Solution to Puzzle #85: Colliding Ants

This was a difficult puzzle overall, and only two people sent correct answers – Aman Singla and Girish Tutakne. I think in the end this was a bit difficult for children!

I am reproducing the answer from Aman, Girish had sent a similar answer:

50 (the number Peter sent) reach Cynthia; 20 (the number Cynthia sent) reach Peter; 1000 collisions.

Since the number of ants headed in each direction is still the same following the collision, mathematically, one can assume that a collision is same as one ant climbing on top of the other and crossing over. So 50 will eventually reach Cynthia, and 20 Peter. And each of the 50 reaching Cynthia will have had to cross over 20 headed towards Peter – hence 1000 crossovers/collisions.

Hope you enjoyed the puzzle!

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Puzzle #85: Colliding Ants

Very nice puzzle from gurmeet.net/puzzles, which has a great set of puzzles on the site. I am taking a puzzle from this site after a long time. This puzzle should be interesting for children as well as adults.

Peter and Cynthia stand at each end of a straight line segment. Peter sends 50 ants towards Cynthia, one after another. Cynthia sends 20 ants towards Peter. All ants travel along the straight line segment. Whenever two ants collide, they simply bounce back and start traveling in the opposite direction. How many ants reach Peter and how many reach Cynthia? How many ant collisions take place?

 

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 colliding!

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Solution to Puzzle #84: Reason Without Equations

Many people sent correct answers, which I had expected. However, very few sent the explanation behind the answers, which is what I was looking for. Some of you used equations to solve the question. Intent was to solve the questions with no equations, and only use logical reasoning.

Ones to send the correct answer included Ajay Pandit (Bangalore), Anshi Aggarwal (Gurgaon), Radhika Goyal (New York), Karan Sharma (Jaipur), Pravin Panchagnula and Rhea Mittal (Mumbai).

Here is a way to think about both the questions without equations:

Answer #1

(i) The number is greater than 9 since there are two digits

(ii) It is less than 23 because 23 x 4 1/2 is greater than 100

(iii) It is an even number because it is an integer when multiplied by 4 1/2

(iv) Since the number is even, its half is an integer. When multiplied by 9, it yields the reverse and hence the reverse is divisible by 9

(v) Since the number has the same digits as the reverse, it implies that the number itself is divisible by 9

(vi) between 9 and 23, only number divisible by 9 is 18 which is the answer

 

Answer #2:

(i) None of the four integers is a 10 or a multiple of 5, if it was the product would end in a 0

(ii) At least one of the digits is less than 10, or the product would be 5 digits

(iii) All four are less than 10 [because of reason #(i)]

(iv) Therefore the only possibilities are 1,2,3,4 or 6,7,8,9. Since 1,2,3,4 yields only 24, 6,7,8,9 is the answer!

Hope you enjoyed the puzzle!

 

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Puzzle #84: Reason Without Equations

Two very cute number problems I found which can be solved without any equations. If adults are trying it, they should keep a time limit of 2 minutes, though children are free to take as much time as they need. The puzzle has been taken from the book “The Moscow Puzzles” by Boris A. Kordemsky (edited by Martin Gardner), Puzzle #281.

1. A two digit number, read from right to left, is 4 1/2 times as large as from left to right. What is it?

2. The product of four consecutive integers is 3,024. What are the integers?

As always, please send your answers directly to me at alokgoyal_2001@yahoo.com. If you have any interesting puzzles to share with others, please send them to me as well.

Happy number crunching!

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Solution to Puzzle #83: Belleview and Hillview Distance

The puzzle was relatively simple and many of you sent the correct answers. The first one to send the correct answer was Girish Tutakne, and the first child to send the correct answer was Anshi from Gurgaon. Well done Anshi! I also got a correct answer from Shray from New Jersey.

I am reproducing the answer from Girish:

In a return trip, distance traveled up and down are equal, because uphill in one direction becomes downhill in the other. As downhill speed is double that of uphill, time spent downhill be half of the time spent while driving uphill. Therefore, out of 6 hours, 4 hours are spent driving uphill and 2 hours driving downhill. Therefore, the total distance back and forth is 60×2 + 30×4 = 240 miles ==> distance between the two places is 120 miles.

Hope you enjoyed the puzzle!

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Puzzle #83: Belleview and Hillview Distance

This is a beautiful puzzle for children from the book Mathematical Circle Diaries, Year 1 (Session 22, last puzzle).

Two mountain villages, Belleview and Hillview, are connected by a road. Not a single stretch of this road is flat: the road always goes either uphill or downhill. A round trip on a bus between these two villages takes 6 hours. The speed of the bus on any uphill section of the road is 30 miles per hour, and on any downhill section it’s 60 miles per hour. Find the distance between the these villages.

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 driving!

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Solution to Puzzle #82: Guess the Number

I got many correct answers for this puzzle. The first one to send the correct answer was Sandeep Verma from California. I also got correct answers from my niece Radhika, and from Girish Tutakne and Anubhav Garg.

I am reproducing the answer from Sandeep Verma, which is only slightly different than the one I am posting on video (BTW, LSD means Least Significant Digit and MSD means Most Significant Digit):

EFGH

X 4

——–

HGFE

Since multiplication of 4 and E in the MSD doesn’t lead to  carry over, E has to be either 1 or 2

Since LSD H*4 is E, E can not be 1. So, E=2

This implies H is either 3, or 8, since H*4 ends in E=2 being the units place. From the MSD, H of the result we know that H>4, so H=8

Now from middle two digits,

(G*4+3) mod 10 = F

LHS has to be odd, hence F is odd.

4*F+Carryover from above=G. Since there is no carryover to the MSD, E (4*2=8),

4*F+CO <= 9. Only values of 1 and 2 can satisfy this inequality for F.

As F has to be odd, F=1

Going back, G*4 + 3 = 11 V 21 V 31. G*4 ends in 8 in the units place. This means G=2 or 7. G*4+3 will be 11 or 31 respectively

For G=2, 4*(F=1)+1=2 is incorrect

For G=7, 4*(F=1)+3=7 is correct; hence G=7

Putting together, EFGH=2178 and HGFE=8712.

Moment of truth, 2178*4=8712

Here is a link to a video that explains the answer as well, which might be easier for children:

Hope you enjoyed the puzzle!

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Puzzle #82: Guess The Number

Thanks to Nandini Dube for contributing this puzzle. She is a class V student in New Jersey, and this is a homework puzzle from her class. Please do not be swayed by “Class V” tag, this is a puzzle worth trying for adults as well.

What are the numerical values of the alphabets E,F,G and H if EFGH x 4 = HGFE?

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 learning EFGH…

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Solution to Puzzle #81: The “Sholay” Puzzle

Surprisingly I did not get any solution to this puzzle, though the access to the blog was the highest that it has ever been. Assuming that many of you tried but thought it to be too easy!

Here are the answers to the puzzle:

1. Between the six slots, the odds are equal of being in the odd numbered or the even numbered slots, and hence the chances of winning are equal no matter which one you choose.

2. You should go first, have 2/3rd chances of winning

3. You should go first, have 3/5th chances of winning.

For a detailed answer in video for children, please see the following:

Hope you enjoyed the puzzle!

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Puzzle #81: The “Sholay” Puzzle

A very interesting probability/ strategy puzzle for children, coming straight out of the movie Sholay, one of the most popular Bollywood movies of all time. All the Indians reading this blog I am sure will vividly remember the scene where Gabbar (the villain) has a revolver in his hand which has a cylinder that can hold 6 slots, he puts bullet into 3, and then rotates the cylinder to a random position and there are three people to the shot.

I got the puzzle from my elder daughter, Anisha, who got this from Alok Mittal’s Mathematical Circles class yesterday.

Lets look at 3 variations of this situation. In all of them assume that two people are playing a game, lets call them A and B. The game is that one person shoots the other starting from a random position of the cylinder. If there is actually a bullet at that position, the other person is dead. If not, then the other person gets the revolver and shoots…and the game goes on until one is dead. Objective, of course, is to be alive. The revolver cylinder has 6 slots.

1. If there is only one bullet in the revolver and we start from a random position, what should be the strategy for A…should he/she shoot first or let the other person shoot first. What are the chances of being alive or winning with this strategy?

2. If instead of 1, there are 2 bullets in continuous slots, what would be A’s strategy?

3. If there are two bullets, but both inserted in random locations, what would the strategy be?

Happy shooting….oops, I meant only for a playful cause!

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