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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Puzzle #184: The Sage and the Pharmacist

This is a beautiful riddle that I am borrowing from Alok Mittal’s Mathematical Circles class.

In a far away land, it was known that if you drank poison, the only way to save yourself is to drink a stronger poison, which neutralizes the weaker poison. The king that ruled the land wanted to make sure that he possessed the strongest poison in the kingdom, in order to ensure his survival, in any situation. So the king called the kingdom’s pharmacist and the kingdom’s sage, he gave each a week to make the strongest poison. Then, each would drink the other one’s poison, then his own, and the one that will survive, will be the one that had the stronger poison. The pharmacist went straight to work, but the sage knew he had no chance, for the pharmacist was much more experienced in this field, so instead, he made up a plan to survive and make sure the pharmacist dies. On the last day the pharmacist suddenly realized that the sage would know he had no chance, so he must have a plan. After a little thought, the pharmacist realized what the sage’s plan must be, and he concocted a counter plan, to make sure he survives and the sage dies. When the time came, the king summoned both of them. They drank the poisons as planned, and the sage died, the pharmacist survived, and the king didn’t get what he wanted.

What exactly happened there?

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 living!

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Solution to Puzzle #182: Weights and Balance Once More

This was a difficult puzzle. I got two correct answers – From Suman Saraf and Anahat (Prakhar’s son). Well done, or rather – excellent!

Here is the solution for 5 weights – copying solution from Suman.

let’s say the weights are w1, w2, w3, w4, w5
w1 and w2, let’s say w1 wins
w3 and w4, let’s say w3 wins
weight w1, w3, let’s say w1 wins
after 3 weighings we know w1 > w3 > w4 and w1> w2
now we compare w5 with w3 and depending on the result, compare with w1 or w4. After these, we would have  w1, w3, w4 and w5 in some rank order, possible options are:
w5 > w1 > w3 > w4
w1 > w5 > w3 > w4
w1 > w3 > w5 > w4
w1 > w3 > w4 > w5
All that now remains is to put w2 in the right place.
In either of the above options, 2 comparisons (like we did in the step above) will lead to the right placement, since we already know w2 < w1
Hope you all enjoyed the puzzle!
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Puzzle #183: 8 Digits

This is another 2-minute problem that kids will enjoy. Found this one again in Martin Gardner’s The Colossal Book of Short Puzzles and Problems.

Look at the figure below, which has 8 empty circles.

Puzzle #183: Place the 8 Digits

You need to place digits 1 through 8 into these circles with no repetition. No two adjacent numbers (e.g. 3 and 4) can be in circles that are connected by a line.

There is only one solution (without mirror images) and there is a method to the madness here.

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 digital scramble for brunch today!

 

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Solution to Puzzle #181: Hidden Cross

This was a simple puzzle, but I did not get as many solutions as I would have expected. Both Suman Saraf and Pratik Poddar sent the right answers, along with three people at home – Karan Sharma (my nephew) and Arushi and Anisha, my 10 and 13 year olds.

The original puzzle BTW is adapted from an optical illusion in Charles H. Paraquin’s Eye Teasers – Optical Illusion Puzzles. It was published in 1978.

Here is the solution, I am taking the liberty of copying the one from Suman Saraf.

Solution to Puzzle #181

Solution to Puzzle #181

Hope you all enjoyed the puzzle!

 

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Puzzle #182: Weights and Balance Once More

Somehow puzzles on weights and balances just never end! I found this great one in Martin Gardner’s The Colossal Book of Short Puzzles and Problems.

Five objects, no two the same weight, are to be ranked in order of increasing weight. You have available a balance scale but no weights. How can you rank the objects correctly in no more than seven separate weighings?

For two objects, of course, only one weighing is required. Three objects require three weighings.

For younger children, please do try the problem with four objects, where you are allowed no more than 5 weighings.

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 weighings!

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