Riddles

Isaac and Albert wanted to take a vacation. They were debating how they could get to their hotel in the fastest manner. Isaac said, "We should go by train." But Albert said, "No, the train reaches the end of the line half way to the hotel -- we would have to walk the rest of the way. We should bike to the hotel instead." Isaac disagreed. So Albert biked the whole way to the hotel, while Isaac took the train for the first half of the journey and walked for the remainder.

The speed of the train turned out to be four times that of the bike's speed. The bike's speed turned out to be two times faster than walking speed. Who got to the hotel first?

Alternate Solution #1

Let s be the total distance of the journey to the hotel. Let v be walking speed. So 2v is the bike's speed, and 8v is the train's speed. Let I be the time it took for Isaac to complete the journey, and A be the time it took for Albert to complete the journey. Since distance equals rate times time, we have two equations, one for I and one for A:

I = (s/2)/8v + (s/2)/v = s/16v + s/2v
A = s/2v Note that I exceeds A by s/16v.
Albert will reach the hotel first.

Alternate Solution #2

The problem may be solved more easily with simple logic. If the bicycle is twice as fast as walking, the time it takes to bike the whole way is equal to the time it takes to walk half the way. So if the train's speed is anything shy of infinite, biking will still be faster.
 
At McDonald's you can order Chicken McNuggets in boxes of 6, 9, and 20. What is the largest number of nuggets that it is not possible to obtain by purchasing some combination of boxes?

You can buy any number of nuggets that is evenly divisible by three except for three just by using combinations of boxes of 6 and boxes of 9. (Use at most one box of 9, then multiples of boxes of 6.) If the number is not divisible by three, use a box of 20. If, after a box of 20 is purchased, the remaining number is divisible by three, you're all set. Otherwise, use a second box of 20. The remaining number will necessarily be divisible by 3, and you're all set.

So the largest number that cannot be purchased would be one that requires two boxes of 20 before the remainder is reduced to a number divisible by three. Since three is the only number evenly divisible by three that cannot be purchased, the largest impossible number is 3 + 20 + 20 = 43.
 
You can buy any number of nuggets that is evenly divisible by three except for three just by using combinations of boxes of 6 and boxes of 9. (Use at most one box of 9, then multiples of boxes of 6.) If the number is not divisible by three, use a box of 20. If, after a box of 20 is purchased, the remaining number is divisible by three, you're all set. Otherwise, use a second box of 20. The remaining number will necessarily be divisible by 3, and you're all set.

So the largest number that cannot be purchased would be one that requires two boxes of 20 before the remainder is reduced to a number divisible by three. Since three is the only number evenly divisible by three that cannot be purchased, the largest impossible number is 3 + 20 + 20 = 43.

Let x equal the number of marbles in a small bag, and y represent the number of marbles in a large bag. We know that the following equation holds:

7x + 18y = 233 Ordinarily, it is not possible to solve for two unknowns with a single equation; however, we also know that x and y are positive integers and that x is less than y. We can narrow the search by solving for the highest and lowest possible values for y:

Assume y = x: 7y + 18y = 233 25y = 233 y = 9.32 Since y must be an integer, y ≥ 10.

Assume y = 1: 7 + 18y = 233 18y = 226 y = 12.555 Since y must be an integer, y ≤ 12.

So the only possible values for y are 10, 11, and 12. Substituting each of these values into the original equation, we can find the corresponding values of x. Respectively, they are 7.57, 5, and 2.42. Since x must be an integer, the correct solution is x = 5 and y = 11. So the small bags of marbles contained five each, and the large bags contained eleven each.
 
A box contains two quarters. One is a double-headed coin, and the other is an ordinary coin, heads on one side, and tails on the other. You draw one of the coins from a box and look at one of the sides. Assuming it is heads, what is the probability that the other side shows heads also?

For starters, the answer is not 1/2. When the coin is drawn, there are four possibilities, each of which is equally likely:

Coin Drawn Side Shown Other Side
Double-Headed Coin Heads Heads
Double-Headed Coin Heads (the other heads) Heads
Ordinary Coin Heads Tails
Ordinary Coin Tails Heads

The problem tells us that the last possibility did not occur. Therefore, there are three remaining possibilities, each of which is equally likely. Of the three, two of the possibilities will show heads on the other side; only one will show tails on the other side. So the probability that the other side of the coin is heads is two thirds.
 
Five pirates raid the ship of a wealthy bureaucrat and steal his trunk of gold pieces. By the time they get the trunk aboard, dusk has fallen, so they agree to split the gold the next morning.

But the pirates are all very greedy. During the night one of the pirates decides to take some of the gold pieces for himself. He sneaks to the trunk and divides the gold pieces into five equal piles, with one gold piece left over. He puts the gold piece in his pile, hides it, puts the other four piles back in the trunk, and sneaks back to bed.

One by one, the remaining pirates do the same. They sneak to the trunk, divide the coins into five piles, with always one coin left over. Each pirate puts the gold coin in his own pile, hides it, and puts the remaining four piles back in the trunk.

What is the smallest number of coins there could have been in the trunk originally?

The original number of coins must be a number such that you can subtract one and multiply by four fifths and get an integer. These numbers are 6, 11, 16, 21, 26, and so on.

But the pile remaining after the first pirate has taken his gold must also have this property. So the possibilities for the original number are 16, 36, 56, 76, 96, and so on.

The pile remaining after the second pirate has taken his gold must also have this property. So the possibilities for the original number are 76, 156, 236, 316, 396, and so on.

The pile remaining after the third pirate has taken his gold must also have this property. So the possibilities for the original number are 316, 636, 956, 1276, 1596, and so on.

The pile remaining after the fourth pirate has taken his gold must also have this property. The smallest possibility for this is 1276.

This number is the number of gold pieces in the chest the fourth pirate left behind (for the fifth pirate to divide). The fourth pirate hid a quarter of this number, plus one extra, just before the fifth pirate got there. So the third pirate left behind 1276 + 1276/4 + 1 = 1596 gold pieces.

The third pirate hid a quarter of this number, plus one extra, just before the fourth pirate got there. So the second pirate left behind 1596 + 1596/4 + 1 = 1996 gold pieces.

The second pirate hid a quarter of this number, plus one extra, just before the third pirate got there. So the first pirate left behind 1996 + 1996/4 + 1 = 2496 gold pieces.

The first pirate hid a quarter of this number, plus one extra. So the original number of coins must have been 2496 + 2496/4 + 1 = 3121 gold pieces.
 
Reason why 30414093201713378043612608166064768844377641568960512078291027000cannot possibly be the value of 50 factorial, without actually performing the calculation.

50 factorial includes, as factors, 10, 20, 30, 40, and 50. Therefore, the value of 50 factorial must end in at least five zeroes. The number given only ends in three zeroes. The correct value of 50 factorial is close to this, however. It's 30414093201713378043612608166064768844377641568960512000000000000.
 
Two boys sell apples. Each sells thirty apples a day. The first boy sells his apples at two for fifty cents (and therefore earns $7.50 per day). The second boy sells his apples at three for fifty cents (and therefore earns $5.00 per day). The total received by both boys each day is therefore $12.50.

One day, the first boy is sick, and the second boy takes over his apple selling duties. To accommodate the differing rates, the boy sells the sixty apples at five for a dollar. But selling sixty apples at five for a dollar yields only $12.00 earnings at the end of the day. What happened to the other fifty cents?

After ten sales of five apples, all the three-for-fifty-cents apples are sold; the remainder is still sold at five for a dollar when they should be sold at two for fifty cents.
 
A drove of sheep and chickens have a total of 99 heads and feet. There are twice as many chickens as sheep. How many of each are there?

Let S be the number of sheep and C be the number of chickens. So:

2S = C
5S + 3C = 99
We can rephrase the first equation thusly:

6S - 3C = 0
And then we can add this to the second equation, which yields:

11S = 99
By solving for S, we find that S equals 9. By substituting back in one of the original equations, we find that C equals 18. So there are nine sheep and eighteen chickens
 
A farmer went to his banker to borrow money to buy livestock. The banker made a deal with the farmer. He told the farmer that he would give him $100.00, and that the farmer would not have to pay back the money if he could buy exactly 100 head of livestock for exactly $100.00.

The only rule was that at least one of each kind of livestock -- cows, pigs, and sheep -- had to be purchased. Each cow cost $10. Each pig cost $3. And each sheep cost $0.50. How many of each kind of livestock did the farmer buy?

Let c be the number of cows purchased, p be the number of pigs purchased, and s be the number of sheep purchased. We know that the total number of animals purchased is 100. Moreover, we know that the total cost of each one adds up to $100. This gives us two equations:

c + p + s = 100
10c + 3p + 0.5s = 100
If we multiply the second equation by two, that gives us:

20c + 6p + s = 200
If we take this equation and subtract the first equation, we eliminate s entirely and wind up with:

19c + 5p = 100
We can rewrite this equation like so:

100 - 19c = 5p Since c and p are constrained to positive integers, we can narrow down the possible solutions pretty quickly. We know that c cannot be 6 or higher, because then p would have to be negative for the equation to hold. So we only need to consider 1, 2, 3, 4, and 5 as possible values for c. Substituting 1, 2, 3, or 4 for c, however, results in the left-hand side of the equation being unevenly divisible by 5, which means p would have to be a non-integer for the equation to hold. So c must be 5. Substituting c allows us to solve for p:

100 - 19c = 5p
100 - 95 = 5p
5 = 5p
p = 1
Now we can substitute the values for c and p in one of the two original equations (the first is easiest) and solve for s:

c + p + s = 100
5 + 1 + s = 100
6 + s = 100 s = 100 - 6 s = 94
The farmer purchased five cows ($50), one pig ($3), and 94 sheep ($47).
 
Isaac and Albert were excitedly describing the result of the Third Annual International Science Fair Extravaganza in Sweden. There were three contestants, Louis, Rene, and Johannes. Isaac reported that Louis won the fair, while Rene came in second. Albert, on the other hand, reported that Johannes won the fair, while Louis came in second.

In fact, neither Isaac nor Albert had given a correct report of the results of the science fair. Each of them had given one correct statement and one false statement. What was the actual placing of the three contestants?

solution: Johannes won; Rene came in second; Louis came in third.
 
At a family reunion were the following people: one grandfather, one grandmother, two fathers, two mothers, four children, three grandchildren, one brother, two sisters, two sons, two daughters, one father-in-law, one mother-in-law, and one daughter-in-law. But not as many people attended as it sounds. How many were there, and who were they?

There were two little girls and a boy, their parents, and their father's parents, totaling seven people.
 
97 baseball teams participate in an annual state tournament. The champion is chosen for this tournament by the usual elimination scheme. That is, the 97 teams are divided into pairs, and the two teams of each pair play against each other. The loser of each pair is eliminated, and the remaining teams are paired up again, etc. How many games must be played to determine a champion?

96. All teams but the champion team will lose a game exactly once.
 
Your sock drawer contains ten pairs of white socks and ten pairs of black socks. If you're only allowed to take one sock from the drawer at a time and you can't see what color sock you're taking until you've taken it, how many socks do you have to take before you're guaranteed to have at least one matching pair?


Three. In the worst case, the first two socks you take out will consist of one black sock and one white sock. The next sock you take out is guaranteed to match one or the other.
 
A man is the owner of a winery who recently passed away. In his will, he left 21 barrels (seven of which are filled with wine, seven of which are half full, and seven of which are empty) to his three sons. However, the wine and barrels must be split so that each son has the same number of full barrels, the same number of half-full barrels, and the same number of empty barrels. Note that there are no measuring devices handy. How can the barrels and wine be evenly divided?

Two half-full barrels are dumped into one of the empty barrels. Two more half-full barrels are dumped into another one of the empty barrels. This results in nine full barrels, three half-full barrels, and nine empty barrels. Each son gets three full barrels, one half-full barrel, and three empty barrels.
 
A mountain goat attempts to scale a cliff sixty feet high. Every minute, the goat bounds upward three feet but slips back two. How long does it take for the goat to reach the top?

Fifty eight minutes. Although his net progress each minute is one foot, he reaches the top on the fifty-eighth minute just before he would normally slip back two feet.
 
You have three boxes of fruit. One contains just apples, one contains just oranges, and one contains a mixture of both. Each box is labeled -- one says "apples," one says "oranges," and one says "apples and oranges." However, it is known that none of the boxes are labeled correctly. How can you label the boxes correctly if you are only allowed to take and look at just one piece of fruit from just one of the boxes?

Take a piece of fruit from the box marked "apples and oranges." Suppose the fruit you take is an apple. Then that box must be the box containing just apples. Therefore, the box marked "oranges" can't be the box containing just apples, and it can't be the box containing just oranges either -- so it must be the box containing apples and oranges. The remaining box is therefore the box containing just oranges.

If the fruit you take out is an orange, the solution is derived in a similar fashion: the box marked "apples and oranges" is the box containing just oranges; the box marked "apples" is the box containing both apples and oranges; and the box marked "oranges" is the one containing just apples
 
An Arab sheikh is old and must will his fortune to one of his two sons. He makes a proposition. His two sons will ride their camels in a race, and whichever camel crosses the finish line last will win the fortune for its owner. During the race, the two brothers wander aimlessly for days, neither willing to cross the finish line. In desperation, they ask a wise man for advice. He tells them something; then the brothers leap onto the camels and charge toward the finish line. What did the wise man say?

The rules of the race were that the owner of the camel that crosses the finish line last wins the fortune. The wise man simply told them to switch camels.
 
You have a jug that holds five gallons, and a jug that holds three gallons. You have no other containers, and there are no markings on the jugs. You need to obtain exactly seven gallons of water from a faucet. How can you do it?

Second Problem: You need exactly four gallons. How do you do it?

To get seven gallons, fill the five gallon jug and dump what you can into the three gallon jug, filling it. There are now two gallons in the five gallon jug. Dump out the three gallon jug, and put the two gallons from the five gallon jug into the three gallon jug. Then fill the five gallon jug. The total is seven gallons.

To get four gallons, fill the three gallon jug, and dump it into the five gallon jug. Fill the three gallon jug again, and dump what you can into the five gallon jug. Now there is one gallon in the three gallon jug. Dump out the five gallon jug, and put the one gallon from the three gallon jug into the five gallon jug. Now fill the three gallon jug. The total is four gallons.
 
You are on a game show. You are shown three closed doors. A prize is hidden behind one, and the game show host knows where it is. You are asked to select a door. You do. Before you open it, the host opens one of the other doors, showing that it is empty, then asks you if you'd like to change your guess. Should you, should you not, or doesn't it matter?

Remember that the host knows where the prize is. When you pick a door, there's a 66.7% chance you're wrong. If you're wrong, the host will always open the one door left that doesn't contain the prize. So if you were wrong (66.7% chance), you're better off switching to the door that the host leaves closed.

What you are really being asked here is whether the odds are better with your first choice, or with both of the other two. All the host really does is prove that it isn't in one of the remaining two -- but we knew that anyway, because there's only one prize.

Another way to look at it is this: consider that there are 1000 doors. You pick one. Before you open it, the host opens 998 of the 999 remaining doors. There's a 0.1% chance you guessed correctly -- if you guessed wrong (99.9% chance), then you know the prize is in the one door the host left shut. So you're better off changing your guess.

Still not convinced? This is Brain Food's most contested answer, but it is correct! Let's break it down a little more explicitly:

There are nine possible scenarios:

Prize is behind door A. You pick door A.
Prize is behind door A. You pick door B.
Prize is behind door A. You pick door C.
Prize is behind door B. You pick door A.
Prize is behind door B. You pick door B.
Prize is behind door B. You pick door C.
Prize is behind door C. You pick door A.
Prize is behind door C. You pick door B.
Prize is behind door C. You pick door C.
Now the host, who knows which door the prize is behind, deliberately chooses a door that the prize is not behind. For the three cases where you chose correctly in the beginning, the host has the choice of which other door to open. It doesn't matter which one he picks. But for the six cases where you chose wrong, he deliberately chooses the wrong guess of the remaining two choices, revealing it to be a wrong guess. It all breaks down as follows:

Prize is behind door A. You pick door A. Host reveals door B or C as empty.
Prize is behind door A. You pick door B. Host reveals door C as empty.
Prize is behind door A. You pick door C. Host reveals door B as empty.
Prize is behind door B. You pick door A. Host reveals door C as empty.
Prize is behind door B. You pick door B. Host reveals door A or C as empty.
Prize is behind door B. You pick door C. Host reveals door A as empty.
Prize is behind door C. You pick door A. Host reveals door B as empty.
Prize is behind door C. You pick door B. Host reveals door A as empty.
Prize is behind door C. You pick door C. Host reveals door A or B as empty.
(We've had several complaints that the above listing is incorrect because scenarios 1, 5, and 9 should be broken out. In fact, it is incorrect to break them out, because what's important here is that each listed scenario have an equal probability of occurring, not that each scenario be broken down to irreducible terms.)

Now we come to the point where we must decide whether to stick to the original guess or to switch:

Prize is behind door A. You pick door A. Host reveals door B or C as empty. Switching loses.
Prize is behind door A. You pick door B. Host reveals door C as empty. Switching wins.
Prize is behind door A. You pick door C. Host reveals door B as empty. Switching wins.
Prize is behind door B. You pick door A. Host reveals door C as empty. Switching wins.
Prize is behind door B. You pick door B. Host reveals door A or C as empty. Switching loses.
Prize is behind door B. You pick door C. Host reveals door A as empty. Switching wins.
Prize is behind door C. You pick door A. Host reveals door B as empty. Switching wins.
Prize is behind door C. You pick door B. Host reveals door A as empty. Switching wins.
Prize is behind door C. You pick door C. Host reveals door A or B as empty. Switching loses.
Final conclusion: switching wins six out of nine times, which is equal to two thirds of the time, or about 66.7%.
 
Travelling to a city, an old man lost his way. He came to a fork in the road and did not know which road to take. Standing at the fork were two men. Next to the men was a sign, which you may assume is correct, which stated that one of the two men always told the truth and one of the men always told lies (but it was not known which was which). The sign went on to say that travellers could only ask one of the men one question.

What question could the old man pose that would give him the information he needs to choose the correct route?

Alternate Solution #1

Ask one of the men what the other man would answer to the question, "Is the left road the correct road?" Then assume the answer you are given is false and act on that knowledge.

If the man you ask is the liar, he'll incorrectly give you the truthful man's answer. If the man you ask is the truthful man, he'll correctly give you the liar's wrong answer.

Alternate Solution #2

Ask one of the men, "If I had ask you which path was the correct path ten minutes ago, what would you have said?" Regardless of which man is asked this question, the answer will be the correct path.
 
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