Wednesday, August 30, 2017

Interesting Geometry problem (6 Rectangles puzzle) - Solved in various ways

I came across this interesting problem and thought to share this with my students...

Students started working on this and after about 5 minutes, one of them was ready with her solution...

I would suggest you to try solving this problem on your own first, and when you are ready with your solution, you can read further to see how these students have 'seen, approached and solved' it....

Method-1
Vaishnavi's 1st attempt

Her approach was to count all the sides of 6 rectangles (222 x 6) and remove the ones that are not to be counted. So according to her, the answer was 999. I think, you would have figured out the mistake she did. 

Yes, she forgot that the overlapping sides need to be removed twice and not just once..

But, I am glad she could quickly think and work out her error after overhearing the final answers of two of her peers, while they were explaining me :)

So this is how she finally presented her solution to the class. Note her work below. See her erased work in the yellowed part. Here, she explained that how the red segments are actually two segments when the rectangles are separated and hence need to be removed twice. 

Vaishnavi's 2nd attempt

If you are wondering what her 222 and 111 represents, then this is what she had done..

She has combined the two red breadths and two red lengths (see the 4 sides with vertical strokes) to form one rectangle (222)......what remains is one length (made up of two pieces) and one breadth which is same as half the perimeter i.e. 111

Method-2

Tanvi's explanation to me


Tanvi's explanation to the class

To explain her approach, let me just make another diagram:

Diagram - to understand Tanvi's approach

Her two 444s represents perimeters of four rectangles..  Now she understands that this sum 'includes' the segments that were not to be included --- the overlapping segments. 

From here, she used the same approach as that of Vaishnavi to get 222 and 111. Thus she finally gets 999 - 333 = 666

Method-3

Siddharth's method
He "traveled along the boundary" of the shape and counted the lengths and breadths of rectangles it included...  He calculated 6 lengths and 6 breadths, thus amounting to thrice the perimeter which leads to 666. 

Method-4

Sahil's method

To understand Sahil's method, let me re-draw the figure again...  

Diagram - to understand Sahil's method

I think the above diagram would be self explanatory...

Yes,.... this is how he beautifully saw and made three rectangles out of all the pieces on boundary to get three equivalent rectangles thus amounting to 666.

I thought at least one of the two remaining two students would solve the problem using my method.. However, Rohit's method was same as that of Sahil and Laxmi too had used the same approach as that of Siddharth....

So I asked them if they wanted to know how I solved this problem...

And there was a loud Yes !  :-)

Method-5




So the perimeter = 6 breadths + 6 lengths = 222 x 3 = 666 

They understood the "sliding" that I had done.... But I thought to interrogate them further....

"How can I be sure that when I slide upper block to the left such that pt. D coincides with C, then F too will exactly coincide with E?

Rohit jumped in...

"Sir its obvious..... that x = a...."

"There is nothing obvious in Maths, Rohit... Can you prove this statement ?"

He began...

k = y + a

k = x + y

k = a + b

So, y + a = x + y = a + b

So,  a = x   and  y = b

---------

Two students were absent when we did this problem.. So they solved it the next day... One of them, Jitu, used the same approach as that of Siddharth......

While the other student, Kanchan, gave me a surprise !!

Method-6

Kanchan just removed the three bricks on the top layer and bottom layer. So according to her. the given shape gets converted into this equivalent shape for finding the perimeter: 
Diagram to explain Kanchan's method

So, perimeter of the given shape = 222 x 3 = 666

--------------------

1) Did you understand her method?

2) How did you solve this problem?

3) How about trying this with your students? Plz do share your classroom experience...

4) Which approach did you like the most? Why?

5) Plz share your views and comments on this article......


Thanks and Regards

Rupesh Gesota

Monday, August 21, 2017

Interesting exploration - Thanks to 18th of August :)

On 8th August, I saw an interesting math post on my fb timeline:

Today's date: August 8 

818 is the smallest palindrome that can be expressed as the sum of squares of two prime numbers.  818 = 17^2 + 23^2

As usual, I forwarded this message to many teachers and parents groups... and finally shared this my young group of maths lovers i.e. my students, reading the same message as above, with the only difference being... I did not write the two prime numbers: 17 and 23  



So naturally they started working on this problem... This is how they all worked together...

1) Since the last digit of sum is 8, the two addends can end in (1,7), (2,6), (3,5), (4,4), (9,9), (0,8) .... However square numbers cannot end in 7,2,8 and 3. Further, squares of primes cannot end in 4 as well. So the only option is (9,9)

2) The two prime numbers should be less than 30 because 30^2 = 900

3) In fact they should be less than 29, because 29^2 will be more than 818.. When I probed them for the reason, they said that the difference between 29^2 and 30^2 will be 29+30 = 57

4)  Since the squares are ending in 9, so the primes will end in 7 or 3 i.e. both ending in 3, both ending in 7 or one ending in 3 and other in 7

5) So they started with the options that can be easily eliminated like --

(i)   (17, 17) : Since 17^2 < 400, so their sum will be < 800

(ii)   (23,23) : since 23^2 = 529, so sum will be > 1000

(iii)  (13,13) and (13,17) : same reason as that of (i)

(iv)  (23,3) and (23,7) :  sum < 600

(v)   (23,13) :  Digital root of 23^2 + 13^2 = 7 + 7 = 5. But the required digital root is 8 (8+1+8) 

If you see the snap, one of them did a mistake here... She multiplied the DRs of the 2 primes instead of adding the DRs of their respective squares... But this was smartly caught by other student while working on the next option i.e. (17,23)

(vi) (17,23)  : They said that since this is the only option left and because my claim is correct, hence (17,23) "has to be" correct.... and we verified this using three ways

a) Estimation: 23^2 = 529 and 225 < 17^2  < 400 .. So their sum is between 850 and 930

b) Digital roots:  23^2 + 17^2 = 7 + 1 = 8 which is same as required DR

c) Actual calculation

When I asked them, as to why they did not use the estimation for (23,13) they said they did not want to add the two big square numbers....(529 + 169)   :-))

----

How would have you or your students solved this type of problem?


Regards

Rupesh Gesota


PS: These students are from marathi-medium government school, class-7 and 8....To know more about the math enrichment program, check the website www.supportmentor.weebly.com

Tuesday, August 15, 2017

Celebrating Pythagorean Triplets Day on our Independence Day


I think almost the whole world must be aware, by now, about the interesting fact about 15/08/17 --


I had received this message over whatsapp couple of days before this date itself... Like for anyone, it was a delightful surprise for me too to face & digest this fantastic fact.. And I did not miss this opportunity to share this message with almost everyone in my circle esp. students, teachers and parents on our Independence Day....

While many replied back with the Ooohs and Aaaahs, but there was one guy who did something more.... He is aware of the crazy math experiments that I do with my bunch of government school students... So he asked me "Do your kids know how to manufacture pythagorean triplets (PT) thereby proving that they are infinite in number?"

I replied - "No, I haven't worked on Pythagoras Theorem with them yet... But I am planning to do that today.."

He -- "My son stumbled upon Euclid's proof of 'infinite triplets' in 8th class.. Try to make that happen with your kids (students)"

And while saying this he also shared the link to his blogpost:

I was planning to read this post later, but luckily I clicked on this link - only to get highly motivated.. I now wanted to try this out with my students immediately... I was confident that my students would love to crack this code....

So I scrapped my plan A i.e. to first work on the geometric interpretation and different proofs of Pythagoras theorem....and zoomed into my class with plan B...

After our daily ritual (meditation and mental maths), I shared with them this interesting fact of the date and how it has beautifully coincided with our Independence day this year...

I was little surprised and even disappointed when I could not see much expressions of wonder on their faces as I had expected ! However, I forged ahead, sharing with them the details of ---

1) pythagoras theorem (the right angled triangle, hypotenuse, etc..)
2) with the conventional example of 3,4,5
3) idea of pythagorean triplets -- how (8, 15, 17) is one of those..
4) and finally the story of Euclid - as how he could prove there are infinitely many PTs...

To this, one of my students exclaimed - Yes, there will be infinitely many such triplets because there are infinitely many natural numbers....

True.. But mathematics demands proof ! How can you do so?

They look puzzled to this. So, I asked them - 

"Ok, you all seem to believe that there will be infinite PTs... Can you give some more other than (3,4,5) and (8,15,17)?"

Now I could see them trapped.... Some of them started scribbling on their books, while others were still staring at these two sets, probably to get some clue (pattern)...

I could not resist but teased them after 2-3 mins....  "So? Infinite, right?"  :-)

And now I could see couple of them trying to escape my sight..  :-)

And soon, there was a guess -

"Sir, what about 30, 40, 50 ?"

"Are you sure or are you guessing this?"

"I am not sure..."

"How about verifying this?"

And in another moment, the other student reasoned -- "Sir, it will obviously work !!"

I found the other students relieved with this solution... 

"Sir, then even 80, 150, 170 will work..."

I wrote this triplet on the board... "More?"


One of them reluctantly guessed -- "How about multiplying 3,4,5 with 2?"

"Means?"

"I am guessing 6,8,10?"

"Okay... Can we all check this case....?"

The class verified this and by now, all of them could figure out the game.... 

"Sir, we can now multiply 3,4,5 and 8,15,17 with any number and get more PTs...."

"Good.... and how do we represent this fact?"

I wrote on the board what he told me --- (3x, 4x, 5x)

"what is x?"

"any number..."

"So can I put 0?"

He understood what I meant....

"No Sir.... x is a natural number..."

-----

Now, this realization made me share with them about "Primitive Pythagorean Triplet"

For those unaware -- A Pythagorean Triplet (a,b,c) is said to be Primitive Pythagorean Triplet if the GCD of (a,b,c) = 1  means there is no other common factor among them other than 1....

for example: (3,4,5) , (8, 15,17) are PPTs but  (30,40,60) and (6,8,10) are Non-primitive PTs

But you know what?  I did not give this 'explanation' to my students...

I just told them --  

"Look... I got to know from you that we can make Infinite PTs from (3,4,5) and (8,15,17).... However, these new sets of triplets formed are not the Primitive Pythagorean Triplets....   But (3,4,5) and (8,15,17) are PPTs.... 

So can you tell me now, what do I mean by PPT?"

And I was so happy that one of them could say this --- 
"the numbers in the triplet should be co-prime!"

So then I asked others - Whats co-prime?  And they responded very well -- "numbers with gcd=1" ... "numbers with only one common factor i.e.1"

-----

They thought that their job was done.... since they could make infinite PTs from the two given PTs... 

But soon, I played the Devil......

"Hey wait guys.... what Euclid proved is that there are infinitely many PPTs and not just PTs... So can you give me some more PPTs....?"

And it was worth watching their faces again !!   :-)

There was silence for almost 5 minutes... All engrossed in research.... It was extremely difficult for me to Not give them any hint... But when I saw that some of them were not able to go ahead at all... I thought to intervene,...

I wrote the expression 3^2 + 4^2 = 5^2 on the board.... and asked them...

"What type of numbers are related by this equation?"

 "Natural numbers!" came an instant answer...   :-) :-)

"ohk true.... but are these numbers 9, 16, 25 something more?"

"yes, they are square numbers...."
"Yes, True..... So do we know anything more about square numbers that can help us here?"

I was unaware that Kanchan had already made this observation,... I understood this when she answered to this question instantly....

"Sir, their differences are odd and hence we can now look for....."

Her enthusiasm conveyed to me that she had made that 'million dollar observation' and so I stopped her immediately, so as to not reveal more clue to other students.... I just gave her thumbs up, signaling her to go ahead...., while others were completely puzzled with our exchange of expressions and sign language...

I was pretty sure by now that - not only will kanchan crack this in next few minutes, but even others will pick up the clue to get some breakthrough.... So I waited for 5 more minutes...


I found Kanchan engrossed in her work, while I found others still struggling.... So I decided to write down the list of square numbers on the board....

1
4
9
16
25
36
..
..
..
..
121

I asked them -- "Do you remember we had done many observations relating to this list of square numbers?"

"yes..."

"That's all then,.... Its just one of those imp. observations that's going to help you solve this problem at hand..."

Kanchan got in -- "Sir, you have given them a good hint now..."

"True.. Lets see what they do now....."

I found few of them interacting with others after this interaction..... and then, she raised her hand...

"yes kanchan... give me the Pyth Triplet you found..."

" (5, 12, 13) "

"Can others verify this?"

And they agreed for its correctness...

"Can you find one more?"

Silence dominated the class again.... when finally she broke it with her another announcement...

" (7, 24, 25) "

I verified this on the board with the class.... the result surprising other students.

"Kanchan, can you make the formula now, that will give you the list of all the PTs?"

This was the first time when I was not asking my student to explain her thought process before asking her to work on the extension of the problem..... The reason for this change in approach was my awareness that she was on the 'right track' and I just wanted her to leverage on her buzzing flow of thoughts/ strategy... I had decided to interrogate her after her stint on the formula (generalization) part....In case she gets stuck, then I will grab that opportunity to probe her about her thought process...

While she was again engrossed in her book, I decided to throw some more hint to other students...

"Can you notice something in the two triplets she has shared?"

Almost all of them replied -- " two out of three are consecutive numbers...."

Kanchan looked up at me with a smile.... Probably she knew what I was doing....

By this, I found one of them - Rohit - started staring at the chart of first 100 square numbers, that we have pinned up on our board since long....  


Its amazing that we all have used this chart so many times till now in various situations... I never knew during my school days, that square numbers are so resourceful...

It was difficult for me to wait... But Patience has always yielded more beautiful results....and this time was no different.....  She approached me with her work after 10 minutes...


Of course, I made her explain/ reason in detail what she had done and how her line of thinking about the previous two pythagorean triplets... and further how she constructed the formula, the terms that she required, etc....

I choose to not share these interesting interactions with you directly... as I first want you to think about this on your own.... (the above hit is anyways there with you now)

But as you see, we realized that the formula has some bug and it does not yield the desired triplet for y=2...   So she went back and started working on this....

Meanwhile, Rohit raised his hand.... He had got this triplet...

"Tell loudly..."

" (9, 40, 41) "

The class verified.... It was correct......

"Good one.... How about one more....? But this time, don't tell the answer aloud... Just write and show it to me....."

Soon he came to me with this ---    (11, 60, 61)

"Hmm.... correct.... So it seems you too have made some observations now...."

He nodded with a smile.... 

"Formula??"

He went and Kanchan came back with the second version of her formula, which she was confident of being correct this time......


The most satisfaction part for me as a maths teacher was when she told me that she could "find out the mistake" she had done in her previous work (formula)...and secondly she could also successfully correct it with the associated reasoning...

While I was congratulating her for this effort, she exclaimed --

"Sir, but this does not give us all the PPTs.... The (8, 15, 17) is left out...."

Isn't this observation worth appreciating again?

"True.... So what will you do now to include even such triplets?"

"Hmmm... May be there will be another formula to include the family of such triplets..."

"May be you are right.... So what do we do?"

"I will work on it....."

"Excellent.... I will wait for your work....."

And meanwhile, Rohit had already queued up.....

"Done?"

"Yes! "

"Have you verified it?"

"Yes...!"


I asked him for the explanation.... and he too gave a very satisfactory one.... the mathematical thinking was evident....

I called up Kanchan for having a look at this.... and she could quickly infer...

"Its same.... He has just used a = 2x +1 in the representations and I have used the term 2x+1 everywhere...."

"Correct... any advantage?"

"This looks neater.", she smiled & further added that, "he could have also written c = b + 1 instead of writing the whole expression for b again in the equation for c..."

I looked at Rohit.... He agreed....

"So now....?"

" We need to get the other PPT included which are missed out...."

"yes.... may be both of you can work together if you wish...."

After few minutes, Jeetu too was ready with his work --- he directly came up with the formula instead of sharing few PPTs like Rohit and Kanchan....

But while explaining that, he figured out there is some mistake in representation, and corrected it on the spot.... And this was a delight to watch.....!!


I had already told them to take the break and continue after that..... However there were some who did not move at all... and one of them was Sahil...  

This is what he came up with.....

What's interesting about his work is --- he arrived at the formula for PPT through the patterns..... and not through logic (like other three)....



So I asked him "what would you have done if you were not given the data of (7,24,25), (9,40,41) ?

"Sir that's when I was not able to do anything with just (3,4,5) and (8,15, 17).... It was possible for me to find the formula only when I got more data...."

"Hmmm.... but then how come your other three peers have constructed the formula for PT.... They did not have the privilege that you had (more data/ pattern observation)..."

He was silent.... I think he was getting what I meant.....

I continued... "They were able to do so, because they have used Logic along with some observations in square numbers..... and not Pattern..."

"Okay Sir.... I will try to find out using the other way now...."

And while going, he further added -- "Sir, even the set (8, 15, 17) does not come out from this formula!"    :-) :-)

Some students had not yet arrived at this....though they had made some observations.... and hence we decided to not discuss this exploration today but in the next session, thus giving them some more time to research at home....

So probably you might hear again from me, on this post :)

1) How about trying this out with your students? Do share your experiences if you do so.. Would love to study...

2) What are your views about the exploration that happened and the way it happened?

3) Could you decipher the rationale behind the approaches of Kanchan/ Rohit and Jeetu?

4) How can you include the missed out PPTs like (8,15,17) ? any change in this formula or new formula?  Will that solve the problem of including all the possible PPTs, or still some might be missed out? How do you know? 

5) Any more questions ?  :)


Waiting for your responses/ comments....

Regards

Rupesh Gesota

PS: These students are from marathi-medium government school, class-7 and 8....To know more about the math enrichment program, check the website www.supportmentor.weebly.com

Sunday, August 13, 2017

Mental Math: 9.8 x 2.5


Me - How much is 9.8 x 2.5 ?


Kanchan says,

I will consider the numbers as 98 and 25. Further, lets round up 98 as 100. So now, 25 times 100 gives 2500. But I have taken 2 times 25 more..

So, 98 x 25 = 2500 - 50 = 2450

Now, since I had multiplied each of the two given numbers by 10, the obtained product will be 100 times more than the desired one. Hence the actual answer will be 2450 / 100 = 24.5

Me - Excellent. Any other way?

Rohit says,

I know that 2.5 x 4 = 10 ... So if I am increasing one factor by 4 , then I will have to decrease the other  by 4.... i.e. I need to find 9.8 / 4

Now., I know that 25 x 4 = 100 and so, 24 x 4 = 96..... 
Hence 24.5 x 4 = 98  
Hence 2.45 x 4 = 9.8  i.e.  9.8 / 4 = 2.45

Now, the problem 9.8 x 2.5 becomes 2.45 x 10 = 24.5

Me - Fantastic. Any other way?

Tanvi says,

Lets ignore the decimal points. So the problem is 98 x 25

Now, 98 x 100 = 9800
So, 98 x 25 = 9800 / 4 = Half of Half of 9800 = Half of 4900 = Half of 4800 + Half of 100 = 2400 + 50 = 2450

Now, with the same reason what Kanchan gave, we will have to decrease the product by factor of 100. 

So, 9.8 x 2.5 = 2450 / 100 = 24.5

-----

1) How would you solve this problem mentally?
2) How about your students/ children?
3) What are your views about the approaches used by these students?
4) What are your views about the language used by them to communicate their reasoning?
5) Why do you feel these students were able to solve this problem, this way?

Waiting for your responses...

Regards

Rupesh Gesota

PS: These students hail from the marathi medium municipal school in navi-mumbai. To know more about the maths enrichment program being run with them, check the website www.supportmentor.weebly.com

Monday, July 31, 2017

Comparison of Confusing Fractions !

Few days back I got to attend the maths teachers' education program, where the facilitator had given us few problems to solve. One of the problems motivated her to encourage the teachers to discuss with their students about the difference and equivalence between the following three expressions.

a /  bc ;   (a/b) / c  ;   a / (b/c)

While my students had already encountered, struggled & tackled these expressions, but I recalled that these experiences were within the context, and in isolation i.e. they had fought with these monsters one at a time and that too on different occasions. So I thought it would interesting to notice what happens when they are made to face these three expressions - all together, without any context and that too without any context.

So the next day itself, I challenged them with this problem..


I asked them, 

"Which of the three are equal or not equal?"

They looked at it for a while and soon, one of them said, 

"Sir, we have solved such expressions..."

Me: Yes, Lets do it again then.

Couple of them could not resist and they shouted out their answer. I just looked at them; and they understood their mistake (that they had disturbed their peers' thought process)

After a minute or so, I asked one of them to explain

Rohit - "Sir, 2 and 3 cannot be equal.

"Why?"

"Because in (2), a is divided by the quotient b/x  and in (3), a is divided by the product bx.

I looked at the class for their views. And could see everyone agreeing with this. I wanted someone to counter this, but none did. So I had to play devil.

"Are you sure?"

He guessed something fishy it seems and soon exclaimed - ."..except for x=1, when they will be equal"  :-))

We smiled at each other and then I picked up another student--

Sidharth - "(1) and (3) are equal"

"How?"

He stood up and slowly wrote this on the board -


"Sir, expression (1) is actually a/bx which is same as (3)"

"But how?"

"I multiplied numerator and denominator by x"

"So?"

"In the numerator, we have b... it is divided by x as well multiplied by x.... So there is no effect of x... Only b remains..... And in the Denominator, we get bx..."

Did you notice the language? your comments?

I was confident that he had understood, but I wanted to take him deeper...

"Ok... So what?"

"Sir, then we can now see that both (1) and (3) are equal...."

"But how do you know that the given expression (1) still remains equal to what you have written? May be you would have changed its value by this manipulation?"

I wanted him to argue with me, but then he remained mum.... and kept staring at the 2 expressions.....

Perhaps, Jeetu could not take this prank I was playing... and he intervened immediately, 

"Sir, they are EQUIVALENT FRACTIONS ! So shouldn't they be equal?"

I looked at Sidharth... And found him smiling at all of us :-)

In fact, I could have even probed him further as to why the values of equivalent fractions are equal? and why do you get equivalent fractions by multiplying the Nr and Dr by the same amount? But then, we had worked on these many times earlier... and hence discussing those again now might disturb the flow of the present problem solving process....

I looked around -- "any other views/method?"

Vaishnavi -- "Sir, in (1), we are first dividing a by b, and then the answer is divided by x... So it clearly means that 'a' is divided by the product of b and x..... and hence its equal to a/bx.."

Everyone agreed...... 

"But then I feel,  if a is divided by b and then by x, then its effectively divided by (b+x).."

And immediately the whole class roared aloud -- "No !! We have already worked on this....."

Let me tell you that this remark made by me now, was the same made by them few months ago.....After working out few examples and even through logical reasoning they had proved that why (a/b)/c is not equal to a/(b+c).... I was happy they still remembered their experience......

"Ok, so can someone tell me what will (2) simplify to?"

Tanvi -- "Sir, (2) will become ax/b"

"How?"

"Same method as what Sidharth did... "

"Okay..."

"So can one of you write the simplified forms of all the three expressions?"

One of them wrote these on the board.....  I forgot to take the snap... but this is what it looked like.....


"So then what's the conclusion?"

This time it was a chorus 

"expression (1) and (3) are equal and are not equal to (2)"

I just looked at Rohit again...

"Sir, they all will be equal when x=1"   :-))

I wanted to take them still deeper on this, but I decided to hold for the moment  --- so that we can re-visit this problem after few days, with some flavor...

What do you feel, can there be some more interesting/ important questions? What are those questions?

How did you solve this problem? With understanding or with 'rules and procedure'? :)

How about trying this problem with your students?  Would love to know what happened in your class.....  :)

Regards

Rupesh Gesota

PS: These students are from marathi-medium municipal school.... To know more about the math enrichment program, check the website www.supportmentor.weebly.com