Deduction 8 - Puzzle Prime

Trolls and Cakes

Section: Brain Teasers Category: Deduction

Difficulty: Medium

You are on your way to visit your Grandma, who lives at the end of the valley. It’s her birthday, and you want to give her the cakes you’ve made.

Between your house and her house, you have to cross 7 bridges, and as it goes in the land of make-believe, there is a troll under every bridge! Each troll, quite rightly, insists that you pay a troll toll. Before you can cross their bridge, you have to give them half of the cakes you are carrying, but as they are kind trolls, they each give you back a single cake.

How many cakes do you have to leave home with to make sure that you arrive at Grandma’s with exactly 2 cakes?

SOLUTION

You leave with 2 cakes. Every time you cross a bridge, you give one of them to a troll, and then get it back. Eventually, you will arrive at your grandma with exactly 2 cakes.

Unknown Author June 27, 2018

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The Candy Shop

Section: Brain Teasers Category: Deduction

Difficulty: Medium

A candy shop lets children exchange 3 chocolate wrappers for a brand new chocolate. Willy is walking around town collecting chocolate wrappers from people. How many wrappers must he collect in order to eat 10 chocolates?

SOLUTION

He needs only 21 wrappers. He uses them to get 7 chocolates. After he eats them, he is left with 7 wrappers, which uses to get 2 more chocolates. After he eats these, he is left with 3 wrappers, which are enough to get one more chocolate and make them 10 in total.

Unknown Author June 18, 2018

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King Octopus and His Servants

Section: Brain Teasers Category: Deduction

Difficulty: Medium

King Octopus has servants with 6, 7, or 8 legs. The servants with 7 legs always lie, and the servants with 6 or 8 legs always tell the truth. One day four of the king’s servants had the following conversation:

“We together have 28 legs,” said the first octopus.
“We together have 27 legs,” said the second octopus.
“We together have 26 legs,” said the third octopus.
“We together have 25 legs,” said the fourth octopus.

Which of the four servants told the truth?

SOLUTION

If all four octopuses lied, then their total number of legs must be 28, which is impossible. Therefore exactly one of them said the truth, and the other three had 7 legs each. Since the truthful octopus must have 6 or 8 legs, and 21 + 8 = 29, we see that it has exactly 6 legs, and therefore it is the second one.

Unknown Author June 5, 2018

4 Comments

Gold and Nickel

Section: Brain Teasers Category: Deduction

Difficulty: Expert Tags: Weighings

You have 15 identical coins – 2 of them made of pure gold and the other 13 made of nickel (covered with thin gold layer to mislead you). You also have a gold detector, with which you can detect if in any group of coins, there is at least one gold coin or not. How can you find the pure gold coins with only 7 uses of the detector?

SOLUTION

First, we note that if we have 1 gold ball only, then we need:

1 measurement in a group of 2 balls
2 measurements in a group of 4 balls
3 measurements in a group of 8 balls

Start by measuring 1, 2, 3, 4, 5.

If there are gold balls in the group, then measure 6, 7, 8, 9, 10, 11.
- If there are gold balls in the group, then measure 5, 6, 7.
  - If there are no gold balls among them, then there is a gold ball among 1, 2, 3, 4, and a gold ball among 8, 9, 10, 11, so we can find the gold balls with the remaining 2 measurements.
  - If there are gold balls in 5, 6, 7, then measure 5, 8, 9. If there are gold balls there, then 5 must be gold, and we can find the other gold ball among 6, 7, 8, 9, 10, 11 with the remaining 3 measurements. If there is no gold ball among 5, 8, 9, then there is a gold ball among 1, 2, 3, 4, and a gold ball among 6, 7, so again we can find them with only 3 measurements.
- If there are no gold balls in the group, then measure 5, 12, 13.
  - If there are no gold balls among them, then measure 14, 15. If none of them is gold, then measure individually 1, 2, and 3 to find which are the 2 gold balls among 1, 2, 3, 4. Otherwise, there is a gold ball among 1, 2, 3, 4, and among 14, 15, and we can find them with the remaining 3 measurements.
  - If there are gold balls among 5, 12, 13, then measure 5, 14, 15. If none of them is gold, then there is a gold ball among 1, 2, 3, 4, and a gold ball among 12, 13, so we can find them with 3 measurements. Otherwise, 5 is gold, and again we can find the other gold ball among 1, 2, 3, 4, 12, 13, 14, 15 with 3 measurements.
If there are no gold balls among 1, 2, 3, 4, 5, then we measure 6, 7, 8.
- If there are gold balls in the group, then measure 9, 10, 11, 12, 13.
  - If there are no gold balls among them, we measure individually 6, 7, 8, 14.
  - If there is a gold ball among 9, 10, 11, 12, 13, then there is another one among 6, 7, 8. We measure 8, 9. If none of them is gold, then we can find the gold among 6, 7, and the gold among 10, 11, 12, 13, with 3 measurements total. If there is a gold ball among 8, 9, then we measure 10, 11, 12, 13. If none of them is gold, then 9 is gold and we find the other gold ball among 6, 7, 8 with 2 more measurements. If there is a gold ball among 10, 11, 12, 13, then we can find it with 2 measurements. The other gold ball must be 8.
- If there are no gold balls in the group, then measure 9, 10.
  - If there are no gold balls among them, then measure individually 11, 12, 13, 14.
  - If there are gold balls among 9, 10, then measure 11, 12, 13, 14. If there is a gold ball among them, then there is another one among 9, 10, and we can find them both with 3 measurements. Otherwise, we measure 9 and 10 individually.

Unknown Author March 24, 2018

11 Comments

Students with Hats

Section: Brain Teasers Category: Deduction

Difficulty: Hard

Professor Vivek decided to test three of his students, Frank, Gary and Henry. The teacher took three hats, wrote on each hat a positive integer, and put the hats on the heads of the students. Each student could see the numbers written on the hats of the other two students but not the number written on his own hat.

The teacher said that one of the numbers is sum of the other two and started asking the students:

— Frank, do you know the number on your hat?
— No, I don’t.
— Gary, do you know the number on your hat?
— No, I don’t.
— Henry, do you know the number on your hat?
— Yes, my number is 5.

What were the numbers which the teacher wrote on the hats?

SOLUTION

The numbers are 2, 3, and 5. First, we check that these numbers work.

Indeed, Frank would not be able to figure out whether his number is 2 or 8. Then, Gary would not be able to figure out whether his number is 3 or 7, since with numbers 2, 7, 5, Frank still would not have been able to figure his number out. Finally, Henry can conclude that his number is 5, because if it was 1, then Gary would have been able to conclude that his number is 3, due to Frank’s inability to figure his number out.

Next, we we check that there are no other solutions. We note that if the numbers are 1, 4, 3, or 3, 2, 1, or 4, 1, 3, neither Frank nor Gary would have been able to figure their number out. Therefore, if the numbers were 1, 4, 5, or 3, 2, 5, or 4, 1, 5, Henry would not have been able to figure his number out. Thus, 5 is not the largest number.

Similarly, if the numbers are X, X + 5, X + 10, or X + 5, X, X + 10, once again, neither Frank nor Gary would have been able to figure their number out. Therefore, if the numbers were X, X + 5, 5, or X + 5, X, 5, Henry would not have been able to figure his number out.

Unknown Author March 20, 2018

2 Comments

The Ping Pong Puzzle

Section: Brain Teasers Category: Deduction

Difficulty: Advanced

Three friends – A, B, and C, are playing ping pong. They play the usual way – two play at a time, the winner stays on the table, the loser lets the third one play. If you know that A played 10 matches in total, B played 15 matches in total, and C played 17 matches in total, who lost the second game?

SOLUTION

A lost it. Since there have been (10 + 15 + 17) / 2 = 21 games played in total, and each player never misses 2 games in a row, the only way for A to play just 10 games is if he plays the 2nd, 4th, 6th, etc. games, and every time loses.

Unknown Author February 26, 2018

1 Comment

The Troll Brothers

Section: Brain Teasers Category: Deduction

Difficulty: Expert

There are four troll brothers – Wudhor, Xhaqan, Yijlob, and Zrowag.

Wudhor always says the truth.
Xhaqan always lies.
Yijlob lies or says the truth unpredictably.
Zrowag is deaf and never answers.

You must ask these brothers four YES/NO questions (one troll per question), and figure out their names. What questions would you ask?

SOLUTION

Coming soon.

Source:

Puzzling StackExchange

Lynn February 25, 2018

7 Comments

Broken Clock

Section: Brain Teasers Category: Deduction

Difficulty: Medium

An old wall clock falls on the ground and breaks into 3 pieces. Describe the pieces, if you know that the sum of the numbers on each of them is the same.

SOLUTION

The total of all numbers on the clock is 1 + 2 + … + 12 = 78. Therefore each piece must contain numbers with total sum 26. The only way for this to happen is if the pieces are broken via 3 parallel lines – {1, 2, 11, 12}, {3, 4, 9, 10}, {5, 6, 7, 8}.

Unknown Author December 10, 2017

1 Comment

Rubik’s Chess

Section: Brain Teasers Category: Deduction

Difficulty: Hard

Last week we found out that Puzzle Pranks Co. have invented a new type of puzzle – Rubik’s Chess. The goal is simple – you get a scrambled cube with various chess pieces on its sides, and you must unscramble it so that on each side there is one mated King, assuming the kings cannot capture the neighboring pieces (Queens, Rooks, Bishops, kNights).

We are usually good with this type of puzzles, but we spent our entire weekend trying to solve this one without any success. We even started wondering if it can be actually solved, so decided to share it with you and see if you can help us figure that out.

Below you can see the way the cube looks when seen from 8 different angles:

Remark: The orientations of the pieces are irrelevant to the final solution, i.e. they don’t need to be consistent on each side.

SOLUTION

The Rubik’s Chess puzzle cannot be solved. You can see a detailed solution HERE.

Puzzle Prime November 1, 2017

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The Dark Bridge

Section: Brain Teasers Category: Deduction

Difficulty: Medium

Four friends are trying to cross a bridge in complete darkness, but have only one flashlight. They need respectively 1, 2, 7, and 10 minutes to cross the bridge, and if any three of them step on the bridge at the same time, it will collapse. How many minutes do they need at least in order for all of them to get to the other side?

SOLUTION

They need 17 minutes. Label the friends A (1min), B (2min), C (7min), D (10min). A and B cross the bridge, then A returns back with the flashlight. C and D cross the bridge, then B returns back with the flashlight. Finally, A and B cross the bridge.

In order to see that this is optimal, notice that when D crosses, he needs at least 10 minutes. If C crosses separately, this will make already 17 minutes in total. Therefore C and D must cross together, and A and B must be at that time on the two opposite sides of the bridge. From here it is easy to conclude that the friends indeed need at least 17 minutes.

Unknown Author October 1, 2017

1 Comment