Solution to Puzzle #100: The 100 Square Rug Puzzle

My apologies for vanishing after posting the prize puzzle – I was traveling for the last 3 weekends and hence did not have a chance to post the answer. Overwhelming response, and I am amazed by the abundance of brilliance – many correct answers. The first one to send the correct answer is Abhinav Jain – congratulations to Abhinav. He gets the following book , one of my favorites, as the prize:

Others to send the correct answer included Suman Saraf, Pratik Poddar and Mohit Khare. Pratik sent a link to a more generic problem, which he had posted on his blog, which can be accessed at:

http://www.cseblog.com/2014/05/cut-polygon-puzzle.html

I am taking the liberty of copying and pasting Mohit Khare’s answer:

Answer to Puzzle #100

Answer to Puzzle #100

Hope you all enjoyed the puzzle – thank you for all your support to help with the first 100, now we move on to the next 100!

 

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Puzzle #100: The 100 square rug puzzle (Prize Puzzle)!

Never imagined when I started 2 years back that I will be posting 100 puzzles at some time…thanks to all readers for continuous engagement and encouragement! Here goes the 100th one…will reveal the original source when I post the answer. I heard this puzzle through Alok Mittal’s Mathematical Circles class. First person to send back the correct answer gets a prize – children from Alok’s classes are not qualified as they have done this puzzle already!

When Nuriya Saradzheva was an adolescent, she was awarded a beautiful Turkmenian rug for being first on her collective farm to use an improved method of picking cotton. The rug was 12 feet by 9 feet in dimensions.

Now Nuriya works as an agronomist. Once, while doing her research, she spilled acid on the rug. After the damaged part was cut out, there was a large rectangular hole, 1 feet x 8 feet.

Nuriya decided to repair the rug. Using straight line cuts, she cut the undamaged part of the rug into 2 parts that, when sewn together, formed a square – 10 feet by 10 feet.

Puzzle 100 graphic

Puzzle 100 graphic

How?

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 repairing!

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Solution to Puzzle #99: World’s Most Famous Puzzle Currently

This was a relatively simple puzzle and I received many correct solutions to this puzzle. Ritesh, one of my colleagues, was the first one to respond in literally 30 seconds. Many others gave correct answers that include – Utkarsh (another office colleague), Anirudh Baddepudi, Bharat Kashyap, Prateek Poddar, PR Kumar and someone anonymous (through comments on the site). Well done everyone!

July 16 is the answer. I am taking the liberty of reproducing the answer from Anirudh Baddepudi:

After the first clue, we can rule out any birthdays in May or June as these have days which only appear once (numbers). Then there are 5 possible dates when the birthday could be.

After the second clue, we can rule out July 14 and August 14, because Bernard would not know which day is the birthday between these 2 if he was given 14. Then there can only be August 15, August 17 and July 16 left. However, if Albert knows the birthday in the last step he must have been given July has his month. This is because if he was given August he would have a choice between August 15 and August 17 and would not know when Cheryl’s birthday was.

Hence, we are left with July 16, which must be Cheryl’s birthday.

Prateek sent a link from New York Times that has a good solution to this puzzle as well:

http://www.nytimes.com/2015/04/15/science/answer-to-the-singapore-math-problem-cheryl-birthday.html

And he also shared a link that his friend sent, which is a more difficult version of the puzzle. The die hards may want to try this one out.

http://www.cseblog.com/2015/04/advanced-cheryls-birthday-puzzle.html

Hope you all enjoyed the puzzle.

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Puzzle #99: World’s Most Famous Puzzle Currently!

I am simply amazed how one single puzzle has proliferated across the social media in the last 10 days – the puzzle from Singapore, popularized by Alex Bellos’ blog in The Guardian. Many of you would have seen this, but I thought it is a nice puzzle, and hence putting it on the blog. This will be an interesting one to go through even with younger children.

Here is the puzzle:

Albert and Bernard Just became friends with Cheryl, and they want to know when her birthday is. Cheryl gives them a list of 10 possible dates:

May 15            May 16               May 19

June 17           June 18

July 14            July 16

August 14       August 15          August 17

Cheryl then tells Albert and Bernard separately the month and the day of her birthday respectively.

Albert: I don’t know when Cheryl’s birthday is, but I know that Bernard does not know too.

Bernard: At first I did not know when Cheryl’s birthday was, but now I know

Albert: Then I also know Cheryl’s birthday.

So when is Cheryl’s birthday?

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 birthday!

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Solution to Puzzle #98: How Many Days to be Disease Free

Thanks again to Pallav for the puzzle – as I mentioned earlier, one of my favorites and a wonderful concept of mathematical induction to be taught to children. Only two people sent correct answers – Mohit Khare and Pratik Poddar – well done!

The answer is 23 days. For such puzzles, one can try a smaller problem first – i.e. what happens if there is only one person will yellow eyes, then for 2 and so on. Here is how it works (copying answer from Pratik Poddar directly):

Everyone knows that there is at least one diseased guy. Lets say that there was 1 diseased guy. That guy would see that there is no one else, and move out on day 1. If there were two diseased guys, no one would move out on day 1, as both think that there is someone else who is diseased. So, on day 2, both realise that both are diseased. Similarly, till day 22, all diseased people would be thinking that there are 22 diseased, and all will leave on day 22. But when no one leaves, all 23 diseased people leave on day 23.

Hope you enjoyed the puzzle!

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Puzzle #98: How Many Days to be Disease Free

Thanks to Pallav for yet another very interesting, and in this case, a classic puzzle. My 11th grade chemistry teacher, Mr. Gopinath, had given it to me, and I love this puzzle. Here it goes:

In a village of 100 people all villagers are infinitely intelligent. One day a strange disease hits the village which affects 23 people. The symptom is that if you’re affected the color of your eyes will turn yellow. The village head orders that all the villagers with disease (yellow eyes) should leave the village. The villagers meet everyday evening but they are not allowed to tell the other guy to leave. Nobody can see the color of their own eyes but moment he knows that his eyes are yellow he leaves the village. In how many days will the village become disease free.

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 disease-free world!

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Solution to Puzzle #97: Pirates and Gold Coins – Part II

I did not get as many correct answers to this puzzle as I had expected. Many people gave an answer which can work, but only Karan Sharma, my nephew, gave me a full answer. Other people to give any correct answer included Anisha Sharma Goyal, Anirudh Baddepudi, Mohit Khare and Abhinav Jain – Everyone – well done!

The answer is that A can give the gold coin to anyone amongst C, D, E and F. Here is an explanation to the solution:

Hope you all enjoyed the puzzle!

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Puzzle #97: Pirates and Gold Coins – Part II

I had posted a puzzle earlier (Puzzle #3) – https://alokgoyal1971.com/2013/02/10/puzzle-3-five-pirates-and-looted-gold-coins/. This puzzle is an extension of the same puzzle, and also attributed to the same family – this time to Alok Mittal from his Mathematical Circles group.

Original puzzles goes as follows:

Five pirates have looted 100 gold coins. Now they have to divide their loot.These pirates are very greedy and cruel,so they want to maximize their profit and they don’t shy to kill their pirates. For distribution they decide that senior most pirate will proposes a distribution of the loot. All the pirates will vote, and if at least half accept the proposal, the loot is divided as proposed. If not, the most senior pirate will be killed, and they start over again with the next senior pirate. What solution does the most senior pirate propose? So that he can get the maximum profit, and of course live to have it!

Variation of this puzzle:

Assume there are 6 pirates now – A, B, C, D, E and F. A is the senior most, F being the most junior in that order. They have only 1 gold coin. Like before, for a proposal to be accepted, at least half need to vote in favor. What should be A’s strategy?

To avoid any doubt about the pirates’ behavior, following defines the order of prioritization for them:

– Desire to live

– Greed (have as many gold coins as possible)

– Cruelty – all else being equal, they would rather see someone else being killed?

What is A’s strategy?

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 #96: Daughter’s Ages

This was a relatively simple puzzle, and I received many correct answers – Santosh Kumar, Kamaljit Dhooria, Karan Sharma, Anisha Goyal, Mohit Khare, Abhinav Jain, Vijay Raghavan, Pratik Poddar, P R Kumar and Suman Saraf + his 10 year old daughter! If I have missed out anyone, apologies as I received a lot of correct answers. (Addendum – as suspected I missed out many – Ayush Tutakne from Manila, Sanjana from Washington DC, Vikas Vats from New Jersey and Ashwin Khandel)

Answer is 6,6,1. Like some other puzzles in the past, one needs to solve this through a process of elimination, beginning with a superset of all the possible ages based on the statement that the product of the ages is 36. (Copying solution from Pratik Poddar)

1, 1, 36

1, 2, 18

1, 3, 12

1, 4, 9

1, 6, 6

2, 2, 9

2, 3, 6

3, 3, 4

With the second hint, sum is a constant that both of them know about.

Sum is all the options:

1, 1, 36=38

1, 2, 18=21

1, 3, 12=16

1, 4, 9=14

1, 6, 6=13

2, 2, 9=13

2, 3, 6=11

3, 3, 4=10

Since even after the sum, the answer is not clear – the room number or the sum is 13.

Since there exists a “youngest” daughter, the ages of the daughters are 1, 6 and 6.

Hope you all enjoyed the puzzle.

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Puzzle #96: Daughter’s Ages

This is a wonderful puzzle contributed by Pallav Pandey – thank you Pallav. This is a nice puzzle that does not require use of any specific concept, and should be fun for children as well.

2 maths prof who used to be room-mates meet after a long time in their college reunion and start chatting about family.

The first guy asks, how many children you have and how old are they. The second says 3 girls and the product of their ages is 36. The first guy says, I’m unable to know how old they are, so the second guy says the sum is the same as our room number in college. The first guy says, beats me.

The second guy says the colour of my youngest daughter’s eyes is blue.

The first guy says – oh I get it now.

What are the ages of the second guy’s daughters.

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 age guessing!

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