Puzzle #171: Buying Dimsums

This is a beauty of a puzzle sent by Suman Saraf – thanks Suman!

A fast food restaurant sells dimsums in boxes of 7 and 3. What’s the greatest number of dimsums a person cannot buy. Generalize it for p and q where p and q are relatively prime.

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 eating!
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Solution to Puzzle #169: Division of a Cake

This puzzle was not as easy as it looked…many people sent solution, but only Suman Saraf and his daughter Mahi sent the correct ones.

To copy a line I really liked from Suman’s solution – The essence of fair or envy free devision is that the only reasons a person could be unhappy are if they themselves made a bad choice or a bad cut. Nothing else should give them a reason to be unhappy.

Here is a simple way of doing it for any number of people (say n), people are named as X(1), X(2), …., (Xn).

Assume they are all sitting in a circle. X(1) makes a cut that he/she thinks is 1/n of the overall cake. The one who made the cut has the “current marker”.

X(2) has the following choice – If (s)he thinks that the cut is unfair, i.e. more than 1/n, then (s)he can trim it further until (s)he thinks it is 1/n, or just let it pass. Therefore, the piece will either reduce in size, or remain what it was. If X(2) trims it, then (s)he gets the “current marker”, if not, the marker remains with X(1). We continue this way until we reach a point where the turn comes back to the person with the “current marker”, at which point, they take the piece they had cut/ trimmed. It is “fair” for them to take it as they cut it themselves, and everyone else in the circle had the opportunity to take it and they chose not to.

After one person it out, we repeat the process for (n-1) people.

Please do think about it, and may be even try it at home, I am sure a lot of cake eating will happen 🙂

Hope you all enjoyed the puzzle!

 

 

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Puzzle #170: Rectangle and the Oil Well

This is a wonderful puzzle that I saw today after a long time. I was asked this puzzle in an interview in Microsoft back in 1994 March – yes it has been a long time! I saw it today in a Martin Gardner book – The Colossal Book of Short Puzzles and Problems.

Here is the puzzle:

An oil well being drilled in a flat prairie country struck pay sand at an underground spot exactly 21,000 feet from one corner of a rectangular plot of farmland, 18,000 feet from the opposite corner, and 6,000 feet from a third corner. How far is the underground spot from the fourth corner?

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 oiling!

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Solution to Puzzle #168: Who Gets the Easter Eggs?

This was a very counterintuitive puzzle, and a relatively difficult one, though one does not need to know any specific mathematical construct for it. Only two people (father – daughter pair) got the correct answer – Mahi and Suman Saraf! Very well done!

Bob will win more.

I am reproducing two different answers, one from the source, and the other one from Mahi. I like both the answers, both very intuitive.

Here is the one from Mahi.

 

Mahi Saraf's Answer to Puzzle #168

Mahi Saraf’s Answer to Puzzle #168

 

Here is the answer from the source, which I find equally appealing:

Label a carton with a “b” (respectively, a “c”) if Bob (respectively, Chris) reaches that carton more quickly, and also record Bob’s (respectively, Chris’s) score upon reaching that carton. Label cartons A and L with “xx” since both players reach those cartons simultaneously.
We obtain:

xx b2 b3 b4
c2 c5 b7 b8
c3 c6 c9 xx

Note that there are five b cartons and five c cartons. So the cases in which Alice selects carton A or carton L are equally split between Bob and Chris. Similarly if Alice selects two b cartons then Bob necessarily wins, but these are balanced out by an equal number of cases in which Alice selects two c cartons and C necessarily wins.

The crucial cases occur when Alice selects one b carton and one c carton. Bob wins if the b carton has a lower score than the c carton:

b2 and (c3 or c5 or c6 or c9)
b3 and (c5 or c6 or c9)
b4 and (c5 or c6 or c9)
b7 and c9
b8 and c9

Chris wins if the c carton has a lower score than the b carton:

c2 and (b3 or b4 or b7 or b8)
c3 and (b4 or b7 or b8)
c5 and (b7 or b8)
c6 and (b7 or b8)

Since this yields 12 cases in Bob’s favor and only 11 cases in Chris’s favor, Bob has the advantage.

Hope you all enjoyed the puzzle!

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Puzzle #169: Division of a Cake

This is a very old puzzle. I remember this puzzle being asked by my Chemistry teacher in Class XI – Dr. Ravi Gopinath, someone was has had a strong influence in my passion for puzzles. I got reminded of this puzzle while having breakfast yesterday with my kids and one of their friends.

Anisha and Arushi need to divide a cake (any shape) between themselves and want the division to be fair, then the process is very simple. One person (say Anisha) divides the cake into two parts which she thinks are equal, and Arushi can pick one. Neither party has a reason to be unhappy and think that it was a fair division. Now they have a friends of their – Nishtha – visiting. They need to divide a red velvet pancake into 3 parts so that each one of them thinks that the division is fair. How do they do it?

What happens when their mom and dad also want a fair share, so there are 5 now? n?

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 divisions!

 

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Solution to Puzzle #167: 4 Rats in a Star

This was a relatively simple puzzle, and I received correct answers from Mahi Saraf, Pratik Poddar, “Mashex” and Prakhar. Well done all!

The answer is 2178 (RATS)

Here is the explanation:

Since the product is four digits long, RATS can be no greater than 9999/4. Thus the largest RATS can be 2499. This tells us that R is either 1 or 2.

Can R be 1? We also know that S x 4 ends in R. Since 4 is even and an even number times anything results in an even number, R must be even. Thus R can’t be 1, so must be 2.

2ATS x 4 = STA2

Since R = 2, we know 2ATS is somewhere between 2000 and 2499. Thus the product is somewhere between 8000 and 9996, and S must be 8 or 9.

Can S be 9? We know that S x 4 results in a number that ends in 2. 9 x 4 = 36, which ends in 6. 8 x 4 = 32, which does end in 2. Thus S = 8.

2AT8 x 4 = 8TA2

When we multiply 8 by 4, we get 32. We write down the 2 and carry the 3. Thus we know that T x 4 + 3 ends in A. Since T x 4 is even, T x 4 + 3 must be odd, so A is odd. Since the product is between 8002 and 8992, 2AT8 must be between 2001 and 2248.  So A is between 0 and 2.  The only odd number in that range is 1, so A = 1.

21T8 x 4 = 8T12

For T x 4 + 3 to end in 1, T x 4 must end in 8. The only values that will work are 2 and 7. R is already 2, so T can’t be 2. Thus T must be 7. Checking our work, we indeed get that

2178 x 4 = 8712

So we see that 2178 times 4 equals its reverse. Furthermore, 2178 x 2 = 4356 and 2178 x 3 = 6534, and those values are the reverse of each other as well.

Hope you all enjoyed the puzzle!

 

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Puzzle #168: Who Gets the Easter Eggs?

This is a delightful and a very counterintuitive puzzle, again from the NSA collection. Original puzzle has been written by Tim C., an Applied Research Mathematician at NSA.

Alice has a dozen cartons, arranged in a 3×4 grid, which for convenience we have labeled A through L:

A B C D
E F G H
I J K L

She has randomly chosen two of the cartons and hidden an Easter egg inside each of them, leaving the remaining ten cartons empty. She gives the dozen cartons to Bob, who opens them in the order A, B, C, D, E, F, G, H, I, J, K, L until he finds one of the Easter eggs, whereupon he stops. The number of cartons that he opens is his score. Alice then reseals the cartons, keeping the eggs where they are, and presents the cartons to Chris, who opens the cartons in the order A, E, I, B, F, J, C, G, K, D, H, L, again stopping as soon as one of the Easter eggs is found, and scoring the number of opened cartons. Whoever scores lower wins the game; if they score the same then it’s a tie.

For example, suppose Alice hides the Easter eggs in cartons H and K. Then Bob will stop after reaching the egg in carton H and will score 8, while Chris will stop after reaching the egg in carton K and will score 9. So Bob wins in this case.

Who is more likely to win this game, Bob or Chris? Or are they equally likely to 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 egg hunting!

 

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