Was a difficult puzzle – I got three correct answers for this. Vivek Shah, Mihir and Aman Singla in that order sent the correct answers – very impressed!
Here is the solution:
Make 2 stacks – one of 10 and the other of 90. 10 will have x heads and 90 will have 10-x heads.
If all the coins of the stack of 10 are reversed, both should have equal number of heads (10-x). However, this will increase the number of coins with head up.
Hope you enjoyed the puzzle!
This one is one of the best puzzles I have come across in recent times…contributed by Vardhan Rajkumar based out of London. I do not know the original source of the puzzle. No special knowledge or skill required, so suitable for all ages.
You have 100 coins. 10 of them are head up and 90 tails up. You have been blindfolded. You need to divide the coins into 2 groups that have the same number of heads.
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.
Enjoy applying the heads 🙂
This was a relatively simple puzzle, many people sent me correct answers. Anirudh Baddepudi, Anavi Baddepudi and Praveen Baddepudi all got back with correct answers with slight variations. Anoop Agrawal, Vivek Shah, Rhea Mittal and Mihir also got back with correct answer. First person to solve this, as I found the puzzle was my eldest cousin, Jeetan bhaiya.
Trick to the puzzle was to look at the 100 upside down, and one can then easily form CAT. Here is a link to the solution.
Hope you enjoyed the puzzle!
This is a wonderful 2-minute teaser I picked up from Martin Gardner’s “The Colossal Book of Short Puzzles and Problems”. In the end of the book he has a chapter titled “Twelve More Brain Teasers”, which has 12 problems he liked after he had stopped writing the monthly column for Scientific American.
Puzzle 69
In the graphic above, 9 toothpicks are arranged to form a 100. Can you change 100 to CAT by altering the position of just two picks?
Happy catting!
The response to this puzzle was overwhelming…got many correct answers. First one to respond with the right answer was Anavi Baddepudi from Singapore – well done. Many others got correct answers – Pratik Sarkar, Vivek Shah, Ramendra Mandal, Anubhav Garg, Ravi Kuppanna, Shruti Sancheti and Rhea Mittal. Finally, happy to see that my elder one, Anisha Goyal, solved it as well!
Some people solved this puzzle the brute force way, i.e. they came up with a distribution of numbers in the boxes that satisfied the two conditions – horizontal row summing to 30 and vertical column summing to 22. An easier and cleaner way to solve for x is the following:
If you add the horizontal row and the vertical column, the number x gets added twice, and all the other numbers get added only once. Since digits 1…9 when added yield 45, this sum should be 45+x, which is also 30+22, which is already given to us. Solving gives us x as ‘7’.
Hope you enjoyed the puzzle!
This is a very nice, quick teaser I got from the site brilliant.org. Should be fun to do for the younger lot, no specific knowledge required for this one.
Number At The Intersection
Place integers from 1 to 9, once each, into the 9 squares above, such that the horizontal row sums to 30 and the vertical column sums to 22.
What number would you place in square X?
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 solving!
I got many answers to this puzzle, and mostly all correct. This was a relatively simple puzzle for those who know algebra and probability, I do not know any way to do this one without them, and hence the younger audience it difficult to participate in this puzzle. Ones who sent correct answer included Amritansh Raghav, Aman Singla, Alok Kuchlous, Anubhav Garg and Sirisha Gadepalli.
Several of you pointed out that the answer to the puzzle is different depending on whether or not the rat has memory…that is, once it traverses down a path that does not lead to exit, will it remember or not. Accordingly the answer is different. Aman Singla was the first one to point this out.
In the scenario there is no memory, assume that E is the escape time. Given equal probability of each of the 4 exits, one can formulate the following equation:
E = (E + 8)/4 + (E + 2)/4 + (3 + 5)/4
Solving for E gives 9.
In the event, there is memory, the following equation will be true:
Escape time =
0.25 x 5 +
0.25 x 3 +
0.25 x (8 + 5/3 + 3/3 + 1/3*(2 + 5/2 + 3/2)) +
0.25 x (2 + 5/3 + 3/3 + 1/3*(8 + 5/2 + 3/2))
Solving, one would get 7 minutes 20 seconds as the answer.
Hope you enjoyed the puzzle!
Very nice probability puzzle…though involves basic knowledge of probability and algebra. Found this again on http://www.midcipher.com.
Consider a lab rat in a maze. There are four paths that can be taken with equal probability. Path A leads to exit in 5 minutes. Path B leads back to start and takes 8 minutes. Path C leads to exit in 3 minutes and last path D is 2 minutes long and leads back to start. What is the expected escape time of this poor rat from the maze?
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 escaping!
Got the most hits to the puzzle site this week for 2014…thanks!
I got many correct answers, though the one I loved the most was from Anavi Baddepudi, a 10 year old from Singapore. From the answer, it seemed that she did not know the binary system, but devised a method whereby she gave a combination of wines to each servant and from the servants which die, you could point to a unique bottle…very well done.
As it so happens, her brother, Anirudh was the only one amongst children to solve, and solved it right. Very well done both of you!
Others who solved…Alok Mittal, Alok Kuchlous, Santosh, Aman Singla and Pankaj Kakkar…some of the brightest minds I know! Reproducing Aman’s answer:
Give each wine bottle a unique number 0 to 1023 (you can do this for1024 bottles) written in binary. Assign each servant a unique location 1 to 10 corresponding to bit location. Servant must take a sip of wine from each bottle where his assigned bit location is 1. Number of poisoned bottle spelled out by dead servants being 1, and live being 0.
Also attaching a video link to a more detailed answer for children:
Hope you all enjoyed the puzzle!
A delightful puzzle, found this one on the site http://www.mindcipher.com.
So there’s this king. Someone breaks into his wine cellar where he stores 1000 bottles of wine. This person proceeds to poison one of the 1000 bottles, but gets away too quickly for the king’s guard to see which one he poisoned or to catch him.
The king needs the remaining 999 safe bottles for his party in 4 weeks. The king has 10 servants who he considers disposable. The poison takes about 3 weeks to take effect, and any amount of it will kill whoever drinks it. How can he figure out which bottle was poisoned in time for the party?
Note: For younger children, try the same puzzle with 10 bottles and 4 servants!
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 wine drinking (only the adults)!
 | Senthil Kumar VS on Solution to Puzzle #200: Achie… |
 | Alok Goyal on Puzzle #200: Achieve your… |
 | Vishal Poddar on Puzzle #200: Achieve your… |
| Alok Goyal on Puzzle #200: Achieve your… |
 | Abdul on Puzzle #200: Achieve your… |
Alok Goyal's Puzzles Page 