Brainteasers/Math problems - Page 2

about the water one...im guessing that you fill up the 5 liter jug using the 3 liter jug until the 5 liter jug is full and 1 liter of water is left in the three liter jug.

5 liter jug = full
3 liter jug = 1 liter

then you empty the 5 liter jug, pour the 3 liter jug into the 5 liter jug and, after filling the 3 liter jug, pour the 3 liters into the 5 liter jug making 4 amirite?

1. fill the 3 liter and throw those 3 liters in the 5 liter. now fill the 3 liter again and fill the 5 liter. there is now one liter in the 3 liter jug. empty the 5. throw the one liter from the 3 into the 5. fill the 3 liter jug and add to the 5 liter jug. you now have 4 liters.

2. take nothing out of bag 1, take 1 coin out of bag 2, 2 out of bag 3, 3 out of bag 4 etc. put all the bags on the scale. check to see how many weight you miss. if you miss nothing then bag 1 is the counterfeit. if you miss 0,1 gram bag 2 is the counterfeit bag, miss 0,2 bag 3 is fake etc.

3. set the 7 and the 11 minute glass. drop in the egg when the 7 runs out. then flip the 11 minute hourglass when it runs out. egg is done when the 11 is done the second time (4+11)

4. 1:07 but you need some trig to solve this, too lazy to do this. should be pretty easy someone in high school did this in ~3rd year.

5. surface area of the disks: PI*1^2 * 25 = 25PI

other disks: PI* 0,5^2 * 100 = 0,25*100*PI = 25PI . yes you can.

It's not 7+11, you begin boiling then egg when the 7 finishes, leaving 4 minutes on the 11 hour glass, when the four minutes is over, you flip it for another 11 minutes. 4+11 is 15

1:
Fill the 5l can, fill the 3 liter with as much as u can from the 5 liter = 2lites left in the 5 liter. empty out the 3l. Pour the 2 liters that are in the 5 liter in to the 3 liter. Fill the five liter and fill the 3 liter with the water from the 5 liter = there will be 4 liters left in the 5 liter can.

2:
take 1 coin from the first bag, 2 from the next etc up to the tenth. If none of them were fake they should weigh 55 grams. But since one of the stacks where fake and only weighed 0.9 grams per coin. They will weigh anything between 54.9- 54 and that is how u determine which one is fake since the fake ones will be in the X stack when the weigh is 55- x*0.1

3: start both hourglasses when 7 minute runs out start boiling the egg. That means that it's 4 min left on the 11 minute hour glass. and when it runs out just flip it over again 4+ 11 = 15.

4:
12 hours later?

5: Reducing the radius of an object means that it reduces it's size both in height and width. this being said if if scale this whole question down to it being only 2 disks in the beginning and that the end result would be 4 times more (8) if you reduce the radius to half of it's current state it's real easy to picture it in you're head. half the radius = 3 more disc fit for each existing disc as one is added to it's width and height and one diagonally as this space is freed up as well = there are 3 more per disc than before that there is room for.

6: mind blown :/ gonna think about it...

7: gonna think about it...

8. light up both ends on one of the strings and just one on the other. When the first one burns out light the other end on the second piece of string.

9: 31 - prime 31mod 30 = 1-not prime.

10: yet another one I don't know .p

since it's a paradox it doesn't have an answer but still it's a nice brain teaser

I got a even more elegant solution.
1)flip both over
2)When 7 min hourglass runs out, flip it over and over again.
3) Let 7 min hourglass finish when 11 min hourglass runs out.
+ Show Spoiler +

9. No, it doesn't have to be prime. 31 is prime, divide by 30, remainder 1. 1 isn't prime

I'm afraid not still. Give any prime who's modulus 30 is 1 and it'll disprove your conjecture. 61 disproves, 151 disproves, 541 disproves, etc. In other words, if (x-1)/30=integer, where x is a prime number, then your idea doesn't work. Maybe exclude 1 and it'll work.

Is it correct to assume that the guru is completely useless and only built in as a distraction?

30 = 2*3*5, which means that the remainder can't have any of those factors, else:
Let p be a prime > 30
p mod 30 = R => p = 30*n + R
if R = m*5
p = 30x + 5m = 5*(6x+m) => 5|p => p can't be a prime if it has 5 as a factor (we assume p>30)
same for R=2m and R=3m
So the only possible remainders are:
1, 7, 11, 13, 17, 19, 23, 29
1 happens to be not a prime.
91 mod 30 = 1
;o

Nope, the guru plays a roll in the solution, it's a rather famous riddle, the numbers could be anything though, the logical conclusion goes for every number of people

1. fill 5, pour 5 into 3, empty 3, pour 5 into 3, fill 5, pour 5 into 3. 5 jug now has 4 in it.

3. flip 7 and 11. when 7 is done, flip 7 again. when 11 is done, flip 7 (with 3 on it) and 11. when 3 is done, neutralize 11 (8 on it.) get egg in pan, flip the 8, when 8 is done, flip 7

4. if both hands are infinitely thin then they will always coincide but the real answer you seek is some point near 1:05 that i am too lazy to calculate. you just have to remember it isn't 1:05 because the hour hand will have moved up 1/12 of the distance between 1 and 2 at that point.

5. yes. 25 * 1 * 3.14 = 78.5 100 * 0.5 * 0.5 * 3.14 = 78.5 its the same.

7. yes. weigh 3 on one side and 3 on the other. if one side weighs less, that advances to step two. if both sides are equal, the unweighed 3 coins advance to step 2. then in step 2, weigh any 2 random coins. if equal, its the unweighted 1. otherwise its on the scale.

Weigh 3 coins on each side. If they weight the same, then none were counterfeit and you have 3 coins left. You would then choose 2 of those coins and weigh them. If they're the same, then the third one is counterfeit. If not, then it's the lighter one.

If the initial weighing gives you that one side is lighter, then you have 3 coins, one of which is counterfeit. You can just choose 2 and weigh them. If they weight the same, then the third one is counterfeit. If not, then it's the lighter one.

The first person calls out red if he sees an odd number of red hats and blue if he sees an even number. This person might die. But he calls out red, the next person looks back and if he sees an odd number, then he must be wearing blue. If not, then he must be wearing red.

This is a simple inductive reasoning problem. This means we have to take it one step at a time, starting with 1 blue-eyed person.

1. If there is 1 person he leaves on the first night because he knows he must be the only one with blue eyes.

2. If there are 2 people, each one of them sees each other. The next day, they wonder why the other person didn't leave, because if they were the only blue-eyed person, they would have left. Both of them being perfectly logical come to this conclusion and leave on the second night.

3. If there are 3 people, each one of them sees each other. The second day, no one left, because if there were 2 people they would have to wait until the second night anyways. The third day, they realize that none of them left because there must be 3 people on the island with blue eyes, because otherwise if there were 2 people then the 2 would have left on the second night. So, they leave on the third night.

... So on and so forth until...

100. If there are 100 people, all of them see 99 people. The 100th day, no one left, because if there were only 99 people on the island, they would have left on the 99th night. So, the 100 people leave on the 100th night.