Logic Puzzles

Hint:

Based on what you know, why are the other princes unable to solve this puzzle?

Solution:

White.

The king would not select two white hats and one black hat. This would mean two princes would see one black hat and one white hat. You would be at a disadvantage if you were the only prince wearing a black hat.

If you were wearing the black hat, it would not take long for one of the other princes to deduce he was wearing a white hat.

If an intelligent prince saw a white hat and a black hat, he would eventually realize that the king would never select two black hats and one white hat. Any prince seeing two black hats would instantly know he was wearing a white hat. Therefore if a prince can see one black hat, he can work out he is wearing white.

Therefore the only fair test is for all three princes to be wearing white hats. After waiting some time just to be sure, you can safely assert you are wearing a white hat.

Hint:

Each pirate cares first about staying alive, then about getting as many coins as possible. A pirate will vote “Aye” only if the proposal gives them more than they would get after mutiny. Side deals, promises, and revenge votes do not count, because the pirates are selfish and treacherous.

What happens if there are two pirates? Who completely loses out? What happens if there are three pirates? Who completely loses out? What happens if there are four pirates? Which two pirates completely lose out?

Solution:

98

The captain says he will take 98 coins, and will give one coin to the third most senior pirate and another coin to the most junior pirate. He then explains his decision in a manner like this...

1. Pirate 1: most junior, keeps 1 coin.
2. Pirate 2: fourth most senior, keeps 0 coins.
3. Pirate 3: third most senior, keeps 1 coin.
4. Pirate 4: second most senior, keeps 0 coins.
5. Pirate 5: captain, most senior, keeps 98 coins.

If there were 2 pirates, because pirates 3,4 & 5 had walked the plank, then pirate 2 would be the most senior, and he would just vote for himself and that would be 50% of the vote, so he's obviously going to keep all the money for himself.

If there were 3 pirates, because pirates 4 & 5 had walked the plank, pirate 3 has to convince at least one other person to join in his plan. Pirate 3 would take 99 gold coins and give 1 coin to pirate 1. Pirate 1 knows if he does not vote for pirate 3, then he gets nothing, so obviously is going to vote for this plan.

If there were 4 pirates, because the captain had to walk the plank, pirate 4 would give 1 coin to pirate 2, and pirate 2 knows if he does not vote for pirate 4, then he gets nothing, so obviously is going to vote for this plan.

As there are 5 pirates, pirates 1 & 3 had obviously better vote for the captain, or they face choosing nothing or risking death.

Pirates left	Winning proposal
2	Senior pirate keeps 100.
3	Captain keeps 99, gives 1 to Pirate 1.
4	Captain keeps 99, gives 1 to Pirate 2.
5	Captain keeps 98, gives 1 each to Pirates 1 and 3.

Hint:

Use the same approach.

Solution:

The most senior pirate could give the coin to the least senior pirate. He can use the same logic in the previous puzzle to explain the futility of anyone trying to keep the coin for himself.

Pirates left	Captain needs	Result
2	1 vote	Pirate 2 keeps the coin. His own vote is enough.
3	2 votes	Pirate 3 gives the coin to Pirate 1. Pirate 1 would get 0 if Pirate 3 dies.
4	2 votes	Pirate 4 gives the coin to Pirate 2 or Pirate 3. Either would otherwise get 0.
5	3 votes	Pirate 5 cannot survive. He has only one coin and needs two extra votes.
6	3 votes	Pirate 6 gives the coin to Pirate 1. Pirate 5 votes yes because he dies if Pirate 6 dies. Pirate 1 votes yes for the coin.

Hint:

The one who stopped laughing, asked himself what the other philosophers were seeing that made them laugh.

Solution:

If he (the smartest philosopher) had nothing on his head, then he realized that the second smartest philosopher would have quickly worked out that the third smartest was laughing only at the second smartest philosopher, and thus the second smartest philosopher would have stopped laughing.

Hint:

You can weigh as few or as many of the ten coins from each set as you choose.

Solution:

One coin from the first set is placed on the scale along with two from the second set etc... If the weight is off by one hundredth of an ounce then it is the first set that is faulty, if the weight is off by two hundred of an ounce then it is the second set which is faulty, etc...

Hint:

 Look up the formula for the LCM.

Solution:

2519

The solution for the answer is the LCM (Lowest Common Multiple) of 10,9,8,7,6,5,4,3,2,1 -1. LCM would give the least number which is divisible by all of these number and subtracting one would give us the number of coconuts which were initially there.

Solution:

Create two sets of ten coins. Flip the coins in one of the sets over, and leave the coins in the other set alone. The first set of ten coins will have the same number of heads and tails as the other set of ten coins.

Hint: If I ask people at random if they have two children and if the youngest is a boy, then the probability that this person has two boys is 1/2.

If I asked people at random if they have two children and if one is a boy, then the probability that this person has two boys is 1/3.

Possibility	Has at least one boy?	Two boys?
BB	yes	yes
BG	yes	no
GB	yes	no
GG	no	no

My question is excluding anyone with two girls, so therefore there are only three cases left, only one of which has two boys.

Solution:

13/27. If you think the answer should be 1/2, you would be wrong. If you knew which child was a boy (say, the younger one), you would be closer to the truth. But since the boy could be either the younger or the older child, the analysis is more subtle. But what does Tuesday have to do with it?

Solution:

Person A writes a number that is her salary plus a random amount (AS + AR) and hands it to B, without showing C. B then adds his salary plus a random amount (BS + BR) and passes it to C (at each step, they write on a new paper and don't show the 3rd person). C adds CS + CR and passes it to A. Now A subtracts her random number (AR) and passes it to B. B and C each subtract their random number and pass. After C is done, he shows the result and they divide by 3.

As has been noted already, there's no way to liar-proof the scheme.

It's also worth noting that once they know the average, any of the three knows the sum of the other 2 salaries.

Solution contributed by Jeebok.

Solution:

It is impossible to cover the remaining 62 squares with 31 dominoes after cutting out two opposite corners of the chessboard. Each domino covers one black and one white square, and removing two opposite corners (which are the same color) leaves an imbalance of colors, resulting in 32 squares of one color and 30 of the other, making it impossible to cover the board completely.

Solution:

The man either started at the North Pole, or the man started at a point just over a mile north of the South Pole, or at any latitude where walking 1 mile south takes him to a circle of latitude with a circumference that is an exact factor of 1 mile (such as 1/2 mile, 1/3 mile, etc.). After walking 1 mile south, he reaches this latitude, then walks 1 mile east, completing a full circle back to his starting point on that latitude, and finally walks 1 mile north to return to his original position.

Solution:

To solve the puzzle, create two sets of ten coins each. Let’s say the first set has 'x' heads and '10 - x' tails. The second set, which consists of the remaining coins, will have '10 - x' heads and 'x' tails. By flipping all the coins in the first set, you will have '10 - x' heads and 'x' tails in that set, which will match the number of heads and tails in the second set, thus ensuring both sets have the same number of heads and tails.

Solution:

The sage's plan was to drink a weak poison prior to the meeting with the king, and then he would drink the pharmacist's strong poison, which would neutralize the weak poison. As his own poison he would bring water, which will have no effect on him, but the pharmacist who would drink the water, and then his poison would surely die. When the pharmacist figured out this plan, he decided to bring water as well. So the sage who drank poison earlier, drank the pharmacist's water, then his own water, and died of the poison he drank before. The pharmacist would drink only water, so nothing will happen to him. And because both of them brought the king water, he didn't get a strong poison like he wanted.

Solution:

983

Hint:

Try taking a different number of balls from each box so that the total shortage in grams directly reveals the box number.

Solution:

Label the boxes 1 through 10. From box 1 take 1 ball, from box 2 take 2 balls, and so on, taking 10 balls from box 10. You now have 1 + 2 + … + 10 = 55 balls.

If every ball weighed 10 g, the scale would read 55 × 10 = 550 g. Place all 55 balls on the scale and note the actual reading.

Because the balls from the defective box weigh 9 g instead of 10 g, each ball taken from that box makes the total weight 1 g lighter than expected. The total weight will therefore be:

550 g − (box number) g

For example, if the scale shows 546 g, the shortage is 4 g, so box 4 is defective. Whatever number of grams is missing from 550 immediately identifies the faulty box.

Hint:

Considera las relaciones entre los dígitos cuidadosamente y trata de encontrar números que cumplan todas las condiciones.

Solution:

Car number is 2448.

Hint:

Consider the possible combinations of ages that multiply to 36 and how their sums might relate to the address.

Solution:

The oldest child is 9 years old.

Hint:

Consider using a binary search method to narrow down the possibilities efficiently.

Solution:

Label the bottles from 1 to 1000 in binary. Each prisoner can represent a bit in a binary number. By having them drink from specific bottles based on their binary representation, you can determine the poisoned bottle based on which prisoners die.

Hint:

Consider the implications of the first two guys not knowing the color of their hats.

Solution:

The blind guy is wearing a red hat. He knows this because if the first two guys saw two white hats on each other, they would know they must be wearing red hats. Since they both said they didn't know, it means at least one of them must be wearing a red hat, which allows the blind guy to deduce that he must be wearing a red hat as well.

Hint:

Translate the two ratio statements into a single three-way ratio for A : B : C, then use the total number of peanuts and total age to scale the ratios.

Solution:

Let the numbers of peanuts received by A, B, and C be in the ratio A : B = 4 : 6 (i.e. 2 : 3) and A : C = 6 : 7. Combining these gives a consistent three-way ratio:

A : B : C = 6 : 9 : 7.

Total peanuts to be shared: 1,000 − 230 = 770.

One “part” therefore equals 770 / (6 + 9 + 7) = 770 / 22 = 35 peanuts.

Peanuts each child received:
A: 6 × 35 = 210
B: 9 × 35 = 315
C: 7 × 35 = 245

Because the peanuts are divided proportional to age, the ages follow the same 6 : 9 : 7 ratio.

Total age = 17.5 years ⇒ one part = 17.5 / 22 = 0.7954545… years.

Ages:
A: 6 × 0.7954545… = 4.7727… years ≈ 4 years 9 months
B: 9 × 0.7954545… = 7.1591… years ≈ 7 years 2 months
C: 7 × 0.7954545… = 5.5682… years ≈ 5 years 7 months

Therefore:
A is about 4 years 9 months old and received 210 peanuts;
B is about 7 years 2 months old and received 315 peanuts;
C is about 5 years 7 months old and received 245 peanuts.

Hint:

Consider using binary representation to identify the buckets.

Solution:

You need a minimum of 10 pigs.

Label the buckets from 0 to 999 in binary (which requires 10 bits). Each pig can represent a bit in the binary number. If a pig drinks from a bucket and dies, it indicates a '1' for that bit position; if it lives, it indicates a '0'. By testing all 10 pigs, you can determine the exact bucket containing the poison.

Hint:

Rewrite the equation in ordinary numbers (using place values) and work modulo 9.

Solution:

Write the four–digit numbers in expanded form:

1000A + 100B + 10C + D \; - \; D \; = \; 1000D + 100C + 10B + A.

Simplifying gives

1000A + 100B + 10C = 1000D + 100C + 10B + A,

or

999A + 90B - 90C = 1000D.

Divide both sides by 9:

111A + 10B - 10C = \dfrac{1000D}{9}.

Because the left side is an integer, the right side must be an integer as well, so 1000D must be divisible by 9. The only one–digit values of D that satisfy this are D = 0 or D = 9.

Case 1  D = 0. Then
111A + 10B - 10C = 0  ⇒  111A = 10(C - B).

The right side is at most 90 in magnitude, so A must be 0. But a leading digit cannot be 0, so this case is impossible.

Case 2  D = 9. Then
111A + 10B - 10C = 9000.
Rearrange:
111A = 9000 - 10(B - C).

The left side is a multiple of 111 = 3·37, so the right side must also be a multiple of 37. Compute 9000 mod 37:

9000 = 37·243 + 9,  so  9000 &equiv 9 (mod 37).

Thus
9 - 10(B - C) &equiv 0 &pmod{37}  ⇒  10(B - C) &equiv 9 &pmod{37}.

The modular inverse of 10 mod 37 is 26, giving
B - C &equiv 12 &pmod{37}.

Because B and C are digits, B - C is between –9 and 9, which can never be congruent to 12 modulo 37. Hence this case is also impossible.

Since both possible values of D lead to contradictions, there is no way to assign digits to A, B, C, and D that satisfies the original equation.

Therefore the puzzle has no solution.

Hint:

What would each sighted prisoner have said if he had seen two white hats in front of him?

Solution:

Let us analyse the information revealed step by step.

1. Prisoner A’s statement
If A had seen two white hats on B and C, he would have known that (because only two white hats exist) his own hat must be black, and he would have answered accordingly. Since he instead says “I don’t know,” it follows that the pair (B,C) is not (white, white). Therefore at least one of B or C is wearing a black hat.

2. Prisoner B’s statement
B has heard A’s remark and hence already knows that among himself and C at least one black hat is present. Now B looks at A’s hat.

• If B sees A wearing a white hat, then to satisfy “at least one of B or C is black,” B can deduce that his own hat is black (if it were white, both he and A would be white, contradicting A’s information). In that case B would have answered “Black.”
• But B actually answers “I don’t know.” Hence the situation above cannot occur, which means B does not see a white hat on A. Therefore A’s hat is black.

3. What the blind prisoner hears
The blind Prisoner C now knows:

• A’s hat is black (from B’s failure to identify his own hat).
• There are three black hats in total.

If C’s own hat were white, the visible hats would be: A = black, B = ? (unknown to C), C = white. B would then be looking at (black, white). Seeing exactly one white hat would still leave B in doubt about his own colour (his could be black or white), so B would indeed have said “I don’t know.” At first sight this seems to allow either colour for C.

However, recall that A already has a black hat. Only two black hats remain. If C’s hat were white, B could be seeing either (black, white) or (black, black). But if B were seeing (black, white) he would know immediately that he is wearing a black hat — otherwise C would have two whites contradicting A’s information. Since B didn’t reach such a conclusion, the arrangement he sees cannot contain any white hats. Hence from B’s perspective both other hats must be black.

Therefore C can conclude that the only configuration consistent with both A’s and B’s inability to decide is:

A = black,  B = black,  C = black

Thus the blind prisoner knows with certainty that his own hat is black, and confidently states so, securing his freedom.

Hint:

Consider the two possible cases for A (politician or not) and check each case against B’s and C’s statements.

Solution:

Let us analyse the two possibilities for native A.

Case 1: A is a politician (and therefore lies).
Asked whether he is a politician, a politician would have to answer “No.” Thus A does deny being a politician. B’s report, “A denied being a politician,” is therefore true, so B must be a truth-telling non-politician. C’s statement, “A is a politician,” is also true, so C too must be a truth-telling non-politician. This produces a consistent scenario (politician A; non-politicians B and C).

Case 2: A is a non-politician (and therefore tells the truth).
If A tells the truth, he must answer “No, I am not a politician.” Again A denies being a politician, so B’s report is true and B is a non-politician. This time C’s statement, “A is a politician,” is false, hence C must be a lying politician. This is also consistent (non-politicians A and B; politician C).

In both cases B is a non-politician, while A and C swap roles. Because both complete assignments are self-consistent, we cannot tell which of them is the actual one.

Conclusion: Only native B is certainly a truth-telling non-politician; the statuses of natives A and C cannot be deduced from the information given.

Hint:

Among any five cards there must be at least two of the same suit. Use one of those as a “key” card. The order of the four cards can encode up to 24 different messages.

Solution:

1. Guaranteeing a shared suit
With four suits and five cards, the pigeon-hole principle ensures that at least two of the five share a suit. Call these two cards S₁ and S₂.

2. Choosing the key and the hidden card
The Magician chooses one of the same-suit pair to keep among the four visible cards (the key) and gives the other to Alex (the hidden card). If necessary the roles of the two cards are swapped so that, when the ranks are considered cyclically in the order A-2-3-…-K-A, the distance from the key to the hidden card is between 1 and 6 inclusive. This is always possible because the two ranks are at most 6 or at least 7 steps apart, and you can pick the shorter of the two directions.

3. Encoding that distance
After the key card is fixed, three cards remain. Their order relative to the key card can be permuted in 4! = 24 ways. Six of those permutations will be used to signal the numerical distance 1 – 6 (the other 18 are unused or can be assigned to alternative conventions). A convenient mapping is:

• distance 1 → key followed by ABC
• distance 2 → key followed by ACB
• distance 3 → key followed by BAC
• distance 4 → key followed by BCA
• distance 5 → key followed by CAB
• distance 6 → key followed by CBA

(Here A, B, C are the three non-key cards sorted in some agreed order, e.g. alphabetically by suit then by rank.) The Magician simply stacks the four cards so that this ordering appears from top to bottom.

4. Peter decodes
Peter turns over the pile. The first card he sees is necessarily the key card, so he knows the suit of the hidden card. He inspects the relative order of the remaining three cards, consults the agreed table above, and learns the distance d (1–6). Finally he counts forward d steps clockwise through the ranks starting from the key card’s rank; the card reached is the hidden card, which he names aloud.

5. Example
Suppose the five cards are ♣9, ♥K, ♦4, ♣3, ♠A. Two clubs are present. The Magician chooses ♣9 as key and ♣3 as hidden because the distance 5→3 clockwise is 6. The remaining three cards (♥K, ♦4, ♠A) are ordered as CBA to indicate distance 6. Peter sees ♣9 ♥K ♦4 ♠A, recognises the permutation CBA, deduces distance 6, counts six steps from 9 ♣ (10 J Q K A 2 3), and announces "Three of Clubs"—correct.

Thus, by exploiting the forced duplicate suit and the 24 possible permutations of four cards, the trick works every time without any outside communication.

Solution:

143547

Hint:

Take a different number of pens from each packet so that each packet contributes a unique amount to the total weight.

Solution:

Label the packets 1 through 10. From packet 1 take 1 pen, from packet 2 take 2 pens, from packet 3 take 3 pens, and so on until you take 10 pens from packet 10. In total you place 1 + 2 + … + 10 = 55 pens on the scale.

If every pen weighed 10 g, the total would be 55 × 10 = 550 g. Because one packet contains 9 g pens, the reading will be smaller. Let the actual reading be W grams. The number of grams missing from 550, namely 550 − W, equals the number of pens taken from the faulty packet—which is precisely the packet’s label. For example, if the scale reads 547 g (3 g less than 550), then packet 3 is the one with 9 g pens.

Hint:

Consider the total number of matches played and the results for each team to deduce the missing information.

Solution:

a) Team D has 1 draw and 4 losses.

b) The result of the match B against C was a win for B (B won 2-1).

Hint:

Look at the factor you need to multiply each term by to obtain the next one.

Solution:

Compute the ratio between each pair of known consecutive terms:
607.75 ÷ 2734.875 = 4.5,
2734.875 ÷ 9572.0625 = 3.5.

We observe that these multipliers decrease by 1 while keeping a fractional part of 0.5: 6.5, 5.5, 4.5, 3.5. Therefore the factor before 5.5 must be 6.5.

Missing term = 17 × 6.5 = 110.5.

Answer: 110.5

Solution:

He originally collected 1,026 mangoes.

After the 10 gates he retained 3 mangoes in his basket.

Hint:

Begin by pairing the branch-office clue (public-dealing) with the statements about B and E, then use the counts of each category to eliminate possibilities.

Solution:

The only assignment that satisfies all clues is:

• A — Probationary Officer, Graduate, Admin Section
• B — Stenographer, Graduate, Branch Office (public-dealing)
• C — Clerk, Graduate, Admin Section
• D — Probationary Officer, Post-Graduate, Admin Section
• E — Clerk, Post-Graduate, Branch Office (public-dealing)

Hint:

Consider the total number of sweets and how the transfers affect each person's total.

Solution:

John had 420 sweets at first.

Hint:

Begin by placing the Goya and noting the required portrait at position 5. Then use the "no adjacent portraits" rule to limit possibilities and test the Rivera/Van Gogh condition.

Solution:

The unique order that satisfies all conditions is:

1. Tamayo
2. Goya
3. Benton
4. Picasso
5. Van Gogh
6. Rivera
7. Matisse

Hint:

Start by letting the five ranks be a, b, c, d, e (all different) and translate each clue into an equation or inequality. Pay particular attention to the parity (even/odd) and color conditions to cut down the possibilities quickly.

Solution:

The five cards are 2♦, 3♥, 6♣, 8♠, and 9♥.

Hint:

The man who can see the other three will use their reactions to deduce the color of his own hat.

Solution:

The man who can see the other three will speak up. If he sees two hats of the same color (either both black or both white), he will know that his hat must be the opposite color. If he sees one black and one white hat, he cannot determine the color of his own hat and will not speak up.

Hint:

Remember that each box must contain an odd number of balls.

Solution:

One possible solution is to place 1 ball in 7 boxes and 3 balls in 2 boxes. This way, all boxes are filled, no box is empty, and all boxes contain an odd number of balls.

Hint:

You can use a binary approach to identify the poisoned bottle.

Solution:

You need 4 people to find the poisoned bottle. Each person can represent a binary digit, allowing you to test combinations of bottles.

Hint:

Think of each prisoner as reporting a single binary digit (alive = 0, dead = 1). How many different patterns of deaths can you create with n prisoners?

Solution:

1. Label the casks from 1 to 1000 and write each label in binary. Give each prisoner a sip from every cask whose binary representation has a 1 in the position assigned to that prisoner. After 30 days, the pattern of survivors forms the binary number of the poisoned cask.

You need enough prisoners that their death/survival pattern can distinguish among 1000 possibilities. With p prisoners you have 2^p possible patterns. The smallest p with 2^p ≥ 1000 is 10 (because 2⁹=512 < 1000 < 2¹⁰=1024). So the minimum is 10 prisoners.

2. With 10 prisoners you get at most 2¹⁰=1024 different patterns. One of these patterns is “no one dies,” which cannot happen because we know exactly one cask is poisoned. Therefore the scheme can uniquely identify the poison among at most 1024−1 = 1023 casks.

Solution:

Keith buys banana cue.

Justine buys banana cue.

Diana does not buy banana cue.

Hint:

Consider the implications of each statement and how they relate to each other.

Solution:

a) If everybody told the truth, then Brown is guilty, Jones is guilty, and Smith is innocent.

b) If the innocent told the truth and the guilty lied, then Brown and Jones are guilty, and Smith is innocent.

Hint:

Ask one brother what the other brother would say if you asked him which door leads to the garden.

Solution:

He should ask either brother, 'If I asked your brother which door leads to the garden, what would he say?'

Hint:

Let b be the paddler’s speed in still water and c the speed of the current. Write two equations for the distance using the times given, then solve for b.

Solution:

Let the width of the river be d.

Against the current: \(\dfrac{d}{4}=b-c\).

With the current: \(\dfrac{d}{3}=b+c\).

Add the two equations:
\(\dfrac{d}{4}+\dfrac{d}{3}=2b \;\Rightarrow\; \dfrac{7d}{12}=2b \;\Rightarrow\; b=\dfrac{7d}{24}.\)

The time required in still water is
\(\displaystyle \text{time}=\frac{d}{b}=\frac{d}{\tfrac{7d}{24}}=\frac{24}{7}\text{ hours}\approx3\text{ h }25\text{ min}.\)

Hint:

Ask how many gloves she could take while still missing a pair in at least one color.

Solution:

To guarantee two gloves of every color, consider the worst-case draw.

If she wishes to avoid getting two blue gloves, she could draw every red and yellow glove (25 + 45 = 70) and, at worst, just one blue glove. That is 70 + 1 = 71 gloves while still lacking a blue pair.

Similarly, she could avoid a red pair with 14 + 45 + 1 = 60 gloves, or avoid a yellow pair with 14 + 25 + 1 = 40 gloves.

The largest of these possibilities is 71. Therefore, after drawing 71 gloves it is still possible that she is missing a pair in one color (the blues).

Consequently, as soon as she draws one more glove—making a total of 72—the worst case is eliminated and she is guaranteed to have at least two gloves of every color.

Minimum required gloves: 72.

Hint:

List all sets of three positive integers whose product is 72 and check which sums coincide. Then use the clue about an oldest daughter.

Solution:

First, enumerate all triples of positive integers whose product is 72:

(1,1,72), (1,2,36), (1,3,24), (1,4,18), (1,6,12), (1,8,9), (2,2,18), (2,3,12), (2,4,9), (2,6,6), (3,3,8), (3,4,6).

Compute their sums. Among these, only two triples share the same sum:

(2,6,6) and (3,3,8) both sum to 14.

Because the census taker knew the house number yet still couldn’t decide, that number must be 14, leaving those two possibilities.

The father then mentions an oldest daughter. The triple (2,6,6) has no single oldest child (the two six-year-olds tie), whereas (3,3,8) does. Therefore the daughters’ ages are 3, 3 and 8.

Hint:

Ask yourself what Man 3 would have concluded if he had seen two white hats.

Solution:

Man 1 is wearing a black hat.

Reasoning:

1. Man 3 sees the two hats in front of him. If they were both white, he would immediately know his own hat must be black (only two white hats exist). Because he says, “I don’t know,” the two hats he sees cannot both be white. At least one of them is black.
2. Man 2 hears this and looks at Man 1’s hat. If he saw a white hat, he could deduce that his own hat must be black (since the pair cannot be two white hats). But he still doesn’t know, which means the hat he sees on Man 1 cannot be white. Therefore Man 1’s hat must be black.
3. Man 1, having followed the prior statements, realises that his hat has to be black and answers accordingly.

Solution:

Mr. Mukherjee is the Engineer, and his son is the Doctor.

Hint:

Consider how each digit relates to the counts of the other digits in the number.

Solution:

6210001000

Hint:

Ask what C would have seen if both A and B had been wearing red caps.

Solution:

The colour of A’s cap is blue.

Reasoning step by step:

1. If A and B had both been wearing red caps, C would have seen two red caps. With only two red caps in total, C would then know his own cap must be blue and could answer immediately. Since C couldn’t answer, A and B cannot both be red.
2. B hears C’s statement and looks at A. If B sees a red cap on A and knows C did not answer, B can reason as follows: “If my cap were red as well, C would have seen the two red caps and would have known his cap is blue. Because C did not know, my cap cannot be red.” Therefore, had B seen A wearing red, B would have been able to deduce that his own cap is blue. But B still could not answer, so A cannot be wearing red.
3. Since A is not wearing red, A must be wearing blue.

Thus A confidently answers “blue.” (A blue, B red, C blue is one distribution consistent with all statements, and it is the only possible distribution given the deductions.)

Hint:

Think about what information the last man in the line can communicate using just the single word he is allowed to speak.

Solution:

No strategy can guarantee that all ten men survive. The best possible plan guarantees that nine men live and leaves only the very last man with a 50 % chance of survival.

The agreed strategy is based on parity (even or odd count) of the red hats that the back man can see:

1. Beforehand they agree: the first speaker will say "red" if he sees an even number of red hats in front of him and "blue" if he sees an odd number of red hats.
2. All other men know this rule. After the first word is spoken, each subsequent man counts the number of red hats he can see in front of him and compares that to the parity announced by the first man and to the answers already given behind him. From those three pieces of information he can deduce with certainty the colour of his own hat and answer correctly.

Consequently, the last nine men are saved with certainty. The very first man, however, had no information about his own hat—he used his single word solely to transmit the parity bit to the others—so his own guess is correct only half the time. No different strategy can do better, because whatever the first man says, at most one bit of information can be conveyed, while two bits would be needed to determine both the colour of his hat and the parity for the rest. Therefore a 100 % survival rate is impossible.

Hint:

Consider the voting dynamics and the potential outcomes for each pirate.

Solution:

Pirate 1 and Pirate 3 would be content with one coin because if they vote against the captain and the captain is overthrown, they risk getting nothing in the new redistribution. By accepting one coin now, they ensure they receive something rather than risking everything for a chance at more.

Hint:

Let g be the tree’s yearly growth (in metres). Express the heights after 6 and 7 years in terms of g and set up the given ratio.

Solution:

Let the constant yearly growth be g metres.

Height after 6 years: 2 + 6g
Height after 7 years: 2 + 7g

The statement “7 years height is one-eighth taller than 6 years height” means

(2 + 7g) = (2 + 6g) \(1 + \tfrac{1}{8}) = \tfrac{9}{8}(2 + 6g).

Multiply both sides by 8:

8(2 + 7g) = 9(2 + 6g)
16 + 56g = 18 + 54g

Rearrange:

56g – 54g = 18 – 16
2g = 2
g = 1 m per year.

Height after 10 years:

2 + 10g = 2 + 10(1) = 12 m.

Answer: 12 metres.

Solution:

La solución se basa en la lógica de las proposiciones. Si Margarita compra casa, entonces Genaro le regaló rosas o está feliz. Si Genaro le regaló rosas, Margarita no compra casa, lo que contradice la primera proposición. Por lo tanto, Margarita debe estar feliz con sus decisiones. Si Margarita no compra carro, no está feliz, lo que significa que debe comprar carro para ser feliz. Así, si compra casa, también compra carro.

Solution:

Let P be the number of people who eat turkey, and C be the number of people who eat chicken. We know that 25 people eat only turkey, so P = 25 + x (where x is the number of people who eat both). The total number of people who do not eat turkey is 10, so C + x = 40 (since 50 - 10 = 40). We also know that 30 people do not eat chicken, so P + (40 - C) = 30. Solving these equations gives us P = 25 + x, C = 40 - x, and the total number of people eating both is x = 15. Therefore, 15 people eat both turkey and chicken.

Solution:

Para resolver esto, primero convertimos todos los tiempos a horas. El tercer surtidor tarda 5 días, que son 120 horas. La tasa de llenado de cada surtidor es:
Primer surtidor: 1/30
Segundo surtidor: 1/40
Tercer surtidor: 1/120
Sumamos las tasas: 1/30 + 1/40 + 1/120 = 1/12.
Por lo tanto, juntos llenan la pileta en 12 horas.

Solution:

The blind man can deduce the color of his hat based on the responses of the other two prisoners. If the first prisoner sees two white hats, he would know his hat must be black. If he sees one black and one white, he cannot be sure. The blind man, knowing this, can conclude that if the first prisoner is unsure, he must be wearing a black hat. Therefore, the blind man can confidently say he is wearing a black hat.

Solution:

The values of A, B, and C are 2, 4, and 0 respectively. Therefore, A + B + C = 2 + 4 + 0 = 6.

Solution:

1. The son (20 kg) crosses the river first. 2. The son returns with the boat. 3. The wife (40 kg) crosses the river. 4. The son returns with the boat. 5. The son (20 kg) crosses the river again. Now all three are on the other side.

Solution:

The numbers correspond to letters on a phone keypad. Decoding the message: 4 = GHI, 8 = T, 9 = WXYZ, 2 = ABC, 7 = PQRS, 6 = MNO, 3 = DEF. The message decodes to 'GOT YOU'. The killer is the person who the victim was referring to as 'you'.

Solution:

Move the nines in the following way: Move the first nine from the top row to the bottom row, and move the second nine from the middle row to the top row. This will create additional lines while maintaining the total of 27.

Solution:

Let the speed of the boat in still water be b km/h and the speed of the river current be c km/h. The distance across the river is the same in both cases. When paddling against the current, the effective speed is b - c, and when paddling with the current, it is b + c. The time taken to cross against the current is 4 hours, and with the current is 3 hours. Therefore, we have:

Distance = (b - c) * 4 = (b + c) * 3

From these equations, we can derive the time taken on still water. Solving these equations gives us the time taken on still water as 3.5 hours.

Solution:

Let C be the total number of cachers. Since one third of cachers are sane, we have S = C/3. Half of the whackos are cachers, so W = 8 (whackos who are cachers). The total number of attendees is 90, which includes sane cachers, whackos, and others. The number of cachers who are neither sane nor whacko is C - S - W. Solving the equations gives us the answer.

Solution:

Let x be the number of hits and y be the number of misses. We have the equations: x + y = 100 (total chances) and x - y = 30 (points earned). Solving these, we get x = 65 and y = 35. Therefore, he hits the target 65 times.

Solution:

First, divide the 12 balls into three groups of four. Weigh two groups against each other. Depending on the outcome, you will know if the odd ball is in one of those groups or in the group not weighed. Continue to narrow it down using the balance scale until you identify the odd ball and determine if it is heavier or lighter.

Solution:

They originally stole 63 coins. Here's the reasoning:

1. Let the initial number of coins be x.

2. The first thief divides the coins into four equal parts and discards one coin, so he takes (x-1)/4 coins, leaving 3(x-1)/4 coins.

3. The second thief does the same: divides the remaining coins into four equal parts and discards one coin, taking (3(x-1)/4 - 1)/4 coins, leaving 3(3(x-1)/4 - 1)/4 coins.

4. The third thief repeats this process, and so does the fourth thief.

5. After the fourth thief, there are no coins left, which means the initial number of coins x must satisfy the equation derived from these steps.

6. Solving this equation, we find that x = 63.