Hello , can you please tell what are the prerequisites for the school research programmes at cheenta
@rockz5678Күн бұрын
Hii could you please share the math books link? Thank you
@CheentaКүн бұрын
Here is the link for Beautiful Books for Mathematics : www.cheenta.com/beautiful-books/
@sayanmistry8222Күн бұрын
❤❤❤
@Aamirdhar4572 күн бұрын
Kaczor ☝☝🐺🐺
@abhinavkumarverma22242 күн бұрын
(i) 4k-1 means the form 4k+3. We can see that 4k is divisible by 2 and k and 4k+2 is also divisible by 2 for k>0. Therefore, all the primes are of the form 4k+1 and 4k-1 (ii)Let us assume that there are finite number of prime numbers p1,p2,p3,... of the form 4k-1 Let N = 4×p1×p2×p3×... - 1 We can see that no p1,p2,p3... can divide N. Hence it is a new prime. Here we get a contradiction. Therefore we can say that there are infinitely many prime numbers of form 4k-1
@AtifAhad-i2n3 күн бұрын
Hi. Vaiya❤❤❤❤❤
@aryanahire47583 күн бұрын
the quality of your videos as well as the specific knowledge required to make this video is phenomenal! I am in class 9 preparing for ioqm and it is my goal to go to imo. Is it possible for anyone with enough practice and hard work to get in imo?
@mwiwabenjamin25693 күн бұрын
I THIS A JOKE OR WHAT
@KrishnenduHazraKrishnendu3 күн бұрын
Thankful to you!❤️😊
@dismyhandle09014 күн бұрын
can we substract the remainders?
@shyamal2454 күн бұрын
Too good explaination❤
@gregoriperalmannn4 күн бұрын
what is your name cheeta
@taniyadas53194 күн бұрын
Cheenta Academy Private limited
@coderpro51415 күн бұрын
Yeah even in linear algebra there is something like this, minimal polynomial divides the anhilating polynomial
@ankush_alonso6 күн бұрын
U r doing great job, hope u expand, this will really help people who love solving maths problems
@ramakrishnakamepalli6 күн бұрын
good cheenta!!
@abhinavkumarverma22246 күн бұрын
We can see that in both (u+v) and (u-v), both are divisible by 5. So ther sum also must be divisible by 5 which is 2v and v is x²+10x+2. We clearly see that 10x is divisible by 5 but x²+2 is not. We can say that x²+2 is not divisible by 5 because the last digit of multiples of 5 are 0 or 5. Therefore x² last digit must be 3 or 8 which cannot be possible. Therefore condition not satisfy.
@HimanshuRaj-hy2ck8 күн бұрын
Outstanding lectures ❤️❤️❤️❤️
@HimanshuRaj-hy2ck8 күн бұрын
Very helpful lectures
@dylanparker1308 күн бұрын
So cool - love this!
@jortor29329 күн бұрын
Could you help me with letter of recommendation and essay please i mean what one have to do in these topics
@renudubey48269 күн бұрын
Sir titus lemma is missing
@AlipRanjanSingha9 күн бұрын
But this a inmo pyq
@AnanyaPawar239 күн бұрын
great video! thank you for making this!
@chandanchauhan751210 күн бұрын
nice method
@ankush_alonso10 күн бұрын
Ohh the idea was very simple, nice
@ankush_alonso10 күн бұрын
Ohh easy
@MatheExtrem10 күн бұрын
7:54 If 5 would divide both u+v and u-v, it would also divide their difference, namely 2v, and therefore v. Since 5 divides both u+v and v, it divides their difference, u. Therefore, let a = u/5, b = v/5. Either 5 divides b or 5 divides a. Therefore neither (a-b) nor (a+b) is congruent to 0 modulo 5. Therefore (a+b)*(a-b) = (u+v)(u-v)/5^2 = 2000/25 = 80 cannot be divisible by 5. A contradiction. Therefore 5 cubed must divide either u+b or u-v.
@LifeIsBeautiful-ki9ky11 күн бұрын
this method works for specific problems only. It doesnt work for : Y^2=x^4+1, y^2=x^4+x^3+6
@Deepan_Dutta10 күн бұрын
That's correct. Actually there isn't any such general method to solve a diophantine equation, for this equation it's one of the strategies and may be for some other equations as well.
Sir ape isi course for only number theory and geometry ke liye kitna charge karta hai please answer
@NiyanshiShukla12 күн бұрын
No not possible the last number must be even
@globalolympiadsacademy411612 күн бұрын
The sum is even for all the numbers from 1 to 1000 is an even nunber. Everytime you delete two nunbers and replace it by the difference the parity of sum remains even, so 11 is not possible in the end.
@Custom_lord1712 күн бұрын
: Observations: 1. Initial Numbers: The sum of the numbers on the board starts as , which is: S = \frac{1000 \cdot (1000 + 1)}{2} = 500500 2. Operation: Each time two numbers and are erased, the difference is added to the board. This effectively reduces the total sum by , leaving: S' = S - 2 \cdot \min(a, b) 3. Parity of : S = 500500 \text{ (even)}. Conclusion: Since 11 is an odd number, it is not possible for the last number on the board to be 11.
@1.6180-c12 күн бұрын
Fantabulous wisdom.❤
@abcdefgabcdefg301712 күн бұрын
Keep on doing such videos. Very good problem and solution.
@1.6180-c13 күн бұрын
Dazzling and insightful session.
@Expirationnear13 күн бұрын
The answer for this question of isi should be square root of n but in answer key in ur site it’s written n
@shivajichakraborty524513 күн бұрын
Sir, I shall be grateful if you could kindly provide an algebric solution to this problem (4/5)*x plus (3/5) *x=1
@danielkhan545213 күн бұрын
Good evening everyone. Can someone advice me in calculas for ISI UGB exam The standard book in my coaching is IA Maron but my question is ia maron sufficient for UGB exam or do I have to do BABY RUDIN ? Is it compulsory to do BABY RUDIN for UGB exam as my questions are out of syllabus ?
@beluga219313 күн бұрын
Well Ive no experience with calc but I.a Maron along tomato is sufficient (aese to nothing is sufficient) If you've nothing to do and done with everything then may try black book
@abcdefgabcdefg301712 күн бұрын
Try kaczor and nowak and Putnam and beyond for real analysis. Also practice all jee advanced problems and north Georgina calculus problems
@ravikant-gq3ej13 күн бұрын
❤
@shubhsharma15013 күн бұрын
Damn, hi Rajdeep
@madhavcs166613 күн бұрын
Hi Shubh
@sonypandey683413 күн бұрын
Thanks sir for these interesting lectures ❤️❤️❤️❤️❤️❤️
@harikishan569013 күн бұрын
this is great❤
@coderpro514113 күн бұрын
the sum of series question is also interesting because not only p divides a but also p^2 divides a, however for that you need just slight multiplicative identity knowledge.
@invincible924013 күн бұрын
Amazing please continue this series
@soumenghosh843013 күн бұрын
🙏
@ArnavAnoopNarayanan15 күн бұрын
i have a doubt in 11:28 cant j equals m as well
@armanavagyan187617 күн бұрын
The BEST)
@AryanSingh-iu9vt17 күн бұрын
Thanks a lot sir for bringing problems from Russian Olympiads!