Brainteasers/Math problems - Page 3

Well the question states "in some way",so i guess we have to follow the assumption that it can somehow fill up the whole rectangle without overlaps.A somewhat unrealistic question if u ask me.

Obviously it will occur after 1:00. The minute hand will be slightly past 5 as someone mentioned earlier in thread. This is just a basic system of equations math problem.

Variables: HA = hand angle, MA = Minute Angle, m = number of minutes past 1

HA = 30 + (30 m / 60), where m is the number of minutes that have passed.
MA = 6 m.

Condition: HA = MA
HA = 30 + MA /12 => MA = 30 + MA / 12
Solving:
11 MA = 360 => MA = 360 / 11

m = MA / 6 => m = 60 / 11

Therefore, they overlap in 1 hour and 60/11 minutes (approx 5 min 27 sec)

Yes, the circles cannot overlap. The solution is to note that you can half each dimension of the original rectangle filled with cirlces, and the circles will have their radius cut to 1/2, with the original orientation preserved. Then construct four of these small rectangles and preserve the orientation of the 1/2 circles. Put the four rectangles together in such a way to make the original rectangle dimensions. Then you have a pattern of 1/2 size circles which will always properly fill the rectangle (4 times the original number of cirlces). No overlap as long as the original had no overlap.

Cool Question.

Since one vizier can be wrong, the first vizier needs to say red or blue in a way that will give information to the rest of them.

The first vizier needs to give information on the parity (odd/even) of the red or blue on the hats that he sees. That way, the rest of the viziers can count the hats, and determine his own hat color based on the parity condition.

If there are an even number of viziers (say 100), the first vizier will see an odd number of total hats (say 99) - that means you have an odd number of one hat color, and an even number of the other hat color. The first vizier to go says the hat color of whichever he sees as odd. Then the rest of the viziers can deduce their own hat color based on what they see.

If there are an odd number of viziers (say 99), the first vizier will see an even number of total hats (say 98). Even+even = even or odd+odd = even. Thus, the viziers agree on a convention - for example: say blue if it is odd plus odd, and say red if it is even+even. With this parity information, the rest of the viziers can determine their own hat color.

The viziers will agree on all this on the night before the hat day.

More elegant solution (same idea): see Murderotica's post a couple posts down.

But the guru only speaks once and tells something that everybody knows allready.

1. Put the two strings side by side.

2. Fold one string in half, folding to the side.

3. Fold that string in half again, folding over the top.

4. This will cause there to be 2 quarters of string side by side and 2 on top of those 2. Unfold the string from the quarter that is adjacent to the second string, and put the unfolded string perpendicularly over the other string.

5. Light the extended string. When the flame ignites the second string which is on top of the first (perpendicularly), the allotted time has passed.

That solution can be simplified. We do not need to know whether the amount of viziers is even or odd. They can assert that the number of blue hats will be odd, whether it is or not. So, if the first vizier sees an odd number of hats, he will say he has a red hat to keep the number odd. The next vizier will know that the guy behind him said that his hat was red because he sees an odd number of blue hats ahead of him, and from this information he can deduce what hat is on his head after viewing if the number of hats ahead of him is odd or even. This is more elegant but you were of course still correct.

For example, if we had 2 blue (A & B) 2 brown (C & D) 1 green (guru):

Everyone knows:
-at least 1 blue (Guru statement is key to this problem)
-Therefore, we can assume that if there is a person who sees no others with blue eyes, he will leave knowing he has blue eyes.

NIGHT 1: Nothing happens
A & B:
See 1 blue 2 brown 1 green. Know that there are at least 1 blue. No conclusions can be made.

C & D:
See 2 blue 1 brown 1 green. Know that there are at least 1 blue. No conclusion can be made.

Guru::
2 blue 2 brown. No conclusion.

NIGHT 2: A & B leave.
A & B:
See 1 blue 2 brown 1 green. Noone left on the 1st night, therefore they can imply there are >1 persons with blue eyes & therefore they must have blue eyes. Both of them leave.

C & D:
See 2 blue 1 brown 1 green. Nobody leaving the island on the first night provides no useful information to them since they have already seen two individuals with blue eyes.

Guru:
2 brown 2 blue. Nobody leaving the island on the first night provides no useful information to them since they have already seen two individuals with blue eyes.

NIGHT 3:
C & D
1 green 1 brown. C & D can conclude that their eyes were not blue since A & B have left together. This doesn't help since both C & D cannot deduce if there are 2 or 1 brown eyed people remaining. For all they know their own eyes are red and as such the information from the 1st and 2nd nights is useless.

Guru:
The guru cannot get any further useful information either since there is no information pertaining to green eyes available to him.

Given this example we can begin to see that if there are X blue eyed people they will leave the island on night X, but only if they know there is at least one blue eyed person. We can then conclude that all of the blue eyed people will leave on the 100th night.

Yep, good answer!

I have seen this before and I think it is a terrible question. It is not a math/logic type brainteaser.

The solution involves keeping one light on for a while and then turning it off so it remains hot when you go in, so you can feel it (and turn on another one before going in). Kinda lame imo

Don't you also need to specify that the light bulbs generate heat when lit?

Turn the first switch on and wait for a few minutes, then turn it off again. Turn the second switch on and enter the room.
If the light is on you know it's switch 2
If the light is off but the bulb is hot it's switch 1
If the light is off and the bulb is cold it's switch 3

Fill the 5ltr jug, use it to fill the 3ltr jug. You now have 2ltr left in the 5 ltr jug. Empty the 3ltr jug, pour the 2ltr from the 5 ltr jug into it. The 3ltr jug now has place for 1ltr. Fill the 5 ltr jug, and then, with it fill the 3ltr jug until it is full. You now have 4ltrs in the 5 ltr jug.

It is possible. You take one coin from the first, 2 coins from the second, three coins from the third, etc until 10 coins from the 10th. If you weigh the coins you took, you will get a weight of 55gr (55 coins), -0.1gr for each counterfit coin. If you have one counterfit coin, it will weigh 54.9, for 2 it'll be 54.8, etc until 10 will be 54gr. According to the number of counterfit coins, you will know from which bag you took them.

Flip the 7 and 11 minute hourglasses simultaneously. Start cooking once the 7min one runs out. That leaves 4 minutes until the 11 minute hourglass is empty. Once it empties 4 minutes later, flip it over. Once it is done, you have been cooking for exactly 4 + 11 = 15 minutes.

Let's assume that they move at an exact uniform speed (won't skip eachother). In the next 12 hours, they will meet 11 times (once at 1pm and a bit, once at 2pm and a little more, etc, until the meet again exactly at 12am). Therefor, in 12 hours they meet 11 times. Since they move at uniform speeds, they must meet once every 12/11 hours, meaning 1 hour, 5 minutes, 27 seconds, and 3/11ths of a second (if my arithmetic is correct).

It can be found in 2 comparisons like this:
Split into 3 piles of 3, compare 2 of them on the scale. If one is lighter, then in contains the coin. Otherwise, it is in the 3rd pile. Hence, the coin is in the lightest pile. Take 2 coins from that pile, compare them on the scale, if one is lighter, that is the coin you are looking for, otherwise, it is the coin that wasn't on the scale.

You take the first string and light it on both sides so it will burn twice as fast. You light the other string on one side at the same time. Once the string lighted on both sides finishes burning, exactly 30 minutes have passed, so start burning the other string on the other side, so it will burn twice as fast. Since the remaining 30 minutes of burn times have been halved, it will finish burning in another 15 minutes. Once it finishes burning, 45 minutes have passed.

Let n be a prime number over 30. Let m be the remainder of dividing n by 30, and p is the result rounded down. m is therefor a non-prime number under 30.

Let's assume that m is not a prime number. Therefor, n = 30*p + m.
Hence, p = (n-m)/30. If m is even, therefor, n has to be even, since an odd number minus an even number is an odd number, and p is by definition a whole number. If m is divisible by 3, then so is n. If m is divisible by 5, so is n, for the same reason.
Hence, m has to be a multiplication of prime numbers greater than 5. Hence, m is equal to or greater than 7*7 = 49, which is impossible as it is the remainder for division by 30.

Hence, the remainder has to be a prime number.

Sure they can. The rectangle can be filled with 25, so a rectangle with half the height and half the width can be filled with circles with half the radius in the exact same way. Therefor, you can put 25 of each of the smaller discs in each quarter of the rectangle using the exact same strategy to fill it. Feel stupid that I didn't get that instantly :/

The first vizier says his hat is red if he counts an odd number of blues on the others. Therefor, if he is correct, everyone knows there is an odd number of blues and can deduce that if he sees an even number of blues his hat is blue, and otherwise it is red. If he is incorrect, therefor there is an even number of blue hats, and the other viziers can (in similar fashion) deduce the color of their hat.

You should switch. At your first choice, you had a 1 in 3 chance to get it right, and a 2 in 3 that is one of the others. Since he opened one of them, it is a 2 in 3 chance it is behind the other door

Rephrasing the question: Can any natural number be uniquely expressed in base 2? Answer: yes, those are called binary numbers.

The first line can't cross any. The second line can at most cross the first. The third can cross the other two at most. And so on. Hence: the maximum number of crossings is 0+1+2+3+...+9, therefor, 45 possible intersections.

Pour 3L into the 5L, pour another 2L in leaving 1L in the 3L container, dump out the 5L, pour the 1L into the 5L, then another 3L for a total of 4L.

1 coin from first bag, 2 from second, 3 from third... number of tenths of a gram you are off is the bag of counterfeit coins.

1 from first bag, 2 from second, 4 from third, and 8 from fourth. Convert tenth gram difference to binary and those are the fake bags.

Start both, flip the 7m when it runs out after 7m, flip it again in 4 minutes when the 11m runs out, it will then run out 4 minutes later for a total of 15 minutes.

In 12 hours this happens 11 times, so one hour, 5 minutes and 300/11 seconds, 1:05:27.2727...

100 disks of radius 1 can cover 4 same sized rectangles, shrinking everything to .5 scale you have 1 rectangle of the original size covered by 100 disks of radius .5

On the first night if someone saw noone else with blue eyes they would know they had the blue eyes and they could leave. But since that doesn't happen everyone knows that noone sees 'no one with blue eyes'; ie. everyone sees at least 1 person with blue eyes. So on the second night someone who saw only one person with blue eyes would know that they also had blue eyes... So one the 100th night everyone who sees 99 people with blue eyes will know that they have blue eyes, so all 100 of them leave that night.

Weigh 3 against 3, if one weighs less it contains the counterfeit. If it balances then the counterfeit is in the 3 not weighed. Weigh 1 vs 1 of the last 3.

Light three of the ends at the same time. The one with 2 lit ends should burn in 30 minutes. Then light the other end of the remaining string and it should burn out in 15 minutes.

If the remainder was a multiple of 2,3,5 then it wouldn't have been prime before dividing by 30. So the only remainders possible when starting with a prime number are 7, 11, 13, 17, 19, 23, 29 which are all prime.

1+2+3+4+5+6+7+8+9=45

If we group the 100 numbers by their mod 7. The difference between members in a group would be a multiple of 7. By pigeon hole, the smallest the largest group could be is 15 members.

12 coins, 1 of which is either heavy or light
EEEEEEEEEEEE (24 possibilities), measure 4v4
they balance, the 4 others are either heavy or light
___EEEE (8 possibilities), measure 3 versus 3 legit
___ they balance, the one you didn't measure is either heavy or light
______E (2 possibilities), measure it versus 1 legit
___ they don't, you have 3 heavy or 3 light
______hhh or lll (3 possibilities), compare 2
______ they balance, the third is the fake
______ they don't, the one that matches heavy or light is the fake

they don't, one of the 4 is heavy or one of the other four is light
___hhhhllll (8 possibilities), measure hhl versus hhl
___ they balance, one of the 2 lights you didn't measure is fake
______ll (2 possibilities), measure versus each other
___ they don't, one of the 2 heavies was heavy or the opposite light was light
______hhl (3 possibilities), measure the heavies versus each other
______ they balance, the light was light
______ they don't, the heavy one is heavy

we know immediately that the hands will touch again sometime after one. so i set up two equations, y=x which is the movement of the minute hand at one, and y=x/12 + 5 which is the hour hand at one. from there i just found out where the two lines intersect, which happens 5.45 minutes after the clock hits one.

so the answer is 65 minutes 27 seconds.