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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Puzzle #186: 4×4 Grid

This is another wonderful puzzle from A Moscow Math Circle, Week by week problem Sets by Sergey Dorichenko.

There are two parts to the problem:

(a) Place seven stars in a 4×4 grid so that, no matter which two rows and which two columns are erased, at least one star will remain.

(b) Prove that if six stars are placed in a 4×4 grid, one can always erase all of them by erasing two rows and two columns.

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 Eid to everyone!

 

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Solution to Puzzle #184: The Sage and the Pharmacist

This was a different kind of puzzle, and I was happy to see the response. I got correct answers from Mahi Saraf, Sahaj Sharma (my nephew from Jaipur) and Pratik Poddar – well done all. Thanks again to Alok Mittal for contributing this puzzle.

Here is the answer:

Sage’s plan: If the sage drinks a very weak poison before coming to the king and carries a random drink saying it is his poison (which actually has no poison) he will not die as first he will first have a weak poison and then the strong one from the pharmacist to survive. Next as his own drink has no poison, and the pharmacist drinks his own strong poison, the pharmacist will die.

Pharmacist’s counter plan: After the pharmacist gets to know of this trick he can easily take another normal drink that has no poison in it. So the sage will die as he would have had some previous poison followed by a normal drink followed by pharmacist’s normal drink. The pharmacist just has the normal drinks and survives.
Hope you all enjoyed the puzzle!
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Puzzle #185: Marker on a 9×9 Board

This is a wonderful puzzle, this time from a relatively lesser used source – A Moscow Math Circle, Week by week Problem Sets, by Sergey Dorichenko.

One marker is placed in the center of a 9×9 board. Anisha and Arushi take turns moving the marker to one of the adjacent squares – one sharing a side – provided that this square has never been occupied by the marker. Anisha goes first. The first player unable to move loses. Which of the players can guarantee a win?

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 father’s day!

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Solution to Puzzle #183: 8 Digits

This was a relatively simple problem. I was happy to see my 10 year old, Arushi, solve this one. I also got correct answers from Karan Sharma, Mahi Saraf, Anahat Prakash, Pooja Goyal, Pratik Poddar and Suman Saraf. Well done everyone!

There are two possible solutions, though both have the same basis and are just vertically flipped solutions. Here is the answer:

Solution to Puzzle #183: 8 Digits

Solution to Puzzle #183: 8 Digits

Here is a quick explanation, copied from Pratik Poddar:

C and F independently are connected to all except one circle. So, C and F cannot have two adjacent numbers. So C is 1 and F is 8. So A is 7 and H is 2.

3,4,5,6 needs to be placed in B,D,E,G

B and D cannot be 6
E and G cannot be 3

Since the image is symmetric across vertical line, lets say B is 3. If E is 6, and D and G become 4 and 5 which is not possible. So G is 6. So D is 4 and E is 5.

So, A is 7, B is 3, C is 1, D is 4, E is 5, F is 8 and G is 6.

Hope you all enjoyed the puzzle!

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