Deduction 7 - Puzzle Prime

Unfaithful

Section: Brain Teasers Category: Deduction

Difficulty: Medium

Jack is looking at Anne, but Anne is looking at George. Jack is married, but George is not. Is a married person looking at an unmarried person?

Yes
No
Cannot be determined

SOLUTION

The answer is YES. If Anne is unmarried, then Jack is married and is looking at an unmarried person. If not, then she is married and is looking at an unmarried person.

Unknown Author August 22, 2019

0 Comments

Vowels and Even Numbers

Section: Brain Teasers Category: Deduction

Difficulty: Medium

“If there is a vowel written on one side of a card, then there is an even number written on the other side.”
How many of these four cards do you need to flip in order to check the validity of this sentence?

What would the answer be if you know that each card contains one letter and one number?

SOLUTION

You need to flip all cards except for the second one. If each card contains one letter and one number, then you need to flip only A and 7.

Peter Wason July 19, 2019

15 Comments

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Three Daughters

Section: Brain Teasers Category: Deduction

Difficulty: Hard Tags: Featured

Two friend mathematicians meet each after a long time and have the following conversation:

– I have 3 daughters, the product of their ages is 36.
– I can’t figure out how old they are, can you tell me more?
– Sure, the sum of their ages is equal to the number of my house.
– I know your house number, but still can’t figure out the ages of your daughters.
– Also, my eldest daughter is called Monica.
– OK, now I know how old your daughters are.

What ages are the three daughters of the mathematician?

SOLUTION

Using the first clue, we find that there are 8 possibilities:

(1, 1, 36) -> sum 38
(1, 2, 18) -> sum 21
(1, 3, 12) -> sum 16
(1, 4, 9) -> sum 14
(1, 6, 6) -> sum 13
(2, 2, 9) -> sum 13
(2, 3, 6) -> sum 11
(3, 3, 4) -> sum 10

Since the second mathematician couldn’t guess the ages even after the second clue, the sum has to be 13. Therefore the only possible options are (1, 6, 6) and (2, 2, 9). However, the third clue suggests that there is an “eldest” daughter and then the correct answer is 2, 2 and 9.

Unknown Author June 26, 2019

1 Comment

The Warden and the Three Doors

Section: Brain Teasers Category: Deduction

Difficulty: Hard

An evil warden holds you as a prisoner but offers you a chance to escape. There are 3 doors A, B, and C. Two of the doors lead to freedom and the third door leads to lifetime imprisonment, but you do not which door is what type. You are allowed to point to a door and ask the warden a single yes-no question. If you point to a door that leads to freedom, the warden does answer your question truthfully. But if you point to the door that leads to imprisonment, the warden answers your question randomly, saying either “YES” or “NO” by chance. Can you figure out a way to escape the prison?

SOLUTION

You can point towards door A and ask whether door B leads to freedom. If the warden says “YES”, then you open door B. It can not lead to imprisonment because this would mean that door A leads to freedom and the warden must have told you the truth. If the warden says “NO”, then you open door C. This is because either the warden lied, and then the imprisonment door is A, or he told you the truth, and then the imprisonment door is B.

Unknown Author March 15, 2019

7 Comments

The Temple of Doom

Section: Brain Teasers Category: Deduction

Difficulty: Advanced

You and eight of your team members are trying to escape the Temple of Doom. You are running through a tunnel away from a deadly smoke and end up in a large hall. There are four paths ahead, and exactly one of them leads to the exit. It takes 20 minutes to explore any of the four paths one way, and your group has only 60 minutes until the deadly smoke suffocates you. The problem is that two of your friends are known to be delirious and it is possible that they do not tell the truth, but nobody knows which ones they are. How should you split the group and explore the tunnels, so that you have enough time to figure out which is the correct path and escape the temple?

SOLUTION

You explore the first path. You send two of your teammates to explore the second path. You send the remaining six teammates in groups of three to explore each of the two remaining paths. If your path leads to the exit, then everything is good. Otherwise, you ask the two groups of three whether their paths lead to the exit. If in both groups everyone answers consistently, then nobody is lying, and you will escape. If in both groups there is a person whose answer is different from the others in the group, then the majority in both groups says the truth. Once again, you will know which path leads to the exit. Finally, if in exactly one of the groups everyone answers consistently, you ask the group of two. If the team members there answer consistently with each other, then they say the truth. You will have two groups which tell the truth and will know which path leads to the exit. If the answers of the teammates in the group of two differ, then in the inconsistent group of three the majority will be saying the truth. Again, you will be able to deduce which path leads to the exit.

Unknown Author February 28, 2019

14 Comments

25 Horses

Section: Brain Teasers Category: Deduction

Difficulty: Advanced

There are 25 horses and you want to find the fastest 3 among them. You can race any 5 of the horses against each other and see the final standing, but not the running times. If all the horses have constant, permanent speeds, how many races do you need to organize in order to find the fastest 3?

SOLUTION

Let us label the horses H1, H2, H3, H4, …, H24, H25.

We race H1 – H5 and (without loss of generality) find that H1 > H2 > H3 > H4 > H5. We conclude that H4, H5 are not among the fastest 3.

We race H6 – H10 and (without loss of generality) find that H6 > H7 > H8 > H9 > H10. We conclude that H9, H10 are not among the fastest 3.

We race H11 – H15 and (without loss of generality) find that H11 > H12 > H13 > H14 > H15. We conclude that H14, H15 are not among the fastest 3.

We race H16 – H20 and (without loss of generality) find that H16 > H17 > H18 > H19 > H20. We conclude that H19, H20 are not among the fastest 3.

We race H21 – H25 and (without loss of generality) find that H21 > H22 > H23 > H24 > H25. We conclude that H24, H25 are not among the fastest 3.

We race H1, H6, H11, H16, H21 and (without loss of generality) find that H1 > H6 > H11 > H16 > H21. We conclude that H16, H21 are not among the fastest 3.

Now we know that H1 is the fastest horse and only H2, H3, H6, H7, H11 could complete the fastest three. We race them against each other and find which are the fastest two among them. We complete the task with only 7 races in total.

Unknown Author January 1, 2019

6 Comments

Calendar Puzzle

Section: Brain Teasers Category: Deduction

Difficulty: Medium

What numbers should you put on the sides of two cubes, so that every date of the month can be expressed using these cubes?

SOLUTION

You put the numbers 0, 1, 2, 3, 4, 5 on the first die and the numbers 0, 1, 2, 7, 8, 6/9 on the second die.

Unknown Author November 4, 2018

0 Comments

Donuts and Candies

Section: Brain Teasers Category: Deduction

Difficulty: Easy

Huey has 3 donuts, Dewey has 5 donuts. Louie comes along and the three of them split the donuts equally. In exchange, Louie offers 8 candies to Huey and Dewey. What is the fair way to split the candies?

SOLUTION

Huey must take 1 chocolate, and Dewey must take 7. This is because each of them ate 8/3 donuts, and therefore Huey gave away 1/3 of a donut, whereas Dewey gave away 7/3 of a donut.

Unknown Author September 21, 2018

2 Comments

False Statements

Section: Brain Teasers Category: Deduction

Difficulty: Medium

None of these statements is correct.
At most 1 of these statements is correct.
At most 2 of these statements are correct.
…
At most 98 of these statements are correct.
At most 99 of these statements are correct.

How many of these statements are correct?

SOLUTION

If the number of true statements is X, then statements 1, 2, … , X are wrong, and the rest are correct. Therefore X = 100 – X and X = 50. Thus, there are 50 correct statements.

Unknown Author September 16, 2018

2 Comments

Mastermind

Section: Brain Teasers Category: Deduction

Difficulty: Easy

You must find a 3-digit number. You know that:

682 shares one digit with the number, and it is well placed;
614 shares one digit with the number, but it is wrongly placed;
296 shares two digits with the number, but they are wrongly placed.

What is the number?

SOLUTION

The first and second clues imply that the 6 is not a part of the number. The third clue then implies that 2 and 9 are parts of the number. Now the first clue implies that 2 is the last digit of the number, and the third clue implies that 9 is the first digit of the number. Finally, the second clue implies that 4 is the second digit of the number. Therefore, the number is 942.

Unknown Author September 7, 2018

5 Comments