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

Here is a quick explanation, copied from Pratik Poddar:

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

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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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.

Hope you all enjoyed the puzzle!

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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.

Happy weighings!

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The answer is “white”. I am borrowing the answer from Mahi.

The bean remaining will always be white because every time we are doing something to the pot we are taking out only 0 or even number of white beans. As we started with an odd number of white beans, the bean that will be left will be white.

Hope you all enjoyed the puzzle!

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The task is to trace in the larger figure a shape geometrically similar to the smaller one shown besides it.

Happy tracing!

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

A=90kg, B=80kg, C=60kg, D=40kg, Supplies=20kg

Here is the sequence:

- D and C go
- C comes back
- A goes
- D comes back
- C and D go
- C comes back
- B and supplies go
- D comes back
- C and D go

Hope you all enjoyed the puzzle!

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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!