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.
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.
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.
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.
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}.
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.
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.
Your grandma’s wall clock chimes the appropriate number of times at every whole hour, and also once every 15 minutes. If you hear the wall clock chime once, how much more time do you need to figure out what the time is, without looking at it?
SOLUTION
1 hour and 30 minutes. The wall clock will chime once at 12:15, 12:30, 12:45, 1:00, 1:15, 1:30, 1:45. If you hear 7 consecutive times just 1 chime, then the time is 1:45. If you hear less than that, you will easily find what the time is once again.
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