Puzzle #155: Bugs on a Square

This is a very famous puzzle that many of you would have seen in the past as well. I saw it recently in one of the Martin Gardner books again, but am reproducing it here from a New York Times article from September 8, 2014.

screenshot-2016-09-24-18-10-50

Please look at the figure above. Four bugs are at the corners of a square which has each side measuring 10 inches. Each bug moves directly towards its neighbour as indicated in the arrows. How long does each bug travel before they meet? Please note that you do not need to know calculus to be able to compute the answer.

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 bugging!

 

 

 

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Solution to Puzzle #153: Get Your Gifts

A very nice puzzle and many people sent me answers. I was disappointed that not too many children tried it. Even though I am posting the solution, I would encourage you to try to puzzle with your children. Many people got it right, that included Suman Saraf, Abhinav Jain, Karan Sharma, Vishal Poddar and Aman Goyal.

The answer is 1, 1, 7. Here is the answer explained through a simple video for children:

For folks who prefer to just read, here is the explanation:

The possible outcomes are 009, 018, 027, 036, 045, 117, 126, 135, 144, 225, 234, 333. Belle was not able to answer, so the number is not 333. Carol was not able to guess so the number is not 009. Nick was not able to guess, so the number is not 018. After Nick was not able to answer the number Carol was able to find the number, this implies that there were 2 possibilities from the second number. As Carol was able to find the number, this means that the second digit is 1 and if it is not 018, then the number is 117.

Hope you all enjoyed the puzzle!

 

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Puzzle #154: Can You Divvy Up the Gold?

I continue my love affair with the NSA puzzles, and here is one more, third week in a row – thanks again Suman!

We have done a few pirates and gold problems in the past. This one is slightly easier. It is a good puzzle to try with middle graders For adults/ avid puzzlers, this will be more of a speed test!

Following their latest trip, the 13 pirates of the ship, SIGINTIA, gather at their favorite tavern to discuss how to divvy up their plunder of gold coins. After much debate, Captain Code Breaker says, “Argggg, it must be evenly distributed amongst all of us. Argggg.” Hence, the captain begins to pass out the coins one by one as each pirate anxiously awaits her reward. However, when the captain gets close to the end of the pile, she realizes there are three extra coins.

After a brief silence, one of the pirates says, “I deserve an extra coin because I loaded the ship while the rest of you slept.” Another pirate states, “Well, I should have an extra coin because I did all the cooking.” Eventually, a brawl ensues over who should get the remaining three coins. The tavern keeper, annoyed by the chaos, kicks out a pirate who has broken a table and who is forced to return her coins. Then the tavern owner yells, “Keep the peace or all of you must go!”

The pirates return to their seats and the captain, left with only 12 total pirates, continues to distribute the coins – “one for you,” “one for you.” Now, as the pile is almost depleted, she realizes that there are five extra coins. Immediately, the pirates again argue over the five extra coins. The captain, fearing that they will be kicked out, grabs the angriest pirate and ushers her out of the tavern with no compensation. With only 11 pirates left, she resumes distribution. As the pile nears depletion, she sees that there won’t be any extra coins. The captain breathes a sigh of relief. No arguments occur and everyone goes to bed in peace.

If there were less than 1,000 coins, how many did the pirates have to divvy up?

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 pirating!

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Solution to Puzzle #152: The Four Weights

You give simple problems to avid puzzlers and mathematicians, and you always find some twists to the problem that you never thought of. This puzzle was one of them. But before I go into the complications, here are the people who responded with correct answers:

  • Amit Mittal, Kaushik Paul, Deepak Yadav and Prakhar Prakash with partially correct answers i.e. they found one correct answer each
  • Suman Saraf (and his daughter), Pratik Poddar, Karan Sharma, Sid Mulherkar and Abhishek Masand gave both the correct answers

There are two answers to the puzzle

(1) 2,4,6,10

(2) 1,5,7,9

When I did the question, I did it more intuitively (and also trial and error and so did some others). I am reproducing Karan’s logic to the problem here:

Two observations that we can make after looking at the weights:
1. All weights are even or odd
2. No weight is repeated
Using brute force further, i observed two possible cases-
1,5,7,9 and 2,4,6,10

Sid Mulherkar, however, very rightly pointed out that we made an assumption that the weights are integers. The problem   does not say that, and theoretically therefore, the numbers could even be 1.5, or 5.5 etc. It just happens to be that the two solutions are both integers only. Reproducing his full solution (very similar to Pratik Poddar as well)

Let the measures of the weights be a,b,c,d and WLOG let a<b<c<d (where < means less than or equal to).

Now using a<b<c<d, we can only have two possibilities. They are:

a) (a+b)<(a+c)<(b+c)<(a+d)<(b+d)<(c+d)

b) (a+b)<(a+c)<(a+d)<(b+c)<(b+d)<(c+d)

In case a) we get the following system

a+b=6….(i)
a+c=8…..(ii)
b+c=10….(iii)
a+d=12….(iv)
b+d=14….(v)
c+d=16….(vi)

Subtracting (ii) from (iii) and adding the resulting equation to (i) we get b=4. From here we can easily calculate value a,c,d. giving us (2,4,6,8)

In case b) we get a similar system as case a) but this time we have

a+d=10….(iii)
b+c=12…..(iv)

Subtracting (ii) from (iv) and adding this to (i) we get b=5. Then we easily get a,c,d. The final answer is (1,5,7,9).

Hope you all enjoyed the puzzle!

 

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Puzzle #153: Get Your Gifts

Another nice puzzle from the NASA collection sent to me by Suman Saraf.

Four friends, Holly, Belle, Carol, and Nick, gather for May birthdays. Holly announces that she has a game before dinner. She hid gifts for each of her friends inside three separate boxes secured with padlocks. She challenges her friends to figure out the combination without consulting each other.

She provides the following information. All the padlocks have the same combination. The padlocks use 3 digits from 0 to 9. She also tells them that the sum of the three digits is equal to nine, and every digit is equal to or greater than the previous digit. Holly tells each of her friends one of the digits in the combination. She states, “I’ve given the first digit to Belle, the second digit to Carol, and the third digit to Nick.” The caveat is that the friends cannot share their numbers with each other or they will forfeit the gifts.

Then Holly gives her friends 30 minutes to open the padlocks while she watches and finishes dinner.

The three friends begin to think of the solution. One by one, they each try their hand at their padlock, but none of them opens the padlock. Seeing that no one has succeeded, suddenly Carol realizes she knows the answer, and successfully opens her box, revealing a new fitness tracker. Following this, Nick opens his padlock, revealing a new tablet; and Belle opens her box to find new pair of headphones.

Having watched this entire event unfold, can you determine the correct combination?

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 gifting!

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Solution to Puzzle #151: Journey from 458 to 14

This was a simple puzzle and I got lots of correct answers. Ones to send correct answers were Suman Saraf’s daughter, Amit Mittal, Ram, Karan Sharma, Pratik Poddar, Ravish Aggarwal, Prakhar Prakash and Smiti Mittal. My elder one, Anisha, also solved this one.

It takes 8 steps to get to the answer. Trick was to get to 9 or 90 (or 18 or 180) and use the doubling function to get to 144 or 1440, and then truncate. There are many possibilities. Here is one:

458->45->90->9->18->36->72->144->14

Hope you all enjoyed the puzzle, and thanks again to Pooja Singh for sending the puzzle!

 

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Puzzle #152: The Four Weights

Thanks to Suman Saraf for recently pointing me to NSA (National Security Agency’s) monthly puzzle blog. Here is a nice and simple problem that they posted in their April blog. Worth trying with your children!

This puzzle created by Andy F., Applied Research Mathematician, NSA

Mel has four weights. He weighs them two at a time in all possible pairs and finds that his pairs of weights total 6, 8, 10, 12, 14, and 16 pounds. How much do they each weigh individually?

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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