Question 1

You are in a dark room with a deck of 52 cards. You know exactly 10 cards are face up. You can't see or feel which way cards face. Split the deck into two piles so each pile has the same number of face-up cards.

Accepted Answer

Take any 10 cards off the top and flip them over. This pile now has the same number of face-up cards as the remaining 42-card pile. If the top 10 cards had k face-up cards, then the bottom pile has 10-k face-up cards. After flipping all 10, the top pile has 10-k face-up cards too.

Question 2

You have 12 balls, one is heavier or lighter. You have a balance scale and 3 weighings. Find the odd ball and whether it's heavier or lighter.

Accepted Answer

Divide into 3 groups of 4. Weigh group A vs group B. Case 1: they balance — the odd ball is in group C. Take 3 from C, weigh against 3 known-good balls. If they balance, the remaining ball is odd — weigh it against a known ball to determine heavier/lighter. If they don't balance, you know whether the odd ball is heavier or lighter; weigh 2 of the 3 suspects against each other to isolate it. Case 2: A and B don't balance — you now know which side is heavier. Relabel and use the third weighing to isolate the odd ball among the suspects.

Question 3

You have 25 horses and can race 5 at a time. What is the minimum number of races needed to find the 3 fastest horses (no stopwatch)?

Accepted Answer

7 races. First, run 5 races of 5 horses each (5 races). Rank the winners and race them against each other (race 6). Now the overall fastest horse is the winner of race 6. The 2nd and 3rd fastest must be among: the 2nd and 3rd place from the winner's original group, the 2nd place from the group whose winner came 2nd in race 6, and the winner who came 3rd in race 6. That's 5 candidates — race them (race 7). The top 2 from race 7 plus the overall winner are your 3 fastest.

Question 4

You are standing outside a closed room with three light bulbs and three switches. You can flip switches as much as you want but can only enter the room once. How do you determine which switch controls which bulb?

Accepted Answer

Turn on switch 1 and wait 5 minutes. Turn it off, turn on switch 2, and enter. The lit bulb is switch 2. The warm but off bulb is switch 1. The cold, off bulb is switch 3.

Question 5

You have a 3-litre jug and a 5-litre jug with unlimited water. How do you measure exactly 4 litres?

Accepted Answer

Fill the 5L jug, pour into the 3L jug (leaving 2L). Empty the 3L jug, pour the 2L into it. Fill the 5L jug again and pour into the 3L jug until full (1L poured). Exactly 4L remain in the 5L jug.

Question 6

Why are manhole covers round rather than square?

Accepted Answer

A round cover cannot fall through its own opening. A circle has constant diameter regardless of orientation, whereas a square cover could be tilted diagonally and fall through since the diagonal is longer than the side.

Question 7

A single lily pad doubles in size every day. It takes 48 days to cover the entire lake. On which day was the lake half-covered?

Accepted Answer

Day 47. Since the lily pad doubles every day, if it covers the full lake on day 48, it must have been half-covered on day 47.

Question 8

100 prisoners in a line each wear a red or blue hat. Starting from the back, each must guess their hat color. They can see all hats in front but not their own. What strategy guarantees 99 are released?

Accepted Answer

The last prisoner counts blue hats in front and says 'Blue' if even, 'Red' if odd (parity check). Each subsequent prisoner tracks the running parity from previous guesses and what they see ahead to deduce their own hat color. All 99 after the first are guaranteed correct.

Question 9

On a standard analog clock, what is the exact angle between the hour and minute hands at 3:15?

Accepted Answer

7.5 degrees. At 3:15, the minute hand is at the 3 (90 degrees). The hour hand moves 0.5 degrees per minute, so in 15 minutes past 3 it has moved 7.5 degrees beyond the 3. The gap is 7.5 degrees.

Question 10

A blind man has 10 black socks and 10 navy blue socks in a drawer. How many must he pull out to guarantee a matching pair?

Accepted Answer

3 socks. By the Pigeonhole Principle, with only 2 colors, the 3rd sock must match one of the first two.

Question 11

You have two ropes, each burns for exactly 1 hour but not uniformly. How do you measure exactly 45 minutes?

Accepted Answer

Light rope 1 from both ends and rope 2 from one end simultaneously. Rope 1 burns out in 30 minutes. At that moment, light the other end of rope 2. Rope 2 had 30 minutes of burn left, but now burning from both ends it takes 15 more minutes. Total: 30 + 15 = 45 minutes.

Question 12

You have 10 people in a room. How many total handshakes if they all shake hands?

Accepted Answer

45. (10 choose 2) = 10! / (2! × 8!) = 45 — this is the total number of ways two people can shake hands.

Question 13

A group of people wants to determine their average salary on the condition that no individual would be able to find out anyone else's salary. Can they accomplish this, and, if so, how?

Accepted Answer

Yes. The first person thinks of a random number X and adds it to their salary. The rest of the group simply adds their salary to the running total. Then, the first person subtracts X and divides by the group size to get the average.

Question 14

You have a 3 gallon jug and 5 gallon jug, how do you measure out exactly 4 gallons?

Accepted Answer

Fill the 3-gallon jug and pour into the 5-gallon. Fill the 3-gallon again and pour until the 5-gallon is full (1 gallon remains). Empty the 5-gallon, pour in the 1 gallon, then fill the 3-gallon and add it. 1 + 3 = 4 gallons.

Question 15

You're in a room with three light switches, each controls one of three bulbs in the next room. All lights are off. You can only inspect the other room once. How do you determine which switch controls which bulb?

Accepted Answer

Turn switch 2 on for ten minutes, then turn it off and quickly turn on switch 3. Go into the room. The lit bulb is switch 3. The off but warm bulb is switch 2. The remaining cold bulb is switch 1.

Question 16

Determine the angle (in degrees) between the minute and hour hands at 9:45 PM on a standard clock.

Accepted Answer

At 9:45 PM, the minute hand is directly on the 9. The hour hand is $ \frac{3}{4} $ of the way between 9 and 10, as $ \frac{3}{4} $ of an hour has passed since 9 o'clock. The angle between the two hands is $ \frac{3}{4} 	imes 30^\circ = 22.5^\circ $. This is because there are 12 equally spaced numbers on the clock, making the total degrees in a circle 360°. Consequently, each hour mark is separated by 30° ($ 360^\circ \div 12 = 30^\circ $). Thus, the angle between the hands at 9:45 PM is 22.5°.

Question 17

A clock is observed to have the hour hand exactly at some minute marker, and the minute hand six minutes ahead of it. Later, the hour hand is observed to be exactly at some other minute mark, and the minute hand seems to be seven minutes ahead of it. How many minutes have passed between the two observations?

Accepted Answer

Define m, p and h where h represents the hour on the clock,

m represents the minute on the clock,

p represents the minute mark the hour hand is positioned at.

As there are five minute marks per hour, the position of the hour hand (p) is given by 5*h + m/12.

For the minute hand to be exactly 6 minutes ahead of the hour hand, we need to satisfy this equation:

m - p = 6,

or

m - 5*h - m/12 = 6.

Solving this equation for m=12 (corresponding to the minute mark '1' on the clock face) gives:

12 - 5h - 12/12 = 6,

or

5 = 5h,

which has a unique solution h=1. Therefore, the first observation must have been at (1:12).

In a similar fashion, for the minute hand to be 7 minutes ahead of the hour hand, we need to satisfy the equation:

m - p = 7,

or

m - 5*h - m/12 = 7.

Solving this equation for m=24 (corresponding to the minute mark '2' on the clock face) gives a unique solution for h=3. Therefore, the second observation must have been at (3:24).

So, the time elapsed between the two observations is 2 hours and 12 minutes, or 132 minutes.

Question 18

Given a conventional clock with an hour and a minute hand that align perfectly at 12:00 midnight, when will they coincide next? How many times a day will they overlap?

Accepted Answer

Let's determine when the hour and minute hands of a clock will next overlap after 12:00 midnight.

The minute hand completes one full cycle ($360$ degrees) every $60$ minutes, while the hour hand completes the same cycle every $12$ hours. This means that for every minute, the minute hand moves $360/60 = 6$ degrees, and the hour hand moves $360/720 = 0.5$ degrees.

If $t$ represents the time in minutes after 12:00, then in $t$ minutes, the minute hand will have moved $6t$ degrees, and the hour hand will have moved $0.5t$ degrees.

For the two hands to overlap, their angular difference should be a multiple of $360$ degrees. So, the relative motion between the two hands is: $$6t - 0.5t = 5.5t.$$ When the minute hand has gained $360$ degrees over the hour hand, they will overlap. Thus: $$5.5t = 360,$$ $$t = 360/5.5 = 65.45.$$ This is $65$ minutes and $27$ seconds. This means the hands will overlap at approximately 1:05:27 after midnight.

Now, for how many times they overlap in a day:

The hands of the clock will overlap $11$ times in a $12$-hour period (they won't overlap at $11/12^{th}$ of the hour, as the minute hand is slightly ahead of the hour hand). Therefore, in a $24$-hour period, the hands will overlap $22$ times.

Question 19

5 people are sitting in an airport waiting to travel on holiday to 1 of 5 different locations. Each of them has a unique departure time, color shoes and item they carry.
1. Adam leaves at 6pm and takes a hat with him.
2. The person with black shoes is in the middle.
3. James leaves at 9pm.
4. Jess wears blue shoes, and is left of the person that is carrying a hat.
5. To the right of the person carrying a watch is a person going to Las Vegas.
6. Wilson is heading for Paris.
7. Next to the person carrying a book is a person that is wearing green shoes.
8. The person going to Thailand leaves at 5pm.
9. Ruby leaves at 7pm and is to the right of the person going to Las Vegas.
10. The person wearing red shoes is going to Dublin.
11. Next to the person leaving at 7pm is a person wearing white shoes.
12. The person on the end carries a laptop.
13. The person wearing black shoes leaves at 8pm.
14. The person carrying a laptop is next to the person carrying a book.
15. The person going to Dublin leaves at 6pm.
Who is going to Singapore?

Accepted Answer

First, let's organize the relevant variables:

People:
 Adam, James, Jess, Wilson, Ruby

Departure times:
 5pm, 6pm, 7pm, 8pm, 9pm

Shoe colors:
 black, blue, green, red, white

Items:
 hat, watch, book, laptop, unspecified fifth item

Destinations:
 Las Vegas, Paris, Thailand, Dublin, Singapore

Next, let us consider what we know about each individual:

Adam:

- leaves at 6pm (cf. 1)

- carries hat (cf. 1)

- goes to Dublin (cf. 15)

- wears red shoes (cf. 10)

James:

- leaves at 9pm (cf. 3)

Jess:

- wears blue shoes (cf. 4)

- left of Adam (cf. 4)

Wilson:

- goes to Paris (cf. 6)

Ruby:

- leaves at 7pm (cf. 9)

From this we can then deduce that either Wilson or Jess departs at 5pm and the other departs at 8pm. Statement #8 then implies that Jess must be the one departing at 5pm and going to Thailand. Since Wilson then departs at 8pm he must also be the one wearing black shoes and is seated in the middle (cf. 8 & 2).

The next step is then to assign some individuals to their positions. We know that the person carrying the laptop is on either end (cf. 12) and the person carrying a book is next to them (cf. 14). This implies that the person carrying the laptop is wearing green shoes (cf. 7) since the person in the middle is known to be wearing black shoes (cf. 2). This allows us to preclude Adam being on the right of Wilson (i.e. the middle) since that would leave no way Jess to be to the left of him (cf. 2).

We now know that the order of the individuals is Jess, Adam, Wilson followed by two others. The person to the right of Wilson is going to Las Vegas (cf. 5) and Ruby is to the right of them (cf. 9).

Since we now know that Jess, Adam, Wilson and James are going to Thailand, Dublin, Paris, and Las Vegas, respectively, it must follow that Ruby is the one going to Singapore.

IMPORTANT:
 we are getting reports from people that the original question does not specify that departure times are unique. We don't see how the problem remains solvable in that case, but please let us know if you do.

Question 20

You have an empty screen and a stopwatch. In one second you may either
(a) type a single "x",
(b) copy the current block of x’s
(c) paste the clipboard.
Starting from nothing, what is the minimum number of seconds needed to produce at least 200 x’s?
What about at least one trillion ($10^{12}$) x’s?

Accepted Answer

Step 1: Model one batch: Copy once and paste k times ⇒ multiplier $k+1$ in $k+1$ seconds.

The efficiency per second is $(k+1)^{1/(k+1)}$, maximised at $k=2$, i.e. Copy–Paste–Paste (tripling in 3 s).

Step 2: Find a factorisation of the target that minimises the sum of batch lengths, using mostly 3’s and at most one 2 or 4.

A) Target $\ge 200$

Choose product $3^3 	imes 4 	imes 2 = 216$.

Time = $3	imes3 + 4 + 2 = 15$ seconds.

B) Target $\ge 10^{12}$

Choose product $3^{24} 	imes 4 \approx 1.13	imes10^{12}$.

Time = $24	imes3 + 4 = 76$ seconds.

No combination of batches can beat these sums, because replacing a 3-batch (3 s) with any longer or shorter batch either reduces the product per second or fails to reach the goal.

$$\boxed{	ext{200 x’s: 15 s}\quad	ext{  1 trillion x’s: 76 s}}$$

Question 21

How many times does the digit $9$ appear when counting from $1$ to $1000$?

Accepted Answer

A fast method is to pad numbers to three digits and count from $000$ to $999$. There are $1000$ strings of length $3$, giving $3000$ digit-positions in total. By symmetry, each of the $10$ digits appears equally often across all positions, so each digit occurs $3000/10 = 300$ times; in particular, $9$ occurs $300$ times on $000$–$999$. Switching back to $1$–$1000$ does not change the count because neither $000$ nor $1000$ contains a $9$. Therefore, the digit $9$ appears $300$ times.

Question 22

A craftsman has a wooden block shaped as a cube, measuring three inches on every side. He plans to slice it into 27 smaller cubes, each one-inch on a side. He knows he can achieve this in six straight cuts while keeping the pieces together. But, if he repositions the pieces after every slice, can he decrease the number of cuts?

Accepted Answer

No, the minimum number of cuts required is six. Consider the cube in the very center of the wooden block. This cube is surrounded on all its six faces. Each face would need a separate cut to be exposed, regardless of how the other pieces are rearranged.

Question 23

A traveler is on his way back from a village fair. With him, he has a cat, a mouse, and a piece of cheese. When he reaches a stream, he must use a tiny raft only big enough for him and one other item. Unfortunately, if the cat is left alone with the mouse, it will eat it, and similarly, the mouse will eat the cheese if left alone with it. How can the traveler cross the stream?

Accepted Answer

The traveler can navigate this tricky situation with a few carefully planned trips across the stream:

Traveler takes the mouse across the stream first.

On the other side of the stream, we now have the traveler and the mouse. On the starting side, we have the cat and the cheese. This is safe because the cat won’t eat the cheese.

Traveler returns alone to the original side.

On the other side of the stream, we have just the mouse. On the original side, we have the traveler, the cat, and the cheese.

Traveler takes the cat across the stream next.

On the other side of the stream, we have the traveler and the cat. The mouse is also on the other side, but since the traveler is with the cat, the cat won't eat the mouse. On the original side, we only have the cheese.

Traveler returns with the mouse.

On the other side of the stream, we have the cat. On the original side, we have the traveler, the mouse, and the cheese.

Traveler takes the cheese across the stream.

On the other side of the stream, we now have the cat and the cheese. On the original side, we have the traveler and the mouse.

Traveler returns alone to the original side.

On the other side of the stream, we have the cat and the cheese. On the original side, we have the traveler and the mouse.

Traveler takes the mouse across the stream.

Now, on the other side of the stream, we have the cat, the mouse, and the cheese, while the traveler is also there ensuring they coexist peacefully.

In this way, the cat, the mouse, and the cheese all make it safely to the other side of the stream without any of them being eaten.

Question 24

Three players take turns marking points on a 1-meter-long cylinder. Each player's goal is to maximize the length of the segment their point belongs to after all points are placed and midpoints between adjacent points are marked, dividing the cylinder.
Which player’s position (first, second, or third) gives the best outcome if all players act optimally?

Accepted Answer

Let the cylinder be represented as a circle of total length 1. When each player places a point, the circle is divided into arcs between consecutive points. The boundary between two players' territories lies at the midpoint of the arc connecting their points. Hence, if the three arcs between the points are $ g_1, g_2, g_3 $ with $ g_1 + g_2 + g_3 = 1 $, the territory of each player equals half the sum of the two adjacent gaps, i.e. $ 	frac{g_i + g_{i+1}}{2} $.

Suppose Player 1 places their point $A$ first at an arbitrary position. Player 2 can then place their point $B$ directly opposite $A$, creating two equal arcs of length $ 	frac{1}{2} $.

Now when Player 3 chooses their point $C$, they can only place it somewhere along one of these two half-arcs. Regardless of where they place it, the segment they control will be exactly half of the larger arc in which they sit. Since the largest arc is $ 	frac{1}{2} $, the maximum territory Player 3 can guarantee is $ 	frac{1}{4} $.

Thus, Player 3 cannot ensure a segment of length $ 	frac{1}{3} $; the best they can do is $ 	frac{1}{4} $. In contrast, Player 2, by choosing second and placing opposite Player 1, can limit Player 3’s options and often secure a larger share themselves (for example, if Player 3 sits midway in one half, Player 1 and Player 2 each get $ 	frac{3}{8} $ while Player 3 gets $ 	frac{1}{4} $).

Therefore, the optimal position is to be the second player.

Question 25

You and your partner are presented with a challenge using a standard deck of cards. Five cards will be drawn randomly from the deck. Your task is to devise a system where you can hand over four of these five cards to your partner. Using only the information conveyed through these four cards, your partner must be able to deduce the identity of the fifth (hidden) card.
Is there a method that guarantees success every time? If so, explain how it works.

Accepted Answer

Yes, there is a foolproof method:

Since you're picking $5$ cards from a deck, by the Pigeonhole Principle, you are guaranteed to have at least two cards of the same suit. The first card you hand over to your partner can be used to signify the suit of the hidden card.

The next step is to encode the denomination (i.e., the number or face value) of the hidden card. Consider a predetermined order of the cards (from Ace to King). The three remaining cards can represent a number between $ 1 $ and $ 6 $ based on their arrangement. For example, if Ace represents $ 1 $ and King represents $ 13 $, the three cards' order can be used to denote a certain value between $ 1 $ and $ 6 $.

When your partner receives the first card, they’ll know the suit of the hidden card. With the next three cards, they can decode the value they represent ($ 1-6 $). To find the exact denomination of the hidden card, your partner will add this number ($ 1-6 $) to the denomination of the first card you provided, taking the result modulo $ 13 $. This ensures they determine the exact card that is hidden, as the maximum distance modulo $ 13 $ between two denominations is $ 6 $.

With this strategy, your partner can always guess the fifth card correctly.

Quantitative Finance

Quantitative Finance

Quantitative Finance

Quantitative Finance