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!

 

Posted in Puzzles | Tagged , | Leave a comment

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!
Posted in Solution | Tagged , | Leave a comment

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!

Posted in Puzzles | Tagged , | 1 Comment

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!

Posted in Solution | Tagged , | Leave a comment

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!

Posted in Puzzles | Tagged , | 1 Comment

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!
Posted in Solution | Tagged , , | Leave a comment

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!

 

Posted in Puzzles | Tagged , | 1 Comment