Puzzle #191: 1×4 Rectangular Stone Plates

This is a gem of a puzzle from A Moscow Math Circle, Week by Week Problem Sets by Sergey Dorichenko.

Can a 10×10 square be paved with 1×4 rectangular stone plates?

As always, please send your answers as comments within the blog (preferred), or send an e-mail to alokgoyal_2001@yahoo.com. Please do share the puzzle with others if you like, and please also send puzzles that you have come across that you think I can share in this blog.

Happy paving!

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Solution to Puzzle #189: Blue Eyed Sisters

This was a simple puzzle, and I got correct solutions from Praneeth, Prartik Poddar and Suman Saraf. Well done all. I am borrowing the answer from Suman.

For this, the most likely answer is 3 blue eyed out of 4 children. (Important to note that there are other solutions too)

b -> blue eyed
t -> total
b/t*(b-1)/(t-1) = 1/2

the sanest solution without involving the wrath of family planning authorities is 3/4 ūüôā

[Next possibility would be 21 children of which 15 are blue eyed!]

Hope you all enjoyed the puzzle!

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Puzzle #190: Escalator Race

I cannot stop loving the book A Moscow Math Circle – Week by week problem sets by Sergey Dorinchenko. So here is another one from this great collection.

Anisha and Arushi are riding down an escalator. Halfway down the escalator, Arushi grabbed Anisha’s hat and threw it on the up escalator (and such things often happen!). Anisha ran up the down escalator in order to run down the up escalator after her hat. Arushi ran down the down escalator in order to run up the up escalator after the hat. Who will be the first to the hat? Assume the speeds of the girls relative to the escalator are equal and do not depend on the direction of motion.

As always, please send your answers as comments within the blog (preferred), or send an e-mail to alokgoyal_2001@yahoo.com. Please do share the puzzle with others if you like, and please also send puzzles that you have come across that you think I can share in this blog.

Happy hatting!

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Solution to Puzzle #188: Jars and Paints

Apologies for late posting of the solution to this puzzle, but better late than never!

This puzzle was similar to one of the first puzzles I posted (Milkman and the Three Jars –¬†https://alokgoyal1971.com/2013/02/03/puzzle-2-milkman-and-the-three-jars/). I received correct answers from Suman Saraf (as always!), Tanmay Ratnaparkhe and Sarthak Mitra. Well done all.

I am taking the liberty of copying the entire solution from Tanmay.

Take 3 JARs – 1 , 2 , 3. I have taken liquids A, B ,C and kept the qty as 20 each for easy calculation. The capacity of the Jars is obviously 30. In multiple rounds, the liquids are poured into different jars.

JAR 1 JAR 2 JAR 3 Comments
20A 20B 20C
Round 1 20A + 10B 0 20C + 10 B Pour B equally into JAR 1 and JAR 3
Round 2 10A + 5B 10 A+ 10 B + 10C 10 C + 5 B Pour Half contents of JAR 1 and JAR 3 into JAR 2
Round 3 10 A + 10 B + 10 C 10 A+ 10 B + 10C 0 Pour all contents of JAR 3 into JAR 1
Round 4 20/3 A + 20/3 B + 20/3 C 20/3 A + 20/3 B + 20/3 C 20/3 A + 20/3 B + 20/3 C Divide equally between all 3 JARs
Hope you all enjoyed the puzzle!
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Puzzle #189: Blue Eyed Sisters

A very interesting two minute puzzle from Martin Gardner’s The Colossal Book of Short Puzzles and Problems. Please do share this with your children as well.

If you happen to meet two of the Jones sisters (this assumes that the two are random selections from the set of all the Jones sisters), it is an exactly even-money (that is 50-50) bet that both the girls will be blue-eyed. What is your best guess as to the total number of blue-eyed Jones sisters?

As always, please send your answers as comments within the blog (preferred), or send an e-mail to alokgoyal_2001@yahoo.com. Please do share the puzzle with others if you like, and please also send puzzles that you have come across that you think I can share in this blog.

Happy Sunday!

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Solution to Puzzle #187: Sum as 100

I am back after a break of two weekends in-between during which I moved from Gurgaon to Bangalore. Lets begin with the answer to this puzzle. I received many correct answers – from Anahat Prakash, Deepak Goyal (my brother) and Pratik Poddar. Well done all!

The answer is “not possible”. ¬†Here is the explanation:

There are 40 pairs of numbers between 10-100 that sum to 100 – (10,90), (11,89)….(49,51). That leaves out the 10 two digit numbers 50, 91, 92…..99. Even if we take all of these 10, and one number each from the 40 pairs, we will still end up with only 50 numbers. If we take any more, we will need to take both the numbers from at least one pair, and they will sum to 100.

I also would like to reproduce the answer from Pratik, who gave the same explanation in another interesting way:

The problem can be changed to:

Is it possible to have 50 numbers from [-40, 49] such that sum of no two numbers is 0.
[41,49] would not matter here

So, Is it possible to have 41 numbers from [-40, 40] such that sum of no two numbers is 0.

Not possible!

Hope you all enjoyed the puzzle!

 

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Puzzle #188: Jars and Paints

I will be taking a break from puzzles for the next two weekends after this post!

My fascination with A Moscow Math Circle (Week by week Problem Sets) by Sergey Dorichenko continues. Here is another one from there.

Each of three identical jars is 2/3 full of paint of different colors. Any part of the paint in one jar can be poured into another jar, and the paints mix homogeneously when this happens. How can the same mixture be obtained in all three jars if paint cannot be poured either out or into any other container?

As always, please send your answers as comments within the blog (preferred), or send an e-mail to alokgoyal_2001@yahoo.com. Please do share the puzzle with others if you like, and please also send puzzles that you have come across that you think I can share in this blog.

Happy painting!

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Solution to Puzzle #186: 4×4 Grid

This was a simple puzzle, a classical application of the pigeon hole principle. I got correct answers from Deepak Goyal (my brother in NY), Mahi Saraf (as always) and Pratik Poddar (as always again).

I am taking the liberty of copying the answer from Mahi.

Part 1
There are multiple answers to this one. I am producing two one from Mahi, and another one from Deepak.
From Mahi:
The plus signs represent the squares with the dots and the ones with the underscore signs represent the empty squares
_ _ _ +
_ + + _
+ _ +  _
+  + _ _
From Deepak
Solution to Puzzle #186

Solution to Puzzle #186

Part 2
When we  arrange 6 stars in a 4 by 4 grid, by pigeon hole principle there will be at least one row with 2 stars so we strike that. If the row we chose had exactly 2 stars then there will be one more row with 2 stars so we strike that to. Now we will be left with 2 stars and 2 strikes hence, we can cut all the stars.
If the first row we chose had 3  stars then we will be left with three stars and 3 strikes so even in this case we will be able to strike out all stars.
The last possibility could be that the first row we chose had 4 stars in it so we will have 2 stars and 3 strikes remaining.
Therefore, We can never arrange 6 stars in a 4 by 4 grid so that by striking 2 rows and 2 columns we will still be left with 1 star.
Hope you all enjoyed the puzzle!
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Puzzle #187: Sum as 100

I seem to be enamoured with this book at the moment, and hence another one from A Moscow Math Circle – Week by Week Problem Sets by Sergey Dorichenko.

Is it possible to write more than 50 two digit numbers on a blackboard without having two numbers on the board whose sum is 100?

As always, please send your answers as comments within the blog (preferred), or send an e-mail to alokgoyal_2001@yahoo.com. Please do share the puzzle with others if you like, and please also send puzzles that you have come across that you think I can share in this blog.

Happy century!

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Solution to Puzzle #185: Marker on a 9×9 Board

This was a great puzzle, and I received correct answers from Suman Saraf and Prakhar – very well done!

Please see the figure below.

Solution to Puzzle #185

Let’s divide the entire board, except for the central square, into 2×1 dominos. For example, it can be done as shown in the diagram. At each move, Anisha shifts the marker into some square of the same domino. If Arushi then moves the marker to the second square of the same domino, she will always be able to move, and thus Arushi will win.

Hope you all enjoyed the puzzle!

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