Solution to Puzzle #75: Clock Reflection

Thanks for the response to this puzzle. I got correct answers from three people – Ashutosh Diwakar, Rajeev Deshpande and Pallav Pandey in that order. Well done!

I am also taking the liberty of replicating Ashotosh’s solution which I thought was very well explained. If you want to use a video for children, you can access it on the following link:

https://www.educreations.com/lesson/view/solution-to-puzzle-75/24663132/?s=7LLAbY&ref=app

The answer is 27 minutes and 41 seconds past the hour after 6 pm.

Puzzle 75 Solution

Puzzle 75 Solution

In one hour or 60 minutes of duration, hour hand is shifted or advanced by 5 minutes.

 

So, let’s suppose after 6 o’clock, it takes another x minutes to reach the condition mentioned in question i.e.  reflection should show the same timing.

Then,

In 60 min., hour hand is advanced by = 5 min.

In x min., hour hand will be advanced by = (5/60)*x min.

 

Now, at this time w.r.t. 6 o’clock, minute hand will be back by = (30-x) min.

 

From the question, the time by which the hour hand is advanced should be equal to the time by which minute hand is back:

=>           (5/60)*x = (30-x)

=>           x = (27+9/13)min.

 

Therefore, (27+9/13)min. after 6 o’clock, the reflection in the mirror will show the same timing.

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Puzzle #75: Clock Reflection

Here is a nice clock problem I found on a web site, originally created by Lewis Carroll.

A clock has hour and minute hands of the same length and no numerals on its face. At what time between 6 and 7 o’clock will the time on the clock appear to be the same as the time read on the reflection of the clock in a mirror?

As always, please send your answers directly to me at alokgoyal_2001@yahoo.com. If you like the puzzle, please share it with others. If you have interesting puzzles to share, please send them to me at my e-mail given above.

 

Happy solving!

 

 

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Solution to Puzzle #74: Overlapping Clock Hands

Apologies for not posting the answer last weekend, I was traveling.

This was perhaps too simple a puzzle, and that is why I did not get a response from anyone!

The correct answer is 22. An easy way to think about this is that from 1 to 2 pm (or am), the hands would overlap once, just after 1.05. Similarly between 2 and 3 pm, they will overlap once, again just after 2.10. This goes on..until between 10-11 pm. However, between 11 pm and 1 am, this happens only once, exactly at 12 am. Therefore in any 12 hour cycle, the clock hands overlap 11 times, hence in 24 hours, the clock hands overlap 22 times.

Hope you enjoyed the puzzle!

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Puzzle #74: Overlapping Clock Hands

This is a relatively simple puzzle, more geared for children than for adults. My friend Alok Mittal recently asked this question in his Mathematical Circles class and also reminded me that he was asked this puzzle (probably 30 years back!) by my father, who was very fond of puzzles.

It’s 3 pm on Saturday afternoon. By 3 pm on Sunday i.e. in exactly 24 hours, how many times would the minute and the hour hand overlap?

As always, please send your answers directly to me at alokgoyal_2001@yahoo.com. If you like the puzzle, please share it with others. If you have interesting puzzles to share, please send them to me at my e-mail given above.

 

Happy counting!

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Solution to Puzzle #73: Find the Two Numbers

This was a difficult puzzle. Many people sent me the answers, though none of them were completely correct. One grade 11 student sent me the correct answer but the reasoning was not completely correct. So unfortunately no names on the honor board this time. Sirisha, who contributed the puzzle, also sent the answer which was very nicely laid out and hence I will take the liberty of posting her solution.

The correct answer is 2 and 2. The best way to do this is to create a matrix of the possibilities of two numbers, both for sum, as well as the product.

Figure 1

Figure 1

 

In step 1, we should be able to eliminate all the possibilities where A would get to know at least one of the numbers, for example, if the sum was ‘0’, then A would know that the two numbers are ‘0’ and ‘0’. Similarly, if the sum is ‘1’, then A would know that the two numbers are ‘0’ and ‘1’. So, by removing these possibilities, one gets the following matrix for ‘A’:

Figure 2

Figure 2

Similarly for B, if the product is such that it can be achieved by a unique combination (e.g. 64 can only be achieved by 8 and 8) or if one of the numbers can be figured out, e.g. if the product is ‘0’, then one knows that at least one of the numbers has to be ‘0’. If we eliminate all such possibilities, one gets the following set of possibilities for B. Note that a lot of combinations get eliminated here and also note that the products left appear at least twice e.g. 36 can be achieved by 9 and 4 and also by 6 and 6.

Figure 3

Figure 3

Now that A says that he knows the answer, it can only be because if we look at the corresponding sums of two numbers left after elimination of so many possibilities, the sum of the two numbers has to be unique. IN the matrix below, note that only those possibilities are represented that correspond to the product combinations left after B said that he cannot guess the answer. Within these sums, only the sum 4, 12 and 13 appear once. For example, if the sum of the two numbers was 11, A would still not have been able to give the answer as it could still be 9 and 2 or 8 and 3. Which means the sum has to be either 4, 12 or 13.

Figure 4

Figure 4

Interestingly, for B, the sum 12 in the above matrix is formed by 6 and 6, which leads to a product of 36, and 13 if formed by 9 and 4, also yielding 36. Since B is able to give the answer in the end as well, it means that the only possibility could have been 2 and 2.

Figure 5

Figure 5

Hope you enjoyed the puzzle!

 

 

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Puzzle #73: Find The Two Numbers

This is a beautiful puzzle that came to me through an IITD friend (Sirisha), and I had not done before – thank you Sirisha!

There are two friends A and B. Their teacher calls them separately and tells A, the sum of two numbers (both between 0 and 9, repetition allowed) and tells B, the product of both numbers

A and B then talk. A says – I dont know the two numbers. B says – Even I dont. A says – Now I do. B says – even I do now

Both A and B are good at mathematics.

What are the two numbers?

As always, please send your answers directly to me at alokgoyal_2001@yahoo.com. If you like the puzzle, please share it with others. If you have interesting puzzles to share, please send them to me at my e-mail given above.

 

Happy number hunting!

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Solution to Puzzle #72: Who Lies?

This was a relatively simple puzzle and lots of people answered correctly. First one to answer correctly was Anisha, my elder one (aged 10), though she had the benefit of getting the question at the time I was posting it. Prasanna Raghavendra, Anirudh & Anvi from Singapore, Vivek Shah, Sanjiv Gupta and Swati Bhargava in that order were the other ones to send correct answers. Well done and thank you all!

The best articulated answer came from Anavi Baddepudi (who I think is also a 10 year old from Singapore), and I am taking the liberty of reproducing her answer verbatim:

The answer is that both of them were lying and the black- haired is the girl where as, the red haired is the boy.

In the case that only the red haired child is lying, that means that he is actually a boy..  But that would also mean that the black haired child is a boy too, and the question states that there is one boy and one girl..

This is  the same if the black haired child is the only one lying.

So this proves that both are lying.

Hence, the  black- haired is the girl where as, the red haired is the boy.

Hope you enjoyed the puzzle!

 

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Puzzle #72: Who Lies?

This is a simple puzzle, hopefully the younger ones will like it too. I got inspired by this puzzle through Alok Mittal’s last Mathematical Circles class on “truth tellers and liars”. This puzzle is taken from Martin Gardner’s “The Colossal Book of Short Puzzles and Problems“, Chapter 13 which has many puzzles on truths and lies. This is an adaptation from an original puzzle by Martin Hollis in Tantalizers.

A boy and a girl are sitting on the front steps of their building.

“I am a boy,” said the one with black hair.

“I am a girl,” said the one with red hair.

If at least one of them is lying, who is which?

As always, please send your answers directly to me at alokgoyal_2001@yahoo.com. If you like the puzzle, please share it with others. If you have interesting puzzles to share, please send them to me at my e-mail given above.

Happy investigating!

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Solution to Puzzle #71: May The Fastest Horses Win

Apologies for not posting the solution last week, I was traveling on vacation on the weekend.

This was a relatively difficult puzzle, but I was encouraged to see the number of people who responded, most of them with correct answers. Anirudh & Anvi from Singapore, Alok Kuchlous from Bangalore, Siddhansh (IIT Kanpur) and Swati, a colleague at Helion, sent the correct answers. Prakash Prakash sent an “almost correct” answer. Well done all!

Reproducing the answer from Alok Kuchlous, and also attaching a link to the video if you would like to use that:

It can be done in 7 races.

Divide into 5 groups and race. (5 races)

Let rankings be:

A1, A2, A3, A4, A5; B1..B5; C1..C5; D1..D5; E1..E5

Race: A1, B1, C1, D1, E1.  (6th race)

let the order be same as above i.e. A1 to E1. So first is A1.

2nd and 3rd could be:

A2, A3

A2, B1

B1, B2

B1, C1

So you need to only race A2, A3, B1, B2, C1. (7th race)

Pick the top two.

Hope you enjoyed the puzzle.

 

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Puzzle #71: May the Fastest Horses Win

A beautiful puzzle, initially given to me by a close friend – Naresh Kumra, while trekking to the Everest Base Camp two years back. Came across this again last week in MindCipher.

The London racetrack needs to submit its top three horses to the Kentucky Derby next month in order to compete for a prize. However in a recent electrical storm, all the racetrack’s previous race history was erased such that no one knows the previous times of any of the horses. To make matters worse, each horse looks identical and it is impossible to remember which ones were the fastest.

London racetrack is home to 25 horses, but their track can only race 5 horses at a time. What is the fewest number of races that can be conducted in order to determine the 3 fastest horses?

As always, please send your answers directly to me at alokgoyal_2001@yahoo.com. If you like the puzzle, please share it with others. If you have interesting puzzles to share, please send them to me at my e-mail given above.

Happy racing!

 

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