http://einstein.et.tudelft.nl/~arlet/puzzles/logic.html
1)Three people check into a hotel. They pay $30 to the manager and go to their room. The manager finds out that the room rate is $25 and gives $5 to the bellboy to return. On the way to the room the bellboy reasons that $5 would be difficult to share among three people so he pockets $2 and gives $1 to each person. Now each person paid $10 and got back $1. So they paid $9 each, totalling $27. The bellboy has $2, totalling $29. Where is the remaining dollar?
ans)
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Each person paid $9, totalling $27. The manager has $25 and the bellboy $2. The bellboy's $2 should be added to the manager's $25 or subtracted from the tenants' $27, not added to the tenants' $27.
ages
Ten years from now Tim will be twice as old as Jane was when Mary was nine times as old as Tim.
Eight years ago, Mary was half as old as Jane will be when Jane is one year older than Tim will be at the time when Mary will be five times as old as Tim will be two years from now.
When Tim was one year old, Mary was three years older than Tim will be when Jane is three times as old as Mary was six years before the time when Jane was half as old as Tim will be when Mary will be ten years older than Mary was when Jane was one-third as old as Tim will be when Mary will be three times as old as she was when Jane was born.
How old are they now?
The solution:
Tim is 3, Jane is 8, and Mary is 15. A little grumbling is in order here, as clue number 1 leads to the situation a year and a half ago, when Tim was 1 1/2, Jane was 6 1/2, and Mary was 13 1/2.
This sort of problem is easy if you write down a set of equations. Let t be the year that Tim was born, j be the year that Jane was born, m be the year that Mary was born, and y be the current year. As indefinite years come up, let y1, y2, ... be the indefinite years. Then the equations are
y + 10 - t = 2 (y1 - j) y1 - m = 9 (y1 - t)
y - 8 - m = 1/2 (y2 - j) y2 - j = 1 + y3 - t y3 - m = 5 (y + 2 - t)
t + 1 - m = 3 + y4 - t y4 - j = 3 (y5 - 6 - m) y5 - j = 1/2 (y6 - t) y6 - m = 10 + y7 - m y7 - j = 1/3 (y8 - t) y8 - m = 3 (j - m)
t = y - 3 j = y - 8 m = y - 15
attribute
All the items in the first list share a particular attribute. The second list is of some items lacking the attribute.
with: battery, key, yeast, bookmark w/out: stapler, match, Rubik's cube, pill bottle
with: Rubik's cube, chess set, electrical wiring, compass needle w/out: clock, rope, tic-tac-toe, pencil sharpener
with: koosh, small intestine, Yorkshire Terrier, Christmas Tree w/out: toothbrush, oak chair, soccer ball, icicle
Points to realize: 1. There may be exceptions to any item on the list, for instance a particular clock may share the properties of the 'with' list of problem two, BUT MOST ORDINARY clocks do not. All the properties apply the vast majority of the the items mentioned. Extraordinary exceptions should be ignored. 2. Pay the most attention to the 'with' list. The 'without' list is only present to eliminate various 'stupid' answers.
The attribute puzzle format is a traditional format in math education. It occurs in logic materials developed in the sixties by EDC in Boston, with visual objects. Example:
these are gloops: A B C D E
these are not gloops: F G J L N
which of these are gloops? O P Q R S
with: battery, key, yeast, bookmark w/out: stapler, match, Rubik's cube, pill bottle
Needs to be placed inside something else when used for its usual purpose.
with: Rubik's cube, chess set, electrical wiring, compass needle w/out: clock, rope, tic-tac-toe, pencil sharpener
Uses color to distinguish between otherwise identical parts.
with: koosh, small intestine, Yorkshire Terrier, Christmas Tree w/out: toothbrush, oak chair, soccer ball, icicle
Villiform.
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bookworm
A bookworm eats from the first page of an encyclopedia to the last page. The bookworm eats in a straight line. The encyclopedia consists of ten 1000-page volumes and is sitting on a bookshelf in the usual order. Not counting covers, title pages, etc., how many pages does the bookworm eat through?
On a book shelf the first page of the first volume is on the "inside"
__ __
B| | | |F
A|1 |...........................|10|R
C| | | |O
K| | | |N
| | | |T
----------------------------------
so the bookworm eats only through the cover of the first volume, then 8 times 1000 pages of Volumes 2 - 9, then through the cover to the 1st page of Vol 10. He eats 8,000 pages.
If the bookworm ate the first page and the last page, it ate 8,004 pages.
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boxes
Which Box Contains the Gold? Two boxes are labeled "A" and "B". A sign on box A says "The sign on box B is true and the gold is in box A". A sign on box B says "The sign on box A is false and the gold is in box A". Assuming there is gold in one of the boxes, which box contains the gold?
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The problem cannot be solved with the information given.
The sign on box A says "The sign on box B is true and the gold is in box A". The sign on box B says "The sign on box A is false and the gold is in box A". The following argument can be made: If the statement on box A is true, then the statement on box B is true, since that is what the statement on box A says. But the statement on box B states that the statement on box A is false, which contradicts the original assumption. Therefore, the statement on box A must be false. This implies that either the statement on box B is false or that the gold is in box B. If the statement on box B is false, then either the statement on box A is true (which it cannot be) or the gold is in box B. Either way, the gold is in box B.
However, there is a hidden assumption in this argument: namely, that each statement must be either true or false. This assumption leads to paradoxes, for example, consider the statement: "This statement is false." If it is true, it is false; if it is false, it is true. The only way out of the paradox is to deny that the statement is either true or false and label it meaningless instead. Both of the statements on the boxes are therefore meaningless and nothing can be concluded from them.
In general, statements about the truth of other statements lead to contradictions. Tarski invented metalanguages to avoid this problem. To avoid paradox, a statement about the truth of a statement in a language must be made in the metalanguage of the language.
Common sense dictates that this problem cannot be solved with the information given. After all, how can we deduce which box contains the gold simply by reading statements written on the outside of the box? Suppose we deduce that the gold is in box B by whatever line of reasoning we choose. What is to stop us from simply putting the gold in box A, regardless of what we deduced? (cf. Smullyan, "What Is the Name of This Book?", Prentice-Hall, 1978, #70)
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camel
An Arab sheikh tells his two sons to race their camels to a distant city to see who will inherit his fortune. The one whose camel is slower will win. The brothers, after wandering aimlessly for days, ask a wise man for advise. After hearing the advice they jump on the camels and race as fast as they can to the city. What does the wise man say? Solution
The wiseman tells them to switch camels.
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centrifuge
You are a biochemist, working with a 12-slot centrifuge. This is a gadget that has 12 equally spaced slots around a central axis, in which you can place chemical samples you want centrifuged. When the machine is turned on, the samples whirl around the central axis and do their thing. To ensure that the samples are evenly mixed, they must be distributed in the 12 slots such that the centrifuge is balanced evenly. For example, if you wanted to mix 4 samples, you could place them in slots 12, 3, 6 and 9 (assuming the slots are numbered from 1 to 12 like a clock). Problem: Can you use the centrifuge to mix 5 samples?
Solution
The superposition of any two solutions is yet another solution, so given that the factors > 1 of 12 (2, 3, 4, 6, 12) are all solutions, the only thing to check about, for example, the proposed solution 2+3 is that not all ways of combining 2 & 3 would have centrifuge tubes from one subsolution occupying the slot for one of the tubes in another solution. For the case 2+3, there is no problem: Place 3 tubes, one in every 4th position, then place the 4th and 5th diametrically opposed (each will end up in a slot adjacent to one of the first 3 tubes). The obvious generalization is, what are the numbers of tubes that cannot be balanced? Observing that there are solutions for 2,3,4,5,6 tubes and that if X has a solution, 12-X has also one (obtained by swapping tubes and holes), it is obvious that 1 and 11 are the only cases without solutions.
Here is how this problem is often solved in practice: A dummy tube is added to produce a total number of tubes that is easy to balance. For example, if you had to centrifuge just one sample, you'd add a second tube opposite it for balance.
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chain
What is the least number of links you can cut in a chain of 21 links to be able to give someone all possible number of links up to 21?
Solution
Two.
OOO C OOOOO C OOOOOOOOOOO
(where Os are chained unbroken links, and the Cs are the unchained broken links)
And equivalently:
OOO C OOOOOO C OOOOOOOOOO
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children
A man walks into a bar, orders a drink, and starts chatting with the bartender. After a while, he learns that the bartender has three children. "How old are your children?" he asks. "Well," replies the bartender, "the product of their ages is 72." The man thinks for a moment and then says, "that's not enough information." "All right," continues the bartender, "if you go outside and look at the building number posted over the door to the bar, you'll see the sum of the ages." The man steps outside, and after a few moments he reenters and declares, "Still not enough!" The bartender smiles and says, "My youngest just loves strawberry ice cream." How old are the children? A variant of the problem is for the sum of the ages to be 13 and the product of the ages to be the number posted over the door. In this case, it is the oldest that loves ice cream. Then how old are they?
Solution
First, determine all the ways that three ages can multiply together to get 72: (quite a feat for the bartender)
72 1 1
36 2 1
24 3 1
18 4 1
18 2 2
12 6 1
12 3 2
9 4 2
9 8 1
8 3 3
6 6 2
6 4 3
As the man says, that's not enough information; there are many possibilities. So the bartender tells him where to find the sum of the ages--the man now knows the sum even though we don't. Yet he still insists that there isn't enough info. This must mean that there are two permutations with the same sum; otherwise the man could have easily deduced the ages.
The only pair of permutations with the same sum are 8 3 3 and 6 6 2, which both add up to 14 (the bar's address). Now the bartender mentions his "youngest"--telling us that there is one child who is younger than the other two. This is impossible with 8 3 3--there are two 3 year olds. Therefore the ages of the children are 6, 6, and 2.
Pedants have objected that the problem is insoluble because there could be a youngest between two three year olds (even twins are not born exactly at the same time). However, the word "age" is frequently used to denote the number of years since birth. For example, I am the same age as my wife, even though technically she is a few months older than I am. And using the word "youngest" to mean "of lesser age" is also in keeping with common parlance. So I think the solution is fine as stated.
In the sum-13 variant, the possibilities are:
11 1 1
10 2 1
9 3 1
9 2 2
8 4 1
8 3 2
7 5 1
7 4 2
7 3 3
6 6 1
6 5 2
6 4 3
The two that remain are 9 2 2 and 6 6 1 (both products equal 36). The final bit of info (oldest child) indicates that there is only one child with the highest age. This cancels out the 6 6 1 combination, leaving the childern with ages of 9, 2, and 2.
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condoms
How can a man have mutually safe sex with three women with only two condoms? How can three men have mutually safe sex with one woman with two condoms?
Solution
Use both condoms on the first woman. Take off the outer condom (turning it inside-out in the process) and set it aside. Use the inner condom alone on the second woman. Put the outer condom back on. Use it on the third woman.
First man uses both condoms. Take off the outer condom (do NOT reverse it) and have second man use it. First man takes off the inner condom (turning it inside-out). Third man puts on this condom, followed by second man's condom.
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dell
How can I solve logic puzzles (e.g., as published by Dell) automatically?
Solution
#include
#define EITHER if (S[1] = S[0], ! setjmp((S++)->jb)) {
#define OR } else EITHER
#define REJECT longjmp((--S)->jb, 1)
#define END_EITHER } else REJECT;
/* values in tmat: */
#define T_UNK 0
#define T_YES 1
#define T_NO 2
#define Val(t1,t2) (S->tmat[t1][t2])
#define CLASS(x) \
(((x) / NUM_ITEM) * NUM_ITEM)
#define EVERY_TOKEN(x) \
(x = 0; x < TOT_TOKEN; x++)
#define EVERY_ITEM(x, class) \
(x = CLASS(class); x < CLASS(class) + NUM_ITEM; x++)
#define BEGIN \
struct state { \
char tmat[TOT_TOKEN][TOT_TOKEN]; \
jmp_buf jb; \
} States[100], *S = States; \
\
main() \
{ \
int token; \
\
for EVERY_TOKEN(token) \
yes(token, token); \
EITHER
/* Here is the problem-specific data */
#define NUM_ITEM 5
#define NUM_CLASS 6
#define TOT_TOKEN (NUM_ITEM * NUM_CLASS)
#define HOUSE_0 0
#define HOUSE_1 1
#define HOUSE_2 2
#define HOUSE_3 3
#define HOUSE_4 4
#define ENGLISH 5
#define SPANISH 6
#define NORWEG 7
#define UKRAIN 8
#define JAPAN 9
#define GREEN 10
#define RED 11
#define IVORY 12
#define YELLOW 13
#define BLUE 14
#define COFFEE 15
#define TEA 16
#define MILK 17
#define OJUICE 18
#define WATER 19
#define DOG 20
#define SNAIL 21
#define FOX 22
#define HORSE 23
#define ZEBRA 24
#define OGOLD 25
#define PLAYER 26
#define CHESTER 27
#define LSTRIKE 28
#define PARLIA 29
char *names[] = {
"HOUSE_0", "HOUSE_1", "HOUSE_2", "HOUSE_3", "HOUSE_4",
"ENGLISH", "SPANISH", "NORWEG", "UKRAIN", "JAPAN",
"GREEN", "RED", "IVORY", "YELLOW", "BLUE",
"COFFEE", "TEA", "MILK", "OJUICE", "WATER",
"DOG", "SNAIL", "FOX", "HORSE", "ZEBRA",
"OGOLD", "PLAYER", "CHESTER", "LSTRIKE", "PARLIA",
};
BEGIN
yes(ENGLISH, RED); /* Clue 1 */
yes(SPANISH, DOG); /* Clue 2 */
yes(COFFEE, GREEN); /* Clue 3 */
yes(UKRAIN, TEA); /* Clue 4 */
EITHER /* Clue 5 */
yes(IVORY, HOUSE_0);
yes(GREEN, HOUSE_1);
OR
yes(IVORY, HOUSE_1);
yes(GREEN, HOUSE_2);
OR
yes(IVORY, HOUSE_2);
yes(GREEN, HOUSE_3);
OR
yes(IVORY, HOUSE_3);
yes(GREEN, HOUSE_4);
END_EITHER
yes(OGOLD, SNAIL); /* Clue 6 */
yes(PLAYER, YELLOW); /* Clue 7 */
yes(MILK, HOUSE_2); /* Clue 8 */
yes(NORWEG, HOUSE_0); /* Clue 9 */
EITHER /* Clue 10 */
yes(CHESTER, HOUSE_0);
yes(FOX, HOUSE_1);
OR
yes(CHESTER, HOUSE_4);
yes(FOX, HOUSE_3);
OR
yes(CHESTER, HOUSE_1);
EITHER yes(FOX, HOUSE_0);
OR yes(FOX, HOUSE_2);
END_EITHER
OR
yes(CHESTER, HOUSE_2);
EITHER yes(FOX, HOUSE_1);
OR yes(FOX, HOUSE_3);
END_EITHER
OR
yes(CHESTER, HOUSE_3);
EITHER yes(FOX, HOUSE_2);
OR yes(FOX, HOUSE_4);
END_EITHER
END_EITHER
EITHER /* Clue 11 */
yes(PLAYER, HOUSE_0);
yes(HORSE, HOUSE_1);
OR
yes(PLAYER, HOUSE_4);
yes(HORSE, HOUSE_3);
OR
yes(PLAYER, HOUSE_1);
EITHER yes(HORSE, HOUSE_0);
OR yes(HORSE, HOUSE_2);
END_EITHER
OR
yes(PLAYER, HOUSE_2);
EITHER yes(HORSE, HOUSE_1);
OR yes(HORSE, HOUSE_3);
END_EITHER
OR
yes(PLAYER, HOUSE_3);
EITHER yes(HORSE, HOUSE_2);
OR yes(HORSE, HOUSE_4);
END_EITHER
END_EITHER
yes(LSTRIKE, OJUICE); /* Clue 12 */
yes(JAPAN, PARLIA); /* Clue 13 */
EITHER /* Clue 14 */
yes(NORWEG, HOUSE_0);
yes(BLUE, HOUSE_1);
OR
yes(NORWEG, HOUSE_4);
yes(BLUE, HOUSE_3);
OR
yes(NORWEG, HOUSE_1);
EITHER yes(BLUE, HOUSE_0);
OR yes(BLUE, HOUSE_2);
END_EITHER
OR
yes(NORWEG, HOUSE_2);
EITHER yes(BLUE, HOUSE_1);
OR yes(BLUE, HOUSE_3);
END_EITHER
OR
yes(NORWEG, HOUSE_3);
EITHER yes(BLUE, HOUSE_2);
OR yes(BLUE, HOUSE_4);
END_EITHER
END_EITHER
/* End of problem-specific data */
solveit();
OR
printf("All solutions found\n");
exit(0);
END_EITHER
}
no(a1, a2)
{
int non1, non2, token;
if (Val(a1, a2) == T_YES)
REJECT;
else if (Val(a1, a2) == T_UNK) {
Val(a1, a2) = T_NO;
no(a2, a1);
non1 = non2 = -1;
for EVERY_ITEM(token, a1)
if (Val(token, a2) != T_NO)
if (non1 == -1)
non1 = token;
else
break;
if (non1 == -1)
REJECT;
else if (token == CLASS(a1) + NUM_ITEM)
yes(non1, a2);
for EVERY_TOKEN(token)
if (Val(token, a1) == T_YES)
no(a2, token);
}
}
yes(a1, a2)
{
int token;
if (Val(a1, a2) == T_NO)
REJECT;
else if (Val(a1, a2) == T_UNK) {
Val(a1, a2) = T_YES;
yes(a2, a1);
for EVERY_ITEM(token, a1)
if (token != a1)
no(token, a2);
for EVERY_TOKEN(token)
if (Val(token, a1) == T_YES)
yes(a2, token);
else if (Val(token, a1) == T_NO)
no(a2, token);
}
}
solveit()
{
int token, tok2;
for EVERY_TOKEN(token)
for (tok2 = token; tok2 < TOT_TOKEN; tok2++)
if (Val(token, tok2) == T_UNK) {
EITHER
yes(token, tok2);
OR
no(token, tok2);
END_EITHER;
return solveit();
}
printf("Solution:\n");
for EVERY_ITEM(token, 0) {
for (tok2 = NUM_ITEM; tok2 < TOT_TOKEN; tok2++)
if (Val(token, tok2) == T_YES)
printf("\t%s %s\n",names[token],names[tok2]);
printf("\n");
}
REJECT;
}
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elimination
97 baseball teams participate in an annual state tournament. The way the champion is chosen for this tournament is by the same old elimination schedule. That is, the 97 teams are to be divided into pairs, and the two teams of each pair play against each other. After a team is eliminated from each pair, the winners would be again divided into pairs, etc. How many games must be played to determine a champion?
Solution
In order to determine a winner all but one team must lose. Therefore there must be at least 96 games.
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flip
How can a toss be called over the phone (without requiring trust)?
Solution
A flips a coin. If the result is heads, A multiplies 2 prime numbers containing about 90 digits; if the result is tails, A multiplies 3 prime numbers containing about 60 digits. A tells B the result of the multiplication. B now calls either heads or tails and tells A. A then supplies B with the original numbers to verify the flip.
Consider what is involved in multiplying 90 digit numbers. Using the method of long multiplication that we all learned in grade school, you have maybe 90 or so strings of 90 characters (or "digits") each. That's no problem for a computer to deal with. The magnitude of the number represented isn't much of a factor; we're only manipulating the string of digits.
Consider what is involved in factoring 90 digit numbers. There are of course, little tricks for determining factorability by small integers which we all learned in grade school. (Is the last digit even? Is the sum of all the digits divisible by 9? And so on.) But these are of little use in factoring large numbers with large factors. In fact, there is no essentially better method than checking every number smaller that the number to be factored and seeing if it one divides the other evenly. We means we could be checking some 100,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000 nummbers. This is very hard to do, even for a supercomputer. Here, the number of digits this number has is of little consequence. It is the magnitude of the number that we have to worry about, and there just isn't enough time in the world to do this properly.
Where does A find a list of 60- and 90- digit prime numbers? Well, that's not to hard to come by. The simplest method to find a few prime numbers is simply to choose a 90-digit number (or 60-digit number, as the case warrants) randomly, and check to see if it is prime. You won't have to wait too long before you stumble across a prime number.
"But wait!" you cry. "I thought you just said it was incredibly difficult to factor large numbers. If that's the case, then how are you going to know if the number you randomly choose is prime?" A good question. Here we enter into the strange an wacky world of number theory. It turns out (and I won't explain how unless someone asks) there are "probabalistic" algorithms, which depend on us choosing numbers at random. There are probablistic algorithms that when given a prime number, will quickly tell us that it is a prime number, and when given a composite number, will either tell us that it is a composite number (with very, very high probability) or will tell us that it is a prime number (with very, very low probability.) What's the use of an algorithm that only returns the right answer "with very, very high probability?" Well, we can make this probability as high as we want, just by running the algorithm longer. In fact, it is within our technological abilities to quickly run this algorithm for 90-digit numbers so that the probability of it giving a wrong answer is less than the probability of a cosmic ray striking a vital part of the computer at some vital time and causing the computer to spit out the wrong answer anyway. That's what I mean by "very, very high."
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flowers
How many flowers do I have if all of them are roses except two, all of them are tulips except two, and all of them are daisies except two? Solution
friends
Any group of 6 or more contains either 3 mutual friends or 3 mutual strangers. Prove it. Solution
hofstadter
In first-order logic, find a predicate P(x) which means "x is a power of 10." Solution
hundred
A sheet of paper has statements numbered from 1 to 100. Statement n says "exactly n of the statements on this sheet are false." Which statements are true and which are false? What if we replace "exactly" by "at least"? Solution
inverter
Can a digital logic circuit with two inverters invert N independent inputs? The circuit may contain any number of AND or OR gates. Solution
josephine
The recent expedition to the lost city of Atlantis discovered scrolls attributted to the great poet, scholar, philosopher Josephine. They number eight in all, and here is the first.
The kingdom of Mamajorca, was ruled by queen Henrietta I. In Mamajorca women have to pass an extensive logic exam before they are allowed to get married. Queens do not have to take this exam. All the women in Mamajorca are loyal to their queen and do whatever she tells them to. The queens of Mamajorca are truthful. All shots fired in Mamajorca can be heard in every house. All above facts are known to be common knowledge. Henrietta was worried about the infidelity of the married men in Mamajorca. She summoned all the wives to the town square, and made the following announcement. "There is at least one unfaithful husband in Mamajorca. All wives know which husbands are unfaithful, but have no knowledge about the fidelity of their own husband. You are forbidden to discuss your husband's faithfulness with any other woman. If you discover that your husband is unfaithful, you must shoot him at precisely midnight of the day you find that out." Thirty-nine silent nights followed the queen's announcement. On the fortieth night, shots were heard. Queen Henrietta I is revered in Mamajorcan history.
As with all philosophers Josephine doesn't provide the question, but leaves it implicit in his document. So figure out the questions - there are two - and answer them. Here is Josephine's second scroll.
Queen Henrietta I was succeeded by daughter queen Henrietta II. After a while Henrietta like her famous mother became worried about the infidelity problem. She decided to act, and sent a letter to her subjects (wives) that contained the exact words of Henrietta I's famous speech. She added that the letters were guarenteed to reach all wives eventually. Queen Henrietta II is remembered as a foolish and unjust queen.
What is the question and answer implied by this scroll? Solution
locks.and.boxes
You want to send a valuable object to a friend. You have a box which is more than large enough to contain the object. You have several locks with keys. The box has a locking ring which is more than large enough to have a lock attached. But your friend does not have the key to any lock that you have. How do you do it? Note that you cannot send a key in an unlocked box, since it might be copied. Solution
min.max
In a rectangular array of people, which will be taller, the tallest of the shortest people in each column, or the shortest of the tallest people in each row? Solution
mixing
Start with a half cup of tea and a half cup of coffee. Take one tablespoon of the tea and mix it in with the coffee. Take one tablespoon of this mixture and mix it back in with the tea. Which of the two cups contains more of its original contents? Solution
monty.52
Monty and Waldo play a game with N closed boxes. Monty hides a dollar in one box; the others are empty. Monty opens the empty boxes one by one. When there are only two boxes left Waldo opens either box; he wins if it contains the dollar. Prior to each of the N-2 box openings Waldo chooses one box and locks it, preventing Monty from opening it next. That box is then unlocked and cannot be so locked twice in a row. What are the optimal strategies for Monty and Waldo and what is the fair price for Waldo to pay to play the game? Solution
number
Mr. S. and Mr. P. are both perfect logicians, being able to correctly deduce any truth from any set of axioms. Two integers (not necessarily unique) are somehow chosen such that each is within some specified range. Mr. S. is given the sum of these two integers; Mr. P. is given the product of these two integers. After receiving these numbers, the two logicians do not have any communication at all except the following dialogue:
Mr. P.: I do not know the two numbers.
Mr. S.: I knew that you didn't know the two numbers.
Mr. P.: Now I know the two numbers.
Mr. S.: Now I know the two numbers.
Given that the above statements are absolutely truthful, what are the two numbers? Solution
river.crossing
Three humans, one big monkey and two small monkeys are to cross a river: a) Only humans and the big monkey can row the boat. b) At all times, the number of human on either side of the river must be GREATER OR EQUAL to the number of monkeys on THAT side. ( Or else the humans will be eaten by the monkeys!) c) The boat only has room for 2 (monkeys or humans) Solution
ropes
Two fifty foot ropes are suspended from a forty foot ceiling, about twenty feet apart. Armed with only a knife, how much of the rope can you steal? Solution
same.street
Sally and Sue have a strong desire to date Sam. They all live on the same street yet neither Sally or Sue know where Sam lives. The houses on this street are numbered 1 to 99. Sally asks Sam "Is your house number a perfect square?". He answers. Then Sally asks "Is is greater than 50?". He answers again. Sally thinks she now knows the address of Sam's house and decides to visit. When she gets there, she finds out she is wrong. This is not surprising, considering Sam answered only the second question truthfully. Sue, unaware of Sally's conversation, asks Sam two questions. Sue asks "Is your house number a perfect cube?". He answers. She then asks "Is it greater than 25?". He answers again. Sue thinks she knows where Sam lives and decides to pay him a visit. She too is mistaken as Sam once again answered only the second question truthfully. If I tell you that Sam's number is less than Sue's or Sally's, and that the sum of their numbers is a perfect square multiplied by two, you should be able to figure out where all three of them live. Solution
self.ref
Find a number ABCDEFGHIJ such that A is the count of how many 0's are in the number, B is the number of 1's, and so on. Solution
smullyan/black.hat
Three logicians, A, B, and C, are wearing hats, which they know are either black or white but not all white. A can see the hats of B and C; B can see the hats of A and C; C is blind. Each is asked in turn if they know the color of their own hat. The answers are: A: "No." B: "No." C: "Yes." What color is C's hat and how does she know? Solution
smullyan/fork.three.men
Three men stand at a fork in the road. One fork leads to Someplaceorother; the other fork leads to Nowheresville. One of these people always answers the truth to any yes/no question which is asked of him. The other always lies when asked any yes/no question. The third person randomly lies and tells the truth. Each man is known to the others, but not to you. What is the least number of yes/no questions you can ask of these men and pick the road to Someplaceorother? Does the answer change if the third man randomly answers? Solution
smullyan/fork.two.men
Two men stand at a fork in the road. One fork leads to Someplaceorother; the other fork leads to Nowheresville. One of these people always answers the truth to any yes/no question which is asked of him. The other always lies when asked any yes/no question. By asking one yes/no question, can you determine the road to Someplaceorother? Solution
smullyan/integers
Two logicians place cards on their foreheads so that what is written on the card is visible only to the other logician. Consecutive positive integers have been written on the cards. The following conversation ensues: A: "I don't know my number." B: "I don't know my number." A: "I don't know my number." B: "I don't know my number." ... n statements of ignorance later ... A or B: "I know my number." What is on the card and how does the logician know it? Solution
smullyan/painted.heads
While three logicians were sleeping under a tree, a malicious child painted their heads red. Upon waking, each logician spies the child's handiwork as it applied to the heads of the other two. Naturally they start laughing. Suddenly one falls silent. Why? Solution
smullyan/priest
In a small town there are N married couples in which one of the pair has committed adultery. Each adulterer has succeeded in keeping their dalliance a secret from their spouse. Since it is a small town, everyone knows about everyone else's infidelity. In other words, each spouse of an adulterer thinks there are N - 1 adulterers, but everyone else thinks there are N adulterers. People of this town have a tradition of denouncing their spouse in church if they are guilty of adultery. So far, of course, no one has been denounced. In addition, people of this town are all amateur logicians of sorts, and can be expected to figure out the implications of anything they know. A priest has heard the confession of all the people in the town, and is troubled by the state of moral turpitude. He cannot break the confessional, but knowing of his flock's logical turn of mind, he hits upon a plan to do God's work. He announces in Mass one Sunday that there is adultery in the town. Is the priest correct? Will this result in every adulterer being denounced? Solution
smullyan/stamps
The moderator takes a set of 8 stamps, 4 red and 4 green, known to the logicians, and loosely affixes two to the forehead of each logician so that each logician can see all the other stamps except those 2 in the moderator's pocket and the two on her own head. He asks them in turn if they know the colors of their own stamps: A: "No" B: "No" C: "No" A: "No B: "Yes" What are the colors of her stamps, and what is the situation? Solution
supertasks
You have an empty urn, and an infinite number of labeled balls. Each has a number written on it corresponding to when it will go in. At a minute to the hour, you take the first ten balls and put them in the urn, and remove the last ball. At the next half interval, you put in the next ten balls, and remove ball number 20. At the next half interval, you put in ten more balls and remove ball 30. This continues for the whole minute.... how many balls are in the urn at this point? (infinite) You have the same urn, and the same set of balls. This time, you put in 10 balls and remove ball number 1. Then you put in another ten balls and remove ball number 2. Then you put in another ten balls and remove ball number 3. After the minute is over, how many balls are left in the urn now? (zero) Are the above answers correct, and why or why not? Solution
timezone
Two people are talking long distance on the phone; one is in an East- Coast state of the US, the other is in a West-Coast state of the US. The first asks the other "What time is it?", hears the answer, and says, "That's funny. It's the same time here!" Solution
unexpected
Swedish civil defense authorities announced that a civil defense drill would be held one day the following week, but the actual day would be a surprise. However, we can prove by induction that the drill cannot be held. Clearly, they cannot wait until Friday, since everyone will know it will be held that day. But if it cannot be held on Friday, then by induction it cannot be held on Thursday, Wednesday, or indeed on any day. What is wrong with this proof? Solution
verger
A very bright and sunny Day
The Priest did to the Verger say:
"Last Monday met I strangers three
None of which were known to Thee.
I ask'd Them of Their Age combin'd
which amounted twice to Thine!
A Riddle now will I give Thee:
Tell Me what Their Ages be!"
So the Verger ask'd the Priest:
"Give to Me a Clue at least!"
"Keep Thy Mind and Ears awake,
And see what Thou of this can make.
Their Ages multiplied make plenty,
Fifty and Ten Dozens Twenty."
The Verger had a sleepless Night
To try to get Their Ages right.
"I almost found the Answer right.
Please shed on it a little Light."
"A little Clue I give to Thee,
I'm older than all Strangers three."
After but a little While
The Verger answered with a Smile:
"Inside my Head has rung a Bell.
Now I know the answer well!"
Now, the question is:
How old is the PRIEST?? Solution
weighing/balance
You are given 12 identical-looking coins, one of which is counterfeit and weighs slightly more or less (you don't know which) than the others. You are given a balance scale which lets you put the same number of coins on each side and observe which side (if either) is heavier. How can you identify the counterfeit and tell whether it is heavy or light, in 3 weighings?
More generally, you are given N coins, one of which is heavy or light. Solution
weighing/box
You have ten boxes; each contains nine balls. The balls in one box weigh 0.9 kg; the rest weigh 1.0 kg. You have one weighing on an accurate scale to find the box containing the light balls. How do you do it? Solution
weighing/find.median
What is the least number of pairwise comparisons needed to find the median of 2n+1 distinct real numbers? Solution
weighing/gummy.bears
Real gummy drop bears have a mass of 10 grams, while imitation gummy drop bears have a mass of 9 grams. Spike has 7 cartons of gummy drop bears, 4 of which contain real gummy drop bears, the others imitation. Using a scale only once and the minimum number of gummy drop bears, how can Spike determine which cartons contain real gummy drop bears? Solution
weighing/optimal.weights
What is the smallest set of weights that allow you to weigh on a balance scale every integer number of kilograms up to some number N? Solution
weighing/weighings
Some of the supervisors of Scandalvania's n mints are producing bogus coins. It would be easy to determine which mints are producing bogus coins but, alas, the only scale in the known world is located in Nastyville, which isn't on very friendly terms with Scandalville. In fact, Nastyville's king will only let you use the scale twice. Your job? You must determine which of the n mints are producing the bogus coins using only two weighings and the minimum number of coins (your king is rather parsimonious, to put it nicely). This is a true scale, i.e. it will tell you the weight of whatever you put on it. Good coins are known to have a weight of 1 ounce and it is also known that all the bogus mints (if any) produce coins that are light or heavy by the same amount.
Some examples: if n=1 then we only need 1 coin, if n=2 then clearly 2 coins suffice, one from each mint.
What are the solutions for n=3,4,5? What can be said for general n? Solution
zoo
I took some nephews and nieces to the Zoo, and we halted at a cage marked
Tovus Slithius, male and female.
Beregovus Mimsius, male and female.
Rathus Momus, male and female.
Jabberwockius Vulgaris, male and female.
The eight animals were asleep in a row, and the children began to guess which was which. "That one at the end is Mr Tove." "No, no! It's Mrs Jabberwock," and so on. I suggested that they should each write down the names in order from left to right, and offered a prize to the one who got most names right.
As the four species were easily distinguished, no mistake would arise in pairing the animals; naturally a child who identified one animal as Mr Tove identified the other animal of the same species as Mrs Tove.
The keeper, who consented to judge the lists, scrutinised them carefully. "Here's a queer thing. I take two of the lists, say, John's and Mary's. The animal which John supposes to be the animal which Mary supposes to be Mr Tove is the animal which Mary supposes to be the animal which John supposes to be Mrs Tove. It is just the same for every pair of lists, and for all four species.
"Curiouser and curiouser! Each boy supposes Mr Tove to be the animal which he supposes to be Mr Tove; but each girl supposes Mr Tove to be the animal which she supposes to be Mrs Tove. And similarly for the other animals. I mean, for instance, that the animal Mary calls Mr Tove is really Mrs Rathe, but the animal she calls Mrs Rathe is really Mrs Tove."
"It seems a little involved," I said, "but I suppose it is a remarkable coincidence."
"Very remarkable," replied Mr Dodgson (whom I had supposed to be the keeper) "and it could not have happened if you had brought any more children."
How many nephews and nieces were there? Was the winner a boy or a girl? And how many names did the winner get right? [by Sir Arthur Eddington]