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Sir, I was not able to solve ques no.1 and 2 but *I solved ques no. 3* . This was the moment I gained so much of confidence as I'm in class 10 😁 Thank you very much sir...😁🙏🏻
In subjective is the proof of last question valid like the following? Let the to digit number be aa so, 10a+a = m^2 11a = m^2 so, a=11(k)^2, where k shouldn't be zero as we need a two digit number. Now, Range of a is from 0 to 9. And substituting any natural no. in place of k will not give the following range, so it is not possible in two digit number. Similarly, three digit number be aaa -> 111a = m^2 a = 111(k)^2 And with same explanation no possibilities here, Furthemore, if same digits are take for 4,5... and till endless times, we will only get a= 1111(k)^2 (a=4) a=11111(k)^2 (a=5) a=111111(k)^2 (a=6) . . . a= m(k)^2 where m represents the number of times 1 must be repeated depending on the value of a. for every case the range of a will not be satisfied and so no integer 'n'> or = 10, can be represented as a perfect square if the number contains same digits.
At 1:07:21 , We are assuming p = 9q+2 or 9q-2 (as per slide, bottom lines), but none of these conditions satisfy p as 59 being of the form 9q-2 or 9q+2 ?? where is the contradiction?
The content will be same but on vedantu there will be more personalized learning like doubt solving , session notes, live classes, study material , choose from different teachers , regular tests to analyze your learning and where you stand , extra sessions for general things like time management, strategy and schedule etc.
If you’re an elite student and have lots of academic achievements and you have what it takes to represent India on the International Stage then VOS is here with a lifetime opportunity to help you crack Olympiads that too free of any cost. Click on the Link and Fill the Form to know more: vdnt.in/vos2021olympiadbatches
Wah kya padaya sir
Sir, I was not able to solve ques no.1 and 2 but *I solved ques no. 3* . This was the moment I gained so much of confidence as I'm in class 10 😁
Thank you very much sir...😁🙏🏻
now what
Sir what a teaching style 🔥🔥🔥❤️
52:36 That moment was very special.. understanding and decrypting that sum was wonderfully explained!!
Hi daamin . do u also study from vedantu me too i am ujjwal. also are you giving IOQM 2024
@@shilpikothari3781 yeah bro
41:52 loved this one
i hate this one
This one was easier one 🙂
In subjective is the proof of last question valid like the following?
Let the to digit number be aa
so, 10a+a = m^2
11a = m^2
so, a=11(k)^2, where k shouldn't be zero as we need a two digit number.
Now, Range of a is from 0 to 9. And substituting any natural no. in place of k will not give the following range, so it is not possible in two digit number.
Similarly, three digit number be aaa
-> 111a = m^2
a = 111(k)^2
And with same explanation no possibilities here,
Furthemore, if same digits are take for 4,5... and till endless times, we will only get
a= 1111(k)^2 (a=4)
a=11111(k)^2 (a=5)
a=111111(k)^2 (a=6)
.
.
.
a= m(k)^2 where m represents the number of times 1 must be repeated depending on the value of a.
for every case the range of a will not be satisfied and so no integer 'n'> or = 10, can be represented as a perfect square if the number contains same digits.
SIR WHERE IS SESSION'S PDF??
Thank you very much sir for your fantastic explanation 🙏🙏
Gajab sir ,thnk u
You are the best sir whole in the world
Who know this give thumb up
Best teacher
Maje aaye 😊😊
Simply awesome Sir 🙏🙏🙏🙏
Can we use this for kvpy
Sir can we do by hit and trial
I have solved many questions by that very easily
life ke decisions me bhi hit and trial karega kya?
At 1:07:21 , We are assuming p = 9q+2 or 9q-2 (as per slide, bottom lines), but none of these conditions satisfy p as 59 being of the form 9q-2 or 9q+2 ?? where is the contradiction?
I think sir made a mistake with the sum
46:35 "isko dekhe darna nahi" me dar nahi raha tha par apne us par point karke daraadiya🤣🤣🤣
Thankyou sirji❤
Bohot accha tha sir
Wonderful sol of RMO
1991
Sir in prmo Delhi 2016 you has not solved 1x9
Kafi amazing explanation
is this theory good or isi?
No bro
No bro this much theory is not enough
Nice Sir
I can't hear your language please. How can you change this video's language to English language?
Thankyou sir for this lecture 😊
The last question was from INMO 1992
nooo
Rmo 1992*
Are sir prmo 2019 Delhi wala jo ques tha usme 1x9 wale form se solve hi nhi kiya aapne
Sir plz reply
@@aqua80hydra41 1_9 ki koi bhi value carry forward nhi krti isliye use neglect krdiya hai
Ohk
Upar se jab answer 1x1 se aa gaya toh 1x9 se karne ki kya zarurat
sir thanku dil se for so wonderful content
Sir please give us notes
Maza aagya
Nice
20:50 can someone please explain how 3 is taken as -1. Please help
Just divide 3 once again by 4 and 3-4 will be -1 . Similarly you can say n≡ 7≡ -1(mod 8)
sir session pdf nahi hai
sir vantantu pai or youtube pai content to same rahaga na for prmo
The content will be same but on vedantu there will be more personalized learning like doubt solving , session notes, live classes, study material , choose from different teachers , regular tests to analyze your learning and where you stand , extra sessions for general things like time management, strategy and schedule etc.
Sir please share pdf of all the slides of this video.
Maza aa gya
sir What about Root 2 ka square it will be 2 which is ending with 2
But √2 is not an integer and that's why 2 is not a perfect square number....
well done you have proved maths wrong
It is not a natural number ❤
Any natural number...
Beta so ja tu Rene de 😂
Sir I am in class 7th and can I see this lecture
Yes you can if you know the basics till class 10
Note pdf is not available
46:26 what a moment . I solved an rmo level question.😆😆😃
no
Sir 9 possibility ;check77
77+77= 154, which is not a perfect square, so It's not possible'
01:47 🤣🤣
Sir how n sq=10a +b sq?
🖒
Hi sir
Sir there are lots of mistakes in seconds last question done by you
Yes n cannot be 59 because answer ladt digit would be 8 and perfect square last digit is not 8
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