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Your 1st riddle/puzzle is incomplete and confusing in the requirements. Your 2nd riddle breaks the rule of having odd coins in 3 glasses. Even if you have 4 or whatever even number in the bottom or even top inserted glass it breaks the rule you gave us to the riddle/puzzle where the odd number of coins in the glasses is the main exercise. Even though glass B is inside glass A, there are still an even number of coins in glass A.
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D2SProductions
No, if today is the girl's eleventh birthday, that is her last birthday, so it saying that her last birthday was her 10th birthday is false and purposefully misleading. What it should say is her previous birthday was her tenth birthday, and her next birthday is her twelveth birthday, that leaves it open grammatically for her 11th birthday to be today. That example is a grammatical logic fail.
Bonus riddle 2 answer: -
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No, if today is the girl's eleventh birthday, that is her last birthday, so it saying that her last birthday was her 10th birthday is false and purposefully misleading. What it should say is her previous birthday was her tenth birthday, and her next birthday is her twelveth birthday, that leaves it open grammatically for her 11th birthday to be today. That example is a grammatical logic fail.
Bonus riddle 2 answer: -
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Cas
Ngl, some of these out of the box solutions are too far out. The number 4 is not a square, it's a square number (hence you read the previous statement and did not read the number 4 as the word or image square, it didn't even come out of your mouth) and the glass solution, putting one glass inside another doesn't transform it into a new whole glass, therefore there was still 4 coins in one of the glasses.
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Ngl, some of these out of the box solutions are too far out. The number 4 is not a square, it's a square number (hence you read the previous statement and did not read the number 4 as the word or image square, it didn't even come out of your mouth) and the glass solution, putting one glass inside another doesn't transform it into a new whole glass, therefore there was still 4 coins in one of the glasses.
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Kirill
For the second one, another interesting solution would be to place the three glasses in an equilateral triangle type formation. Three coins in each glass, and the last coin in the center of the formation, just so that a third of the coin rests on a section of each glass's rim. But that is only viable if you don't care about whether the number is whole or not.
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For the second one, another interesting solution would be to place the three glasses in an equilateral triangle type formation. Three coins in each glass, and the last coin in the center of the formation, just so that a third of the coin rests on a section of each glass's rim. But that is only viable if you don't care about whether the number is whole or not.
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torgo4ever
the coin and glass riddle is an incorrect solution by definition. The question states that - EACH - glass must contain an odd number of coins. Even if you put the glass with 4 coins INTO the glass with 3 coins, that first glass still contains an even number of coins. It doesn't matter where you put IT, it still contains an even number.
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the coin and glass riddle is an incorrect solution by definition. The question states that - EACH - glass must contain an odd number of coins. Even if you put the glass with 4 coins INTO the glass with 3 coins, that first glass still contains an even number of coins. It doesn't matter where you put IT, it still contains an even number.
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mrmacross
For the second-to-last one, I misread the instructions. I flipped the paper upside down so it'd read XI. Then I covered the lower half with that piece of tape, so it'd look like VI instead of XI. I guess the tape was there so you don't move the paper, though, but the instructions never said you couldn't do that.
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For the second-to-last one, I misread the instructions. I flipped the paper upside down so it'd read XI. Then I covered the lower half with that piece of tape, so it'd look like VI instead of XI. I guess the tape was there so you don't move the paper, though, but the instructions never said you couldn't do that.
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Bob
Bonus answer for the perfect square. Cut a match in two equal parts then put them at a right angle near the intersection of two untouched matches so that each part of the cut match are also forming a right angle with one the untouched matches. He didn-t say we were not allowed to break or cut the match we are moving.
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Bonus answer for the perfect square. Cut a match in two equal parts then put them at a right angle near the intersection of two untouched matches so that each part of the cut match are also forming a right angle with one the untouched matches. He didn-t say we were not allowed to break or cut the match we are moving.
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Fidi987
Riddle 4: What does -one- mean in this case? Couldn't you take one matchstick, break it into two pieces and place them against two others to create a square (with parts of the unbroken sticks sticking out? Then, you would have moved one, but placed its two parts. So, a small way of cheating?
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Riddle 4: What does -one- mean in this case? Couldn't you take one matchstick, break it into two pieces and place them against two others to create a square (with parts of the unbroken sticks sticking out? Then, you would have moved one, but placed its two parts. So, a small way of cheating?
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TheMisterNebo
For Riddle 3, a bonus answer: Since the numerical system isn't specified, one can assume any suitable numerical system, for instance, Base33. in the Base33 numerical system, 10 is thirty-three (decimal) coins, meaning you simply place eleven (decimal) coins in each of the 3 glasses.
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For Riddle 3, a bonus answer: Since the numerical system isn't specified, one can assume any suitable numerical system, for instance, Base33. in the Base33 numerical system, 10 is thirty-three (decimal) coins, meaning you simply place eleven (decimal) coins in each of the 3 glasses.
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Paul-Stelian
My solutions:
1-2: I got the solution
3: Nope
4: Only the -in- one, the -4- one is very funny.
5: I got the right answer but I overcomplicated it by actually trying to do the math.
6: Got stumped.
7: I got it.
Bonus 1: Nope
Bonus 2: Just do! = (unequal)
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My solutions:
1-2: I got the solution
3: Nope
4: Only the -in- one, the -4- one is very funny.
5: I got the right answer but I overcomplicated it by actually trying to do the math.
6: Got stumped.
7: I got it.
Bonus 1: Nope
Bonus 2: Just do! = (unequal)
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Ernest
On the bonus question, follow 3 steps:
a) lift up the pen and write nothing on the paper.
b) lay the pen down so that is crosses the IX horizontally leaving the bottom half of the IX exposed
c) walk around to the opposite side of the table and a VI is displayed above the pen
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On the bonus question, follow 3 steps:
a) lift up the pen and write nothing on the paper.
b) lay the pen down so that is crosses the IX horizontally leaving the bottom half of the IX exposed
c) walk around to the opposite side of the table and a VI is displayed above the pen
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John
Question 3 is BS, it specifically asks to put an odd number of coins in 3 glasses, nowhere are you given the indication you can combine the glasses, otherwise the answer is obvious. Also if you combine the glasses they don't become 1 glass so it's wrong.
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Question 3 is BS, it specifically asks to put an odd number of coins in 3 glasses, nowhere are you given the indication you can combine the glasses, otherwise the answer is obvious. Also if you combine the glasses they don't become 1 glass so it's wrong.
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jason
so the bonus question asks to substract 3 from 9 without lifting the pen, then he lifts the pen to write an s? the actual answer to this is to erase (or cover by folding the paper or cuting the paper) the top part of the Ix and flip it to give VI.
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so the bonus question asks to substract 3 from 9 without lifting the pen, then he lifts the pen to write an s? the actual answer to this is to erase (or cover by folding the paper or cuting the paper) the top part of the Ix and flip it to give VI.
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Samuel
For question 7, just add the total of all the barrels (119) and subtraction each barrel. If the remaining quantity is divisible by 3 then the subtracted barrel is the milk. If you minus 119-20=99, 99 is divisible by 3 hence the answer is 20.
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For question 7, just add the total of all the barrels (119) and subtraction each barrel. If the remaining quantity is divisible by 3 then the subtracted barrel is the milk. If you minus 119-20=99, 99 is divisible by 3 hence the answer is 20.
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Easier way for Riddle #1, without case analysis:
The 3 statements: >=1, =5
Since only one statement is true, and one of =5 must be true. then >=1 must false.
That means # of coins cannot >=1, i. e. < 1. Then I have 0 coins.
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Easier way for Riddle #1, without case analysis:
The 3 statements: >=1, =5
Since only one statement is true, and one of =5 must be true. then >=1 must false.
That means # of coins cannot >=1, i. e. < 1. Then I have 0 coins.
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Carlton
Last riddle is lost depending on region.
9. 50 makes sense in some areas but not the US.
This would be read as: 9. 50 or 9h +. 5h
. 5h is 30 min.
So. 9: 30
I assume other areas read time that way
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Last riddle is lost depending on region.
9. 50 makes sense in some areas but not the US.
This would be read as: 9. 50 or 9h +. 5h
. 5h is 30 min.
So. 9: 30
I assume other areas read time that way
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Jacob
First one you can have between 1 and 5 coins
If only one statement is true, then the other 2 are false, which makes the statement you-re on true, which is the current number of coins. I think it should be reworded
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First one you can have between 1 and 5 coins
If only one statement is true, then the other 2 are false, which makes the statement you-re on true, which is the current number of coins. I think it should be reworded
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Ruby
Riddle 3 - that answer is not correct because your statement says each glass should have an odd number. Your A has 4 coins whether you put B glass in A or not. Each glass has an odd number. You have not done it
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Riddle 3 - that answer is not correct because your statement says each glass should have an odd number. Your A has 4 coins whether you put B glass in A or not. Each glass has an odd number. You have not done it
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Paul
They were all easy. And especially after being asked to think outside the box, I saw another solution to the match sticks: No one told me that it is forbidden to break the one stick I move into half.
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They were all easy. And especially after being asked to think outside the box, I saw another solution to the match sticks: No one told me that it is forbidden to break the one stick I move into half.
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PhilL
In the 5 barrels. I added the liters = 119 so that is 2 more than 117 which is divisible by 3 and only 20 is 2 more than a number divisible by 3 and can be removed from the calc for milk
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In the 5 barrels. I added the liters = 119 so that is 2 more than 117 which is divisible by 3 and only 20 is 2 more than a number divisible by 3 and can be removed from the calc for milk
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kannam
for 3rd riddle even though we place one glass in other, we see each glass seperately and count no. of coins in each glass seperately, so what is the matter of keeping one glass in other
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for 3rd riddle even though we place one glass in other, we see each glass seperately and count no. of coins in each glass seperately, so what is the matter of keeping one glass in other
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John
The odd number of coins in three glasses is WRONG. Putting one glass into another does not make two glasses. There are still three glasses and one of them contains four coins.
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The odd number of coins in three glasses is WRONG. Putting one glass into another does not make two glasses. There are still three glasses and one of them contains four coins.
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Michael
i got all of these correct. and then i threw my couch across the living room with just my mind. i am a god now. i'm off to find a supervillain. be well.
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i got all of these correct. and then i threw my couch across the living room with just my mind. i am a god now. i'm off to find a supervillain. be well.
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Gary
Riddle 5: bobs speed is 90% of Alice-s. So if she runs 10 meters back that-s 110 meters he-s running 90% of so
110 -. 90 = 99 so Alice wins
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Riddle 5: bobs speed is 90% of Alice-s. So if she runs 10 meters back that-s 110 meters he-s running 90% of so
110 -. 90 = 99 so Alice wins
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