Riddles

Continuing from the hint, you must be the one to say 20 in order to win. This pattern continues — you must also be the one to say 16, 12, 8 and 4. This means that you should go second; no matter whether your friend starts with 1, 2 or 3, you can jump to 4 on your turn. After that, continue to jump to the next multiple of 4 on each of your turns until you reach 24, forcing your friend to say 25.

Starting from the hint, we know that E will vote with C as long as he gets at least 1 coin. If C was to offer E nothing, E would vote to kill C and get nothing from D instead. So in this scenario, C would get 99 coins, and E would get 1 coin.

Now we proceed from the case where B, C, D and E are alive. In this case, D will vote with B as long as he gets at least 1 coin, as he knows he'll get nothing if C is in charge. B can therefore suggest 99 coins for himself and 1 coin for D, giving him the 50% required to approve the split.

Finally, we can deduce how A should offer to split the coins. C and E both want to avoid B being in charge, and so will vote with A as long as they get at least 1 coin. Therefore, A should suggest a split of 98 coins for himself, and 1 coin each for C and E.

They should proceed as follows:

1. Put the treasure into an empty basket, which falls to the ground.
2. The prince gets into the basket outside the window, and descends. The treasure climbs back up.
3. The queen swaps with the treasure and descends, while the prince comes up again.
4. The queen and prince vacate their baskets, and the king puts the treasure into an empty basket, which falls to the ground.
5. The queen picks up the treasure and gets into the basket; they start to rise as the king comes down in the other basket.
6. The king and queen get out of their baskets, leaving the treasure to drop down again.
7. The prince now gets into the basket outside the tower window and descends, while the treasure goes up.
8. The queen again swaps with the treasure and descends, while the prince rises again.
9. The prince lets the treasure drop, with an empty basket rising.
10. Finally, the prince gets into the empty basket and descends safely, as the treasure goes back up.

All that remains is to dodge the falling treasure.

Add an extra pill to the mix from bottle A, and then split each of the pills in half. Consume one half of each of the pills in the mix, which adds up to two halves of pill A and two halves of pill B, the correct dosage. You can then consume the remaining halves in two hours time, followed by the remaining whole pills two hours later.

You have to think in 3D — form a tetrahedron (otherwise known as a triangular based pyramid), so that the 4 faces are the triangles you require!

You might like to see Wikipedia's page on tetrahedrons.

It's easiest to think about this problem in terms of the amount of time the drone will be flying for — which will be equivalent to the amount of time that it takes the trains to collide. Since each train will collide after travelling 50km (which will take 1 hour), to find the distance the drone has travelled we simply have to multiply the speed of the drone by the time it was travelling; which yields a distance of 75km.

You are shopping for house-numbers.

When you place a book on a library shelf, the 1st page is not on the leftmost side of the book, it is on the rightmost side! This means that when the worm starts from the 1st page of the 1st book, it only has the cover left to eat its way through before it reaches the second book, so 0.5cm. Then, it can eat its way through all the books from 2 to 9 (a further 8 books, so another 32cm) and then finally eat its way through the back cover of the 10th book to reach the final page of the 10th book. This means that the worm has travelled a total distance of 33cm.

Start both sand-timers at the same time. When the 4 minute sand-timer has run out, flip it back over. Once the 7 minute sand-timer runs out, there will be 1 minute left on the 4 minute sand-timer. Flip the 7 minute sand-timer over as soon as it finishes, and then once the 4 minute sand-timer runs out (leaving an amount of sand equivalent to 1 minute in the bottom of the 7 minute sand-timer), flip the 7 minute sand-timer back over to complete the 9 minutes.

The following table details the state of the timers minute-by-minute.

The teacher should tell the students that she is going to give a detention to the first student on the register who wants to skip the test. This means that the first student will sit the test, but because the second person is now at risk of a detention, they will also take the test, and so on.

The person would make the following journey:

1. Cross with the goat and leave it on the opposite bank.
2. Cross back alone.
3. Cross with the wolf and leave it on the opposite bank.
4. Cross back with the goat and leave it on the starting bank.
5. Cross with the cabbage and leave it on the opposite bank with the wolf.
6. Cross back alone.
7. Finally, cross with the goat, and join the wolf and the cabbage on the opposite bank.

Each knight mounted the horse of the other knight; by doing this, if they got the horse that they were riding to the tower first, this would mean that their own horse would have come last and therefore they would win.

The queen divides the nine stones into three piles of three. Then, the queen will place two of the piles of stones on the scales, with one pile of stones on each side. If the scales tip to one side, then the fake stone must be in the pile that is heavier; if they are balanced, the fake stone must be in the pile that was not placed on the scales. Having identified which pile the stone is in, the queen must then take two of the stones from that pile and place them on the scales in a similar fashion — if one side of the scales tips, then the fake stone must be on the side that has made it tip; if they are balanced, the fake stone must be the remaining stone from the pile of three.

The statement you should make is "at least one prisoner has green eyes." The prisoners will then escape on day 100. To understand why this works, consider an island with only two prisoners, Alice and Bob. Alice and Bob look at each other's eyes, and see green. Since both of them saw green, they cannot be sure that their own eyes are green, and so neither of them leave on day 1. By day 2, since neither of them left on day 1, they both know that the other person must have seen green eyes, otherwise they would have left the island. They therefore both declare their eye colour is green on day 2. For three prisoners, this process would take an additonal day, meaning the prisoners would leave on day 3. Extending this pattern to 100 prisoners means that the prisoners leave on day 100.

It is possible to be kinder to the prisoners — by declaring "at least 99 prisoners have green eyes", no additional information is contained in your statement, but the prisoners will leave on day 2 (since no prisoner can be sure about their eye colour on day 1).

To be guaranteed to win, you should choose to go first, and place your coin at the centre of the table. From then on, wherever your opponent places a coin, you should place a coin directly opposite. This means that whenever your opponent is able to place a coin, you will also be able to place a coin, and thus you will take the final spot for the coin on the table.

Only 10 slaves are needed. In order to determine which bottles each slave should drink from, the king first numbers all of the bottles in binary, starting from 0. Then, he numbers the 10 slaves with powers of 2 (i.e. 1, 2, 4, 8, 16, 32, 64, 128, 256, 512), thus allocating each of them a distinct binary digit. Then the king goes through the bottles, and considers the digits in their binary representations. Each slave is forced to drink from every bottle that has a 1 in their number's position, for example the slaves corresponding to 1 and 4 will drink from bottle 5. It will then be possible to deduce which bottle was poisoned from which slaves die in 24 hours time. The following table is in the case of 8 bottles:

So if slaves A and B were to die, bottle 3 would be the poisoned bottle; if slaves A, B, and C were to die, bottle 7 would be the poisoned bottle; and if none of the slaves were to die then bottle 0 would be the poisoned bottle. This approach can be extended to all 1000 bottles (and could be extended up to 1024 bottles if necessary).

If you place 1 white marble in one bowl, and the remaining marbles in the other bowl, this gives you a 74.7% (3 sf) chance of survival. This is because you have a 50% chance of picking the first bowl, but once you've picked this bowl, you've got a 100% chance of survival. For the other bowl, you have a 50% chance of picking it, but once you've picked it, you only have a \(\frac{49}{99}\) chance of surviving. This means that overall, you have a \(\frac{1}{2}+\frac{49}{198} = 0.747\) (3 sf) chance of survival.

Number the balls 1–12. As each weighing has three possible outcomes (the left side is either heavier, lighter than, or equal to the right hand side) I will address each possible outcome separately. To start, weigh 1,2,3,4 against 5,6,7,8. The following cases could then occur:

1. It balances. This means that the odd marble out must be contained within the group 9,10,11,12. Now weigh 6,7,8 (or any other 3 known good marbles) against 9,10,11. Now the following cases could occur:
1. It balances. In this case, 12 must be the odd marble out. Simply weigh 12 against any other known marble to determine whether it is heavy or light.
2. 9,10,11 is the lighter side. As any one of these three must be a light marble, weigh 9 against 10; if they balance, 11 is the light marble, otherwise the lighter of 9 and 10 is the light marble.
3. 9,10,11 is the heavier side. As any one of these three must be a heavy marble, weigh 9 against 10; if they balance, 11 is the heavy marble, otherwise the heavier of 9 and 10 is the heavy marble.
2. 1,2,3,4 is lighter than 5,6,7,8. This means that any one of 1,2,3,4 could be the light marble, or any one of 5,6,7,8 could be the heavy marble. Weigh 1,2,5 against 3,6,12 (or any other known good marble in the place of 12). Now the following cases could occur:
1. It balances. This means that none of 1,2,3,5,6 is the odd marble out, leaving the possibility of 4 being a light marble, or 7 or 8 being a heavy marble. Weigh 7 against 8, if they balance, then 4 is the light marble, otherwise the heavier of 7 and 8 is the heavy marble.
2. 3,6,12 is heavier than 1,2,5. This means that 3 cannot be the odd marble out, as it couldn't be on both the lighter and heavier side of a weighing; leaving either 6 as a heavy marble, or 1 or 2 as a light marble. Now weigh 1 against 2, if they balance then 6 is the heavy marble, otherwise the lighter of 1 and 2 is the light marble.
3. 3,6,12 is lighter than 1,2,5. This means that 1, 2 and 6 cannot be the odd marble out, as they couldn't be on both the lighter and heavier side of a weighing; leaving either 3 as a light marble, or 5 as a heavy marble. Weigh 3 against 12 (or any other known good marble), if they balance then 5 is the heavy marble, otherwise 3 is the light marble.
3. 1,2,3,4 is heavier than 5,6,7,8. Similarly to case 2, this means that any one of 1,2,3,4 could be the heavy marble, or any one of 5,6,7,8 could be the light marble. Weigh 1,2,5 against 3,6,12 (or any other known good marble in the place of 12). Now the following cases could occur:
1. It balances. This means that none of 1,2,3,5,6 is the odd marble out, leaving the possibility of 4 being a heavy marble, or 7 or 8 being a light marble. Weigh 7 against 8, if they balance, then 4 is the heavy marble, otherwise the lighter of 7 and 8 is the light marble.
2. 3,6,12 is heavier than 1,2,5. This means that 1, 2 and 6 cannot be the odd marble out, as they couldn't be on both the lighter and heavier side of a weighing; leaving either 3 as a heavy marble, or 5 as a light marble. Weigh 3 against 12 (or any other known good marble), if they balance then 5 is the light marble, otherwise 3 is the heavy marble.
3. 3,6,12 is lighter than 1,2,5. This means that 3 cannot be the odd marble out, as it couldn't be on both the lighter and heavier side of a weighing; leaving either 6 as a light marble, or 1 or 2 as a heavy marble. Weigh 1 against 2, if they balance then 6 is the light marble, otherwise the heavier of 1 or 2 is the heavy marble.

The prisoner at the back of the queue should count up all of the red hats in front of them, and declare "red" if the total is even, and "blue" if the total is odd. This means that the prisoner in front of him can count all the red hats that they can see, and deduce the colour of their own hat. For example, if the prisoner at the back declares "red," but the prisoner in front of them can only see an odd number of red hats, that prisoner must be wearing a red hat. This process is then continued, ensuring that the 99 remaining prisoners all get the colour of their hat correct, but the prisoner at the back of the queue still only has a 50% chance of declaring the colour of his own hat correctly. This gives a 99.5% survival rate.

The commander should order his troops to start a fire in front of them. As the two fires will move at the same speed due to the wind, the fire in front of them will burn down the trees ahead of the troops at the same rate as the fire behind them. This means that the troops will be able to ride at their normal speed in the empty path left behind by fire in front of them, and thus they will be able to escape unharmed.

The three solutions are given below, in the same order as the hints:

1. The prisoner could ask a guard "which door would the other guard tell me was the way out?" The prisoner would then go through whichever door the guard did not indicate.
2. The prisoner could ask a guard "is the guard who tells the truth standing in front of the door that leads to the way out?" If the answer is yes, the prisoner must go through the door of the guard he asked, if the answer is no, the prisoner must go through the door of the guard he did not ask.
3. The prisoner could ask a guard "what would be your answer if I had asked you earlier in the day which door leads to the way out?" The prisoner must then go through whichever door the guard indicates.

The queen must give each of the twelve boxes of pills a number from left to right. Then, she must take one pill from box one, two pills from box two, three pills from box three, etc., twelve pills from box twelve. The queen must then measure the mass of the 78 pills that she has removed from the boxes. If all the pills of all the boxes were normal, then the total mass of the pills removed would be 78 x 10 = 780g. Knowing however that a poisoned pill weighs 1g less, all the queen has to do is calculate the difference between the total mass that the pills should have and the mass that the pills actually have to tell her the number of the box which has had its pills substituted. For example, if the queen was to measure the mass of the pills that she removed and found that they weighed 777g, she would know that 3 of the pills had a lower mass (and were therefore poisonous), and therefore the box that had had its pills replaced was box 3.