Solution to Puzzle #60: Hats and Survival

Posted on May 4, 2014 by Alok Goyal

I got a few correct answer, the first one came from Akash Tayal from New Jersey. Well done! Am taking the liberty of reproducing his answer directly:

Out of total 8 permutations, There are 6 ways that the cap colors are split 2 of a kind and 1 of a kind. If the prisoners go with this bet, then anyone who sees 2 of the same color on the others’ head should guess the opposite color; the others should pass. 75% chance to be set. Also have put the answer on a video in more detail:

Hope you enjoyed the puzzle!

 

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Puzzle #60: Hats and Survival

Posted on April 26, 2014 by Alok Goyal

I just heard this puzzle 15 minutes back at Alok Mittal’s Mathematical Circles where he gave this puzzle to all the children. Great puzzle – will encourage all children as well as adults to try this.

There is an unlimited supply of red and black hats, out of which, at random, 3 hats are pulled out and placed on the heads of three prisoners (call them A, B and C). All of them can see each other (and of course the hat color on top of the other folks), but cannot see themselves. At one simultaneous instant, all three of them are supposed to guess the color of their hat, and they are allowed three answers – Red, Black and Pass. Pass does not count as a correct answer, but is not a wrong answer either. All three of them are now allowed to say a Pass.

If there is anyone who gives a wrong answer, then all of them will be killed. If no one is wrong, and at least one is right, then they will all be set free. They are allowed to discuss their strategy before this event occurs. What should their answering strategy to maximize the chances of survival.

As an illustration, if all of them decide to guess Red, then the chances of survival are 12.5% as all of them being Red has a probability of 12.5%.

As always, please send your answers directly to me at alokgoyal_2001@yahoo.com. If you like the puzzle, please share it with others. If you have interesting puzzles to share, please send them to me at my e-mail given above.

Happy strategizing!

 

 

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Solution to Puzzle #59: Can You Move the Knights

Posted on April 26, 2014 by Alok Goyal

This was probably the first time I did not get any answer…felt sad 😦 In hindsight, I think the puzzle was a bit difficult. A group of children with whom I tried this puzzle all reached the conclusion that it is not possible. However, they could not prove it.

The trick to the puzzle is use of graph theory. Solution is attached on the link below:

Hope you enjoyed the answer!

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Puzzle #59: Can You Move the Knights?

Posted on April 19, 2014 by Alok Goyal

This is a very interesting problem I discovered today while trying to prepare for a Mathematical Circles lesson organized by Alok Mittal, where he is teaching a group of children mathematics through puzzles. The puzzle is also from the book “Mathematical Circles”, by Dmitri Fomin, Sergey Genkin and Ilia Itenberg. Chapter 5.

Starting Point

Starting Point

Destination Point

Destination Point

Several knights are situated on a 3×3 chessboard as shown in Starting Point. Can they move, using the usual chess knight’s move, to the position shown in Destination Point?

For those looking for a hint….think graph theory!

As always, please send your answers directly to me at alokgoyal_2001@yahoo.com. If you like the puzzle, please share it with others. If you have interesting puzzles to share, please send them to me at my e-mail given above.

 

Happy knighting!

 

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Solution to Puzzle #58: The Survivor

Posted on April 19, 2014 by Alok Goyal

Thanks for an outstanding response to the puzzle!

This was not an easy puzzle. I got the correct answer from The trio of Ashish+Arushi+Ishir, as well as Shruti Mittal, who did a great job of solving the puzzle with the required trick. Anisha and Arushi also solved the puzzle but with a lot of hints…but well done all of you!

The basic trick to cracking the puzzle is that whenever you have 2, 4, 8, 16, or any power of 2, it is the first one who wins. When you have 1500 people, you look for the closest power of 2, which is 1024. Difference between 1500 and 1024 is 476, which means once 476 people have been killed, whoever has the sword will win. 476th person to be killed will be 476 x 2 = 952, which means at that point 953rd person will have the sword and will win. Answer is explained for children in more detail on this video:

Hope you enjoyed the puzzle!

 

 

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Puzzle #58: The Survivor

Posted on April 12, 2014 by Alok Goyal

This is a fantastic puzzle…those of you who have not seen this before and like puzzles will have a ball of a time! I took the puzzle from a site called http://www.mindcipher.com

Here is the puzzle:

There are 1500 people sitting around a circular table. They are numbered in a regular manner (1 to 1500 along the table).

Now, the 1st person gets a sword and kills the 2nd person. He then gives the sword to 3rd person, who then kills the 4th person and gives the sword to the 5th. This continues so that 1499th person kills the 1500th person and gives the sword back to 1st person.

Now the 1st person kills the 3rd and gives the sword to 5th person. This process continues until only one person remains. Which person remains?

As always, please send your answers directly to me at alokgoyal_2001@yahoo.com. If you like the puzzle, please share it with others. If you have interesting puzzles to share, please send them to me at my e-mail given above.

Happy surviving!

 

 

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Solution to Puzzle #57: Sam Lloyd’s Puzzling Scales

Posted on April 12, 2014 by Alok Goyal

I got relatively fewer responses to this puzzle, perhaps people found it to be too simple. First person with the correct answer this time was Ethan Stock from California and then Shruti Mittal from Mumbai. Well done.

My iPad is not with me at the moment, the answer would have been easier to explain. In general, it is tempting to use algebra, but lets try without it.

If you look at the first two figures, one realizes that there is a top on the left side in both the balances. In the second figure, therefore, where we have the top on the left and a block along with 8 marbles on the right, one can add 3 blocks on both sides, and the scale should still be balanced. On the left now you will have the top along with 3 blocks, which is the same as the first figure. Therefore, the right sides of both the balances should match as well, implying that 4 blocks + 8 marbles should be the same as 12 marbles, therefore leading to the conclusion that a block weighs the same as a marble. Once we have established this, the rest is easy – in the second figure we see that a top is equal to a block + 8 marbles. If we replace the block with a marble (since they weigh the same), we can figure out that a top is equal to 9 marbles!

Hope you enjoyed the puzzle!

 

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Puzzle #57: Sam Lloyd’s Puzzling Scales

Posted on April 6, 2014 by Alok Goyal

This is a puzzle by a famous American “puzzler” Sam Lloyd. Sam was a famous chess player, ranked 15th in the world at his best. He is also the creator of many famous puzzles. The following puzzle is in the form of a graphic, which says it all:

Puzzle #57

Puzzle #57

It is important that no algebra is used to solve this puzzle, and I think the younger children will enjoy this puzzle.

As always, please send your answers directly to me at alokgoyal_2001@yahoo.com. If you like the puzzle, please share it with others. If you have interesting puzzles to share, please send them to me at my e-mail given above.

Happy balancing!

 

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Solution to Puzzle #56: Toothpick Giraffe Problem

Posted on April 6, 2014 by Alok Goyal

This was a relatively simple puzzle, and most people told me that they were able to do this in less than a minute. Pawan Nijhawan from Udaipur was the first one to come back with a correct answers. Others included Muskaan Mittal, Ival Sallay and Deepak Goyal. Well done!

Solution is to take the right (or the back) leg of the giraffe and putting it at the bottom. If you rotate the new image clockwise by 90 degrees, it will be a mirror image of the original image. Here is the solution graphic which has been picked up directly from the original New York Times answer:

Puzzle 56 solution

Puzzle 56 solution

Hope you enjoyed the puzzle!

 

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Puzzle #56: Toothpick Giraffe Puzzle

Posted on March 31, 2014 by Alok Goyal

Found this cute brain teaser from a website (http://mesosyn.com/), though the original is a Martin Gardner puzzle that appeared in New York Times.

Puzzle 56: Toothpick Giraffe

Puzzle 56: Toothpick Giraffe

Five toothpicks form the giraffe shown above. Change the position of just one pick and leave the giraffe in exactly the same form as before. The re-formed animal may alter its orientation or be mirror reversed but must have its pattern unchanged.

As always, please send your answers directly to me at alokgoyal_2001@yahoo.com. If you like the puzzle, please share it with others. If you have interesting puzzles to share, please send them to me at my e-mail given above.

Happy moving the Giraffe!

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