I thought that this was a relatively simple problem – Most people gave me the correct answer, but I did not receive the correct proof for the answer from anyone.
The answer is – the shortest of the tallest in the rows is taller than the tallest of the shortest in the columns. Here is a proof for the same. I have stayed away from algebraic notations to make it simple for children to follow the solution.
Hope you enjoyed this!
This is a relatively simple puzzle after a rather difficult one, I would encourage people to try this one with kids.
In a rectangular array of people, who will be taller – the tallest of the shortest person in each column or the shortest of the tallest person in each row?
Source: World’s Greatest Riddles and Brain Teasers, by Edward Scarzi
Age Group: For all ages
Happy puzzling!
Before I give the solution, I owe it everyone to reveal the original source, which appeared in Scientific American. Here is the exact source:
Martin Gardner,
Mathematical Games,
Scientific American, April 1958.
Martin Gardner,
The Second Scientific American Book of Mathematical Games and Diversions,
Chapter 9, The Monkey and the Coconuts,
Simon and Schuster, pages 104-111
I also would like to complement Sandeep Singh, who happens to be the only person to respond correctly to the puzzle. It was fun trying the puzzle with 2 sailors with my 8 year old, and the 3 sailor version even with adults.
The answer does not lend itself to a video solution, and hence only a text version this time. While there are more correct and technical solutions to the problem, here is one I thought is easier to follow (taken from BBC site):
The reasoning is as follows. Given that the pile of coconuts is divided 6 times into 5 piles, it is then clear that 5 × 5 × 5 × 5 × 5 × 5 = 15625 coconuts can be added to any answer for the number of coconuts to give the next higher answer. Indeed, any integer multiple of 15625 can be added to or subtracted from any answer to obtain another answer. This of course gives us infinitely many negative answers. These all satisfy the equation (11). While there is no small positive integer n that gives rise to a solution to the equation (11), a little trial and error with negative values of n comes up with a solution n = −4, with f = −1. Adding 15625 to this negative solution leads to the smallest positive solution n = 15621.
Imagine the first man A waking up to find a pile of −4 coconuts. He tosses a positive coconut to the monkey, so there is now a pile of −5 coconuts. He hides his share of −1 coconut, and leaves a pile of −4 coconuts for the second man B to discover. The same thing then happens to B, C, D and E in succession, so there is now a pile of −4 coconuts in the morning. After tossing a positive coconut to the monkey, there is now a pile of −5 coconuts, so each of them has −1 coconut more!
As I am finding with many of these puzzles, as you start to explore the internet, you find a lot more theory behind each of these puzzles. Here are some links for the more mathematically oriented readers:
http://www.bbc.co.uk/dna/onthefuture/A882498
Thanks to Vikas Vats for contributing this puzzle, and was originally given to us by our class teacher in Class XI, Ravi Gopinath. This is an interesting one, though could be tough for children below 12 years of age.
Five shipwrecked sailors arrive at a deserted island that has only coconuts. They also see one monkey. The sailors collect all the coconuts and agree to divide up the coconuts into equal shares the next morning before looking for help. during the night each sailor wakes up one at a time and takes 1/5 of the coconuts and hides it. Each time there is one coconut left over and the sailor gives that to the monkey. In the morning they divide what is left of the pile into equal shares and there is still one coconut left for the monkey.
How many coconuts were there to begin with.
Suggestion: For children between 8 and 12 – ask them to solve the puzzle with just 2 sailors or 3 sailors.
This one was a relatively simple puzzle, though not really for children below 12 years as they are not as familiar with the probability concept. I am not posting a video solution to this as there is not much value in a video solution for this one.
The way to maximize the probability of getting a red ball is to put just one red ball into one of the containers and all the other 99 balls (49 reds and 50 whites) in the other container. The probability of getting a red ball is as follows:
(1/2)*1 + (1/2)*49/99 = ~0.75
I do not have any proof that this it the best way to do it. But if anyone has a proof, I would love to see it and post it for other people’s benefit.
Here is a cute problem given to me by a friend, who mentioned that this was part of the Martin Gardner collection.
There are two identical urns, and there are 50 identical red balls and 50 identical white balls. All the balls are distributed between the two urns. You will perform a two step process as follows:
1. Choose one of the urns at random
2. Choose a ball within the chosen urn at random
Your task is to maximize the probability of getting a red ball. How would you distribute the balls between the two urns?
I need to begin with an apology – I have not been able to create a video solution to this puzzle as I seem to have misplaced the writing instrument for my iPad. Hence a written solution to this one, hopefully it is simple enough for everyone to follow.
Solution to Part (1) – All one needs to do is to take different number of balls from each of the machine and weigh them once. For example, if one takes 1, 2 and 3 balls respectively from the three machines, then the “fair” weight would be 60 grams. If the weight is 62 grams, it means that it is the second machine which is faulty, and is producing balls which are 1 grams heavier. Similarly, if the weight is 57 grams, then it is the 3rd machine which is producing balls which are 1 gram lighter.
Solution to part (2) is similar to (1) – in this case at a minimum one need 1,2,3,….10 balls from machine 1, machine 2, …., machine 10 respectively. Faulty machine can be identified by variation from the “fair weight” of 550 grams.
The basic principle for part (3) is similar. Only Alok Kuchlous sent me the correct answer for this. In this case, since the fault can be either 1 or 2 grams, one needs to ensure that if one has selected n(k) balls for machine(k), then one cannot chose either n(k) or 2n(k) balls from any other machine as there could be a possible overlap. The minimum such sequence that one can select is as follows:
1,3,4,5,7,9,11,12,13,15……which adds up to 80 balls.
Hope you all enjoyed the problem and the variation!
Lets take a slightly different version of the faulty balls – You have multiple machines producing table tennis balls. Each ball is supposed to weigh 10 grams. One of the machines is faulty and is producing balls of either 9 grams or 11 grams, i.e. it is off by 1 gram. You have a very accurate digital weighing machine which will tell you the weight in grams of whatever you put into it.
1. (For children up to 12 years) – Assume you have only 3 machines, and are allowed only one weighing. How will you identify the faulty machine.
2. (for children above 12 years) – Assume you have 10 machines, and again are allowed only one weighing. How will you identify the faulty machine.
3. (For high schoolers and adults) – Now there is a twist in the problem – the faulty machine is off by either 1 gram or 2 grams, i.e. it produces balls of one of the following weights (in grams) – 8, 9, 11, or 12. We do not know which one is the case. Again, you can do only one weighing. What is the minimum number of balls you will need to weigh to arrive at the answer for the faulty machine (and how many grams is it faulty by).
Last part of the question was created with some friends on a trek last year around 6 months back to whom credit is due – Alok Mittal, Parul Mittal, Neeraj Aggarwal, Aman Singla & Naresh Kumra.
Happy Puzzling!
I was very happy to see many new “puzzle solvers” for this puzzle. I was particularly impressed with Aadi Kuchlous and Sahil Kuchlous who solved the 6 and the 9 balls puzzle respectively. Same is true for Arushi and Anisha Goyal. Many adults gave the correct answer for the 12 balls puzzle.
Answer to the first two parts is simple and is in the video link.
For the 12 balls puzzle, there are many different ways to solve. I have given one way to solve in the video link, but I would highly encourage you to look at the different ways. I did not realize that there is so much science in this puzzle! Look at the following link for a more complete treatment of this puzzle.
http://www.cut-the-knot.org/blue/OddCoinProblems.shtml
Hope you enjoyed this!
There are many puzzles many of you would have come across which provide you a balance and ask you to identify a ball from a set of balls that is of a different weight than the rest of them. Many of you have urged me to post these problems. Unlike previous times, I am posting a set of puzzles this time, since they are all variations of each other and target different age groups. All the questions here assume that you have a traditional balance with two sides and no weights.
1. (Only for children below 8 years) – You have 3 balls, of which one is heavier than the others, and you do not know which one is heavier as they all look like. Using the balance only once, can you find out which one is heavy?
2. (For children up to 12 years) – You have 9 balls of which one is heavier. Using the balance twice, can you find out which one is the heavier ball.
3. (For Adults) – You have 12 balls, of which one is either heavier or lighter, and you do not know which of the two situations it is. Using the balance thrice, can you find out which is the faulty ball?
Happy balancing!
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