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@faner8653
@faner8653 Күн бұрын
Hello , can you please tell what are the prerequisites for the school research programmes at cheenta
@rockz5678
@rockz5678 Күн бұрын
Hii could you please share the math books link? Thank you
@Cheenta
@Cheenta Күн бұрын
Here is the link for Beautiful Books for Mathematics : www.cheenta.com/beautiful-books/
@sayanmistry8222
@sayanmistry8222 Күн бұрын
❤❤❤
@Aamirdhar457
@Aamirdhar457 2 күн бұрын
Kaczor ☝☝🐺🐺
@abhinavkumarverma2224
@abhinavkumarverma2224 2 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-i2n
@AtifAhad-i2n 3 күн бұрын
Hi. Vaiya❤❤❤❤❤
@aryanahire4758
@aryanahire4758 3 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?
@mwiwabenjamin2569
@mwiwabenjamin2569 3 күн бұрын
I THIS A JOKE OR WHAT
@KrishnenduHazraKrishnendu
@KrishnenduHazraKrishnendu 3 күн бұрын
Thankful to you!❤️😊
@dismyhandle0901
@dismyhandle0901 4 күн бұрын
can we substract the remainders?
@shyamal245
@shyamal245 4 күн бұрын
Too good explaination❤
@gregoriperalmannn
@gregoriperalmannn 4 күн бұрын
what is your name cheeta
@taniyadas5319
@taniyadas5319 4 күн бұрын
Cheenta Academy Private limited
@coderpro5141
@coderpro5141 5 күн бұрын
Yeah even in linear algebra there is something like this, minimal polynomial divides the anhilating polynomial
@ankush_alonso
@ankush_alonso 6 күн бұрын
U r doing great job, hope u expand, this will really help people who love solving maths problems
@ramakrishnakamepalli
@ramakrishnakamepalli 6 күн бұрын
good cheenta!!
@abhinavkumarverma2224
@abhinavkumarverma2224 6 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-hy2ck
@HimanshuRaj-hy2ck 8 күн бұрын
Outstanding lectures ❤️❤️❤️❤️
@HimanshuRaj-hy2ck
@HimanshuRaj-hy2ck 8 күн бұрын
Very helpful lectures
@dylanparker130
@dylanparker130 8 күн бұрын
So cool - love this!
@jortor2932
@jortor2932 9 күн бұрын
Could you help me with letter of recommendation and essay please i mean what one have to do in these topics
@renudubey4826
@renudubey4826 9 күн бұрын
Sir titus lemma is missing
@AlipRanjanSingha
@AlipRanjanSingha 9 күн бұрын
But this a inmo pyq
@AnanyaPawar23
@AnanyaPawar23 9 күн бұрын
great video! thank you for making this!
@chandanchauhan7512
@chandanchauhan7512 10 күн бұрын
nice method
@ankush_alonso
@ankush_alonso 10 күн бұрын
Ohh the idea was very simple, nice
@ankush_alonso
@ankush_alonso 10 күн бұрын
Ohh easy
@MatheExtrem
@MatheExtrem 10 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-ki9ky
@LifeIsBeautiful-ki9ky 11 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_Dutta
@Deepan_Dutta 10 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.
@CombatSports-ug1nk
@CombatSports-ug1nk 11 күн бұрын
PUTNAM AND BEYOND? bhencho yeh to danger hai
@OrelHumphrey-k4n
@OrelHumphrey-k4n 11 күн бұрын
心累了不想再碰了,okx钱包还剩89usdt懒得提了,这是okx钱包的助记词(<pride>-<pole>-<obtain>-<together>-<second>-<when>-<future>-<mask>-<review>-<nature>-<potato>-<bulb>,送各位有缘刷到的朋友。
@太陽-m3u
@太陽-m3u 11 күн бұрын
Sir ape isi course for only number theory and geometry ke liye kitna charge karta hai please answer
@NiyanshiShukla
@NiyanshiShukla 12 күн бұрын
No not possible the last number must be even
@globalolympiadsacademy4116
@globalolympiadsacademy4116 12 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_lord17
@Custom_lord17 12 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-c
@1.6180-c 12 күн бұрын
Fantabulous wisdom.❤
@abcdefgabcdefg3017
@abcdefgabcdefg3017 12 күн бұрын
Keep on doing such videos. Very good problem and solution.
@1.6180-c
@1.6180-c 13 күн бұрын
Dazzling and insightful session.
@Expirationnear
@Expirationnear 13 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
@shivajichakraborty5245
@shivajichakraborty5245 13 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
@danielkhan5452
@danielkhan5452 13 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 ?
@beluga2193
@beluga2193 13 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
@abcdefgabcdefg3017
@abcdefgabcdefg3017 12 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-gq3ej
@ravikant-gq3ej 13 күн бұрын
@shubhsharma150
@shubhsharma150 13 күн бұрын
Damn, hi Rajdeep
@madhavcs1666
@madhavcs1666 13 күн бұрын
Hi Shubh
@sonypandey6834
@sonypandey6834 13 күн бұрын
Thanks sir for these interesting lectures ❤️❤️❤️❤️❤️❤️
@harikishan5690
@harikishan5690 13 күн бұрын
this is great❤
@coderpro5141
@coderpro5141 13 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.
@invincible9240
@invincible9240 13 күн бұрын
Amazing please continue this series
@soumenghosh8430
@soumenghosh8430 13 күн бұрын
🙏
@ArnavAnoopNarayanan
@ArnavAnoopNarayanan 15 күн бұрын
i have a doubt in 11:28 cant j equals m as well
@armanavagyan1876
@armanavagyan1876 17 күн бұрын
The BEST)
@AryanSingh-iu9vt
@AryanSingh-iu9vt 17 күн бұрын
Thanks a lot sir for bringing problems from Russian Olympiads!