Mathematical Puzzles

Mathematical puzzles have always been a source of joy, mental stimulation, and problem-solving challenges. They not only enhance logical thinking but also build problem-solving skills.

List of Mathematical Puzzles

The puzzles listed below are designed to be engaging, thought-provoking, and often include a twist that requires both creative and analytical thinking.

1. Ages of the daughters
2. Total distance travelled by the bee
3. Ways to reach the bottom right
4. Measuring 45 minutes with two burning wires
5. Torch and Bridge
6. Two Eggs and Hundred Floors
7. Probability of White Ball
8. Poison and Rat
9. Hourglasses Puzzle
10. The boy preference ratio riddle
11. Car Wheel
12. Maximum Chocolates
13. Splitting a Cake
14. Rs 500 Note
15. Girl or Boy
16. Know Average Salary without Disclosing
17. Maximum run in cricket
18. Completion of Task
19. Missing Row in Excel
20. Four People on a Rickety Bridge
21. Man fell in well
22. Fifty red marbles and Fifty blue marbles
23. Three Equilateral Triangles
24. Ten identical bottles of pills
25. The Circle cutting puzzle
26. Chain Link
27. The fake note
28. The two Egg drop
29. Snail and Wall
30. Thousand light bulbs switched on/off
31. Four Alternating Knights
32. The Nine-Dot Continuous Path
33. Hundred Cows And Milk
34. One Mile on the Globe
35. The Three-Square Matchstick Challenge
36. The Counters and Board
37. Camel and Banana
38. Six Matches , Right Foot Forward
39. Initial Money
40. Two Creepers Climbing a Tree

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5 Most Common Mathematical Puzzles

Given below are the some famous puzzles

1) The Monty Hall Problem

Suppose you're on a game show, and you're given the choice of three doors: Behind one door is a car; behind the others, goats. You pick a door, say No. 1, and the host, who knows what's behind the doors, opens another door, say No. 3, which has a goat. He then says to you, "Do you want to pick door No. 2?" Is it to your advantage to switch your choice?

Solution: There are two cases here :

1. Switching
2. Not switching

Case 1 :

If we know we are switching, we need to select a door which has a goat in order to win the car. As we select a door having a goat, the host should only open the door that have the another goat so the remaining door has a car which we get by switching.

So, probability of selecting a door which has a goat is 2/3 as 2 doors out of 3 have goats. Therefore probability of winning a car by switching is 2/3.

Case 2 :

If we know we are not switching, we need to select a door which has a car in order to win the car. 

So probability of selecting a door which has a car is 1/3 as 1 door out of 3 has car. Therefore probability of winning a car by not switching is 1/3.

As probability of winning a car by switching is higher than not switching. It is advantage to switch.

2) Measure 4 Liter with 3 and 4 Liter Bottle

You have two bottles: one that can hold exactly 3 liters of water and another that can hold exactly 5 liters.

• Your goal is to measure out precisely 4 liters of water using only these two bottles, with no additional measuring devices or markings available on the bottles.
• You can fill the bottles to their full capacity, empty them completely, or pour water from one bottle to the other until one is full or the other is empty.

By relying solely on clever pouring strategies and logical thinking, how can you achieve exactly 4 liters of water in one of the bottles?

Measuring out exactly 4 liters of water using a 3-liter and a 5-liter container and a tap involves the following steps:

Step 1: Fill the 5-liter container from the tap.

Step 2: Pour the 5 liters into the 3-liter container until it's full, leaving 2 liters in the 5-liter container.

Step 3: Empty the 3-liter container.

Step 4: Pour the remaining 2 liters from the 5-liter container into the 3-liter container.

Step 5: Fill the 5-liter container again.

Step 6: Pour water from the 5-liter container into the 3-liter container until it's full. Since only 1 liter was needed, the 5-liter container will be left with 4 liters of water.

This is the best approach with the least number of steps to measure the water in the container to be exact, 4 liters.

We can measure 4 liters using another approach, which is discussed here → [Check here!].

3) Hundred Doors

There are 100 doors in a row, and all doors are initially closed. A person walks through all doors multiple times and toggles (if open, then close; if closed, then open)

• In the first walk, the person toggles (or opens) every door.
• In the second walk, toggles (or closes) every second door (i.e., 2nd, 4th, 6th, 8th, and so on).
• In the third walk, toggles every third door (i.e., 3rd, 6th, 9th, etc.).

This pattern continues, and in the 100th walk, the person toggles only the 100th door.

Solution: 

• A door is toggled in the i-th walk if i divides the door number, for example: Door number 45 is toggled during the 1st, 3rd, 5th, 9th, 15th, and 45th walks.
• Each door is toggled once for every divisor of its number, and divisors typically come in pairs (e.g., for 45: (1,45), (3,15), (5,9)).
• Each pair of divisors cancels out the toggle effect (open - close or close -open), therefore, doors with an even number of divisors return to their initial closed state.
• Perfect square numbers (e.g., 16) have one unpaired divisor (like 4 in 4×4), resulting in an odd number of divisors.
• An odd number of toggles leaves the door in the open position; hence, only perfect square-numbered doors (e.g., 1, 4, 9, 16, ..., 100) remain open.
• Prime numbers (e.g., 2, 3, 5, 7) have exactly two divisors (1 and itself), which is a pair - the door remains closed.
• Non-square composite numbers (e.g., 15) have divisor pairs, so they are also closed at the end.

⁛So the answer is 1, 4, 9, 16, 25, 36, 49, 64, 81 and 100. 

4) Torch and Bridge

There are 4 people (A, B, C, and D) who want to cross a bridge at night.

• A, B, C and D take 1, 2, 5 and 8 minutes respectively to cross the bridge.
• There is only one torch with them, and the bridge cannot be crossed without the torch.
• There cannot be more than two people on the bridge at any time, and when two people cross the bridge together, they must move at the slower person's pace.

Can they all cross the bridge in 15 minutes?

Solution: They must cross the bridge in the following way:

Step 1: A and B cross the bridge. A comes back. Time taken is 3 minutes. Now B is on the other side.

Step 2: C and D cross the bridge. B comes back. Time taken 8 + 2 = 10 minutes. Now C and D are on the other side.

Step 3: A and B cross the bridge. Time taken is 2 minutes. All are on the other side.

Total time spent: 3 + 10 + 2 = 15 minutes.

To minimize the time:

The trick here is that the persons with the fastest speeds only should come back (and that too only if there is a need to come back, as here we need to bring back the torch). A comes back in step-1 and B comes back in step-2. And, finally reduce the number of traveling back, like C, D does not come back.

5) Birthday Paradox

On a New Year's day, two old friends (A and B) meet at a party. As they met after a long time, person B wanted person A to guess his birthday. As both friends have not been in touch for a long time, person A was unable to guess his birthday. So person B decided to give some hints. Below are the hints:

1. The day before yesterday I was 25 and next year I will be 28, then person B asked person A to find his birthday. 
2. The above condition can be true for only one day in the year.

Solution: Person B's birthday is on December 31.

Explanation:

Person A met his old friend on New Year's, which is on January 1. So, the day before yesterday was December 30 at that time, person B was 25 years old, and on the present Day(January 1), B is 26 years old. This year, on December 31, person B will be 27; his age will be 28.

Therefore, Person B's birthday is on December 31.