Reminder of Remainder

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  • เผยแพร่เมื่อ 14 ม.ค. 2025

ความคิดเห็น • 135

  • @jeesimplified-subject
    @jeesimplified-subject  2 หลายเดือนก่อน +22

    We hope that you learned how fundamental properties can occur at unexpected times, a mix of such unique illustrations must be practices by you in case aiming for top ranks. for that only we offer {SO60} to those who struggle with problem solving
    Here's the link for more details
    courses.jeesimplified.com/courses

    • @uttu2008
      @uttu2008 2 หลายเดือนก่อน +2

      Sir we can solve it by binomial method by breaking 25 as 17+8, 19 as 17+2, 2^some as (17-1)^some. At the end we only get multiple of 17..
      Is my method correct?

    • @dikshasbotique6172
      @dikshasbotique6172 11 วันที่ผ่านมา

      Bhaiya I am from class 9th I have just studied factor theorem I have an Idea to proof-
      Consider
      x^n-1 is divisible from x+1 only when n is even so I guess we have to do something with y^n.Please give a solution and if I errored so please correct me.

  • @jeesimplified
    @jeesimplified 2 หลายเดือนก่อน +84

    How was it 👀?

    • @AdarshMourya410
      @AdarshMourya410 2 หลายเดือนก่อน +5

      @@jeesimplified Bhaiya a^(2n) - b^(2n) ko we can rewrite as (a^n)² - (b^n)² fir x²-y²= (x-y)(x+y) property use krke (a^n-b^n)(a^n+b^n) likh skte hai and (a^n-b^n) is divisible by a-b and hence pura a^(2n) - b^(2n) is divisible by a-b

    • @Naitik_Barnwal
      @Naitik_Barnwal 2 หลายเดือนก่อน +5

      Aye, fake account, Original channel pe comment karte hue😂

    • @AdarshMourya410
      @AdarshMourya410 2 หลายเดือนก่อน +1

      @@Naitik_Barnwal 😭😭😭🙏🏻

    • @Naitik_Barnwal
      @Naitik_Barnwal 2 หลายเดือนก่อน +6

      @@jeesimplified bhai, fake account leke, students attract mat karo, hume original channel ka naam pata hai😆

    • @uttu2008
      @uttu2008 2 หลายเดือนก่อน

      Sir we can solve it by binomial method by breaking 25 as 17+8, 19 as 17+2, 2^some as (17-1)^some. At the end we only get multiple of 17..
      Is my method correct?

  • @anshugoel1765
    @anshugoel1765 2 หลายเดือนก่อน +24

    3:01 proof
    By remainder theorem
    Let f(a,b)=a^2n-b^2n
    Put a=b so the function becomes zero which shows that a-b is a factor of this expression
    Put a=-b
    Again the function becomes zero so a+b is also a factor

    • @dheerendrayadav8335
      @dheerendrayadav8335 2 หลายเดือนก่อน +9

      Everybody is probably just copying from Google and using that modulo thingy/lengthy binomial expression
      But here is a person who used a basic method to prove this logically

    • @Quartzite
      @Quartzite 2 หลายเดือนก่อน +2

      ​@@dheerendrayadav8335 yes , see first forget about the big 190 power. Start with 1, the answer is 0. Then put 2 as power for 25, 19, 8 and 2. You'll get 204, which is divisible by 17, so it is also divisible by 34, you can check further by 3, but then you'll reach the same conclusion. The above
      I don't like to get into abstraction cause I'm lazy, anyone can derive the formula or something. But this is just as easy as that.

    • @xninja2369
      @xninja2369 2 หลายเดือนก่อน +1

      Bro but this is verfication not proof 😅 , there is difference between those two​@@dheerendrayadav8335

  • @GamerzInfinite
    @GamerzInfinite 2 หลายเดือนก่อน +43

    it can be proved using binomial theorem:
    we write a^n = (a - b + b)^n
    expanding using binomial thm:
    a^n = k^n + nC1.k^n-1.b + ...... + b^n (where k=(a-b))
    so, a^n - b^n = k {k^n-1 + nC1.k^n-2.b + .....}
    as combination operator always result in integer the whole bracket on RHS is an integer (say p)
    so, a^n - b^n = (a - b) . p
    thus a^n - b^n is always divisible by (a - b)

    • @crazymathematician88
      @crazymathematician88 2 หลายเดือนก่อน

      I have some confusion here hope you could resolve. In last expansion of binomial there is only 2nC2nb²^n then how did you take k as common from it??????????

    • @AAlok_Yadav_AA
      @AAlok_Yadav_AA 2 หลายเดือนก่อน +1

      It is copied from chatgpt😂

    • @strangeboysam2594
      @strangeboysam2594 2 หลายเดือนก่อน

      @@crazymathematician88 last term se K common nahi nikla hai i.e
      a^n = k {k^n-1 + nC1.k^n-2.b + .....} + b^n
      then usne B^n ko left hand side laya hai

    • @crazymathematician88
      @crazymathematician88 หลายเดือนก่อน

      @@strangeboysam2594 got it. Matlab left side lake k common (jo ki (a-b) hain) common liya hain. 👍

  • @dread8478
    @dread8478 2 หลายเดือนก่อน +10

    Congruence Modulo joins the chat 👽

  • @muskanbansal4909
    @muskanbansal4909 2 หลายเดือนก่อน +15

    (a^n)²-(b^n)²=(a^n-b^n)(a^n+b^n)
    (a^n-b^n) can easily be proved as divisible by (a-b) by long division
    While dividing we get few terms of quotient as
    a^(n-1)b⁰ + a^(n-2) b^1 + a^(n-3)b^2 ........a¹b^(n-2)+a⁰b^(n-1)
    And remainder as zero!

  • @jatin2008-d6u
    @jatin2008-d6u 2 หลายเดือนก่อน +5

    We can write b^n as ( a-(a-b))^n and then if we open the binomial , considering a = x and (a-b) = y then this goes as
    b^n = a^n -nC1.a^n-1.(a-b) + ...
    Multiply both sides by - and shift a^n on the left side then this becomes
    a^n - b^n = nC1.a^n-1.(a-b)..
    Take a-b as common from the binomial then we get
    a^n - b^n = (a-b)k where k is some integer
    Therefore hence proved a^n - b^n is divisible by a-b , ( I know there might be some error in the proof but still gave it a fair try though😅)

  • @Naitik_Barnwal
    @Naitik_Barnwal 2 หลายเดือนก่อน +6

    6:51 bhaiya ka whatsapp profile description, battery about to die😂

  • @Anant-c2c
    @Anant-c2c 2 หลายเดือนก่อน +40

    Here we ioqm student get ahead. Congurence modulo🔥

    • @vamshitarun4399
      @vamshitarun4399 2 หลายเดือนก่อน

      tu tg pe famous hogya

    • @Anant-c2c
      @Anant-c2c 2 หลายเดือนก่อน

      @vamshitarun4399 kha par?

    • @nitinpandey4442
      @nitinpandey4442 2 หลายเดือนก่อน

      Sahi kaha 😅

    • @shwetasinha5953
      @shwetasinha5953 2 หลายเดือนก่อน +1

      Wahi, congruence se aasani se ho jayega

    • @arjunsahni8975
      @arjunsahni8975 2 หลายเดือนก่อน

      Mai bhi yahi soch raha tha

  • @kavishkumar9407
    @kavishkumar9407 2 หลายเดือนก่อน +3

    Here we can take pairs satisfying (x^2n - y^2n) which will be divisible by x-y hence following this method for each possible pair we will get that the number will be divisible by 17,6 and their LCM(17,6)=106 will be the least number for which we will have to check divisibility hence answer A
    This was the solution I found to be shorter.

  • @RubyKumari-rp6sr
    @RubyKumari-rp6sr 2 หลายเดือนก่อน +6

    dimag me ek approach aa raha hai ki sayad summation of gp ke kuch ho isse ache se dekhta hun

    • @jeesimplified-subject
      @jeesimplified-subject  2 หลายเดือนก่อน +1

      yes, absolutely correct 💯

    • @RubyKumari-rp6sr
      @RubyKumari-rp6sr 2 หลายเดือนก่อน +1

      @@jeesimplified-subject Thank you bhaiya

  • @n1rvan383
    @n1rvan383 2 หลายเดือนก่อน +2

    Bhai poori calculation galat hai, 2^190 cancel nahi hoga. - sign bhi hai 19^190 pe. In fact 2^191 ho jaayega

  • @strangeboysam2594
    @strangeboysam2594 2 หลายเดือนก่อน +1

    1:40 but n ki value Even ya odd dono ho sakti hai, example n = 3 leke bhi a^3-b^3 divisible ho raha hai a-b se
    also a^n + b^n is divisible by a+b for n belongs to odd number, this can easily be proved by expanding a^n +b^n/a+b as sum of gp
    i.e a^n-1[1-(-b/a)^n]/1-(-b/a)

  • @curious.mathematical.physics
    @curious.mathematical.physics 2 หลายเดือนก่อน +1

    Proof:
    Consider
    f(x)=x^2n-b^2n
    It is divisible by x-b
    Put x=a and we proved it

    • @diteshsingh9588
      @diteshsingh9588 2 หลายเดือนก่อน

      Umm brother.. we have to prove that how your f(x) = x^2n - b^2n is divisible by x-b..
      I think.

  • @sakshamsrivastava4710
    @sakshamsrivastava4710 2 หลายเดือนก่อน +1

    i am 11th grader jee aspirant what i found was ki a^2n-b^2n ek tarah se aise kuch hoga (a-b)(a+b)(a^2+b^2)(a^4+b^4) by simple algebraic identity of a^2-b^2 and from here we see this expression is divisible by (a-b)

  • @expmaths
    @expmaths 2 หลายเดือนก่อน +3

    Same approach just using congruent modulo or direct writing the remainders is faster.
    Like for checking with 17 the expression will be congruent to:
    (-8)^(190) -(-2) ^190 - 8^190 +2^190. U can see Everything cancels out and u get 0 as remainder.
    Similar approach for 7 gives:
    4^190 - (-2)^190 - 1^190 +2 ^190
    So we are left with only 4^190 -1
    And from there same approach as yours to get the remainder as 3.

  • @vatsalvarenya349
    @vatsalvarenya349 2 หลายเดือนก่อน +7

    Mera (D) aaya
    Pehle 25 and 8 aur 19 and 2 mein a^2-b^2 = (a+b)(a-b) lagaya
    isse 17 ka multiple aa gaya
    then second factor mein a^odd-b^odd is divisible by a-b lagaya

  • @ChinmayBarsaiyan
    @ChinmayBarsaiyan 2 หลายเดือนก่อน +3

    Bhaiya conics concept cruch when ?

  • @AjaySingh126aZ
    @AjaySingh126aZ 2 หลายเดือนก่อน +4

    Can be easily prooved by congruent modullo

  • @PriyanshuChhillar-rt6jz
    @PriyanshuChhillar-rt6jz 2 หลายเดือนก่อน +3

    Bro your given proof is done with 2 methods and thankx for such a op ques bro your ques literally open brain

  • @dcttournaments5262
    @dcttournaments5262 2 หลายเดือนก่อน +2

    when we divide apower2n-bpowe2n by a-b. a^2n-b^2n= (a-b)(a^(2n-1)+ b^(2n-1)) + ab(a^(2n-2) - b^(2n-2)....now i replaced 2n with 2n-2 to obtain a^(2n-2) - b^(2n-2)=(a-b)(a^(2n-3)+b^(2n-3) + ab(a^(2n-4) - b^(2n-4)).substitue this in prev eqn.a series is formed with 2 power dec(hence even power required) till power o.adding one component of a-b each time...

  • @bhavyasinghal8976
    @bhavyasinghal8976 2 หลายเดือนก่อน +5

    Bhaiya 4:02 pe (-19¹⁹⁰) hai aur fir jab use 21-2 likhenge to bahar ke minus ki vajah se ultimately (+2¹⁹⁰) rahega Aur firr +2¹⁹⁰ cross cancel nahi hoga

    • @kanishkabose8834
      @kanishkabose8834 5 วันที่ผ่านมา

      (21-2)^190 ke expansion me last term (-2)^190 hoga and 190 being even, the negative sign disappears. so finally the negative sign outside makes it -2^190 which gets cancelled by +2^190

  • @Shubhu519
    @Shubhu519 2 หลายเดือนก่อน +2

    3:59 pe + 2^190 hoga na sir?

  • @lonehunter007
    @lonehunter007 2 หลายเดือนก่อน +1

    4:23 kaise bhaiya??? wo to 2^190 hi hoga...bahar ek minus tha...aapne 4:28 me bhi to -21Y bithaya hai
    aise kaise chalega😭😭

  • @AdityaRaj-h8q
    @AdityaRaj-h8q 2 หลายเดือนก่อน +2

    ANS IS (D)

  • @ridhamsheth4546
    @ridhamsheth4546 2 หลายเดือนก่อน +53

    Congruence modulo

  • @VedantDikshit
    @VedantDikshit 2 หลายเดือนก่อน +1

    3:02 (a²)^n-(b²)^n=(a²-b²)[a^2(n-1)+a^2(n-2)b^2+.......+b^2(n-1)]
    And since (a²-b²)=(a-b)(a+b) thus (a^2n-b^2n) is divisible by (a-b)....
    Correct me if I am wrong somewhere.

  • @shivamchouhan5077
    @shivamchouhan5077 2 หลายเดือนก่อน +1

    If you use congruence modulo,then this question is pretty easy

  • @ojosshiroy8544
    @ojosshiroy8544 2 หลายเดือนก่อน +1

    a^2n-b^2n is divisible by a-b can be proved by binomial theorem. It's probably one of the core things you're taught in this chapter.

  • @saileshachutha7951
    @saileshachutha7951 2 หลายเดือนก่อน +1

    What the hell is this modulo??????
    Please explain it!!!!!!!

    • @dakshhooda9925
      @dakshhooda9925 2 หลายเดือนก่อน

      It is a part of number theory which is not taught
      in routine or jee maths it is taught in Olympiad preparation it is a very fast method for finding remainders I am in class 10 and this q in less than 2 mints

  • @akaalkiratsingh
    @akaalkiratsingh 2 หลายเดือนก่อน +7

    We know that
    A^n-1 * B^1 + A^n-2 * B^2 ...... + A^1 * B^n-1 = (A^n - B^n) /( A - B) [by gp sum]
    Hence (A^n - B^n) is divisible by (A - B) [ LHS is just a some of integers i.e. also integer]

    • @Baby-gp2fw
      @Baby-gp2fw 18 วันที่ผ่านมา

      yeah I did the same too

  • @arkitech1969
    @arkitech1969 2 หลายเดือนก่อน +1

    divisible by 25-8, -19+2,

  • @kabirsingh4155
    @kabirsingh4155 2 หลายเดือนก่อน +1

    If you are a jee asirant ek important theorem yaad karlo aise kisi bhi question mai kam aayega agar tumhe ek p prime se divide hota hai ya nahi ye nikalna hai to jo power hain usko us number se replace kardo jo uska remaider aayega (p-1) se or base number ko remainder of p se replace kardo jaise is question mai 7 ka check karne ke liye above operation ke baad(190=4:6) 256-625-1+16=-354 which is not divisible by 7 ye ek approach ye sare remainder wale q tackle ho jaenge aur tumhe binmial ki shayad hi jarurat padegi

  • @AdarshMourya410
    @AdarshMourya410 2 หลายเดือนก่อน +4

    Bhaiya a^(2n) - b^(2n) ko we can rewrite as (a^n)² - (b^n)² fir x²-y²= (x-y)(x+y) property use krke (a^n-b^n)(a^n+b^n) likh skte hai and (a^n-b^n) is divisible by a-b and hence pura a^(2n) - b^(2n) is divisible by a-b

    • @jeesimplified
      @jeesimplified 2 หลายเดือนก่อน +4

      areee bhai a^n-b*n is divisible by a-b isko kese assume kariya, isko bhi u’ll have to prove.

    • @AdarshMourya410
      @AdarshMourya410 2 หลายเดือนก่อน

      @@jeesimplified got it , will try it again

    • @debadritoduttaedits
      @debadritoduttaedits 2 หลายเดือนก่อน +3

      a^n = (a-b+b)^n
      Now, applying binomial theorem
      a^n = b^n + (a-b)k, where k is an integer
      Subtracting b^n from both sides
      a^n-b^n = (a-b)k
      Which implies a^n-b^n will be divisible by a-b
      Is the proof okay?

  • @mokshjain7403
    @mokshjain7403 2 หลายเดือนก่อน +1

    Lekin bhaiya hum aise bhi kar sakte kya ki kyoki kisis bhi number ka divisibility toh hum check kar hi sakte hai toh isliye woh unsure waale options toh nikal hi Gaye aur phir kyoki b bhi nhi hoga toh directly answer d hi deduce ho jata hai
    Isliye answer d. Ye within less than 15 sec ho jata.

  • @shravan8292
    @shravan8292 หลายเดือนก่อน

    a^2n-b^2n=(a^n+b^n)(a^n-b^n)
    a^n-b^n can be further simplified into (a^n/2-b^n/2)(a^n/2+b^n/2) so on and so forth
    so if n is even eventually this will simplify to (a+b)(a-b) and hence u ll get a long list of all the addition terms times (a-b) so and a and b are integers and hence it will be some k(a-b)
    hence proved
    hope this makes sense lol hard to explain via comments

  • @Gyan-fx9zx
    @Gyan-fx9zx 2 หลายเดือนก่อน +3

    Shortest method use these theorems:
    1) Fermat's little theorem
    2) Euler's totient theorem
    Of course not in jee syllabi

    • @foreveradirectioner5985
      @foreveradirectioner5985 2 หลายเดือนก่อน

      Bro ek baar karke bhej sakte ho, modulo se?

    • @foreveradirectioner5985
      @foreveradirectioner5985 2 หลายเดือนก่อน

      Bro ek baar kar sakte modulo se?

    • @nilusingh377
      @nilusingh377 2 หลายเดือนก่อน

      ​@@foreveradirectioner5985take mod17 entire equation, 17 is prime so from fermat little theorem
      a^16=1mod17
      Followed by simplification.

    • @foreveradirectioner5985
      @foreveradirectioner5985 2 หลายเดือนก่อน

      @@nilusingh377 Ohk, thanks so much!

  • @pushkarnarware
    @pushkarnarware 2 หลายเดือนก่อน +2

    Woh 17 and 34 wale toh ho gye the pr 28 wala nhi hua tha

  • @Mr.Moody-d2e
    @Mr.Moody-d2e 2 หลายเดือนก่อน

    If n=1 then it will become a^2-b^2 which will be (a-b)(a+b) which is dividible by a-b and clearly all even power will be divisible the same way

  • @abhinavgupta1521
    @abhinavgupta1521 2 หลายเดือนก่อน

    for shorter method u can use modulo method, it is a property generally used in olympiads but that property is basically derived from binomial expansions and it is just what u said but by property it is considered to be known already
    basically u can write any number as the no it will leave remainder and leave the power as it is and try converting all numbers into +1,+2,-1,-2 to ease down calculation and u can solve almost every remainder question within 2 min, this one i especially solved before u started solving without pausing vid

    • @naqihaider849
      @naqihaider849 2 หลายเดือนก่อน

      exactly congurence modulo se easily ho jaega but uske liye number theory padhna padega

  • @pranavnarwade9670
    @pranavnarwade9670 2 หลายเดือนก่อน +1

    Sir pls pls explain more of revision ... i mean explain more about revision course in how much time will it over and all
    I wanna buy it but just a little confused...

  • @ankitttt1
    @ankitttt1 2 หลายเดือนก่อน +3

    You are a good question😂

  • @collectregularly
    @collectregularly 2 หลายเดือนก่อน +1

    Idk I enjoyed? Why this happened as I See u didn't edit the blundered part. I think u were fully immersed into it That's why u made it soo smooth for us.

  • @diteshsingh9588
    @diteshsingh9588 2 หลายเดือนก่อน +3

    Can we prove it this way?
    We know, a²-b² = (a+b)(a-b)
    Similarly, a⁴-b⁴ = (a²+b²)(a²-b²)
    Hence.. since in a^2n - b^2n
    The 2n part always gives an even number.. which in turn gives an even factor in terms like this.. like say for example
    In this, a^2n - b^2n .. n=5
    then, a¹⁰ - b¹⁰ = (a⁵-b⁵)(a⁵+b⁵)
    This a⁵-b⁵ can be further simplified into (a-b)(a⁴ + a³b + a²b² + ab³ + b⁴)
    Thus.. we will always get (a-b) as a factor of a^2n - b^2n

  • @AdityaRanawat-b2x
    @AdityaRanawat-b2x 23 วันที่ผ่านมา

    Solved by nv sir method bhot easily

  • @FAUXVIKINGIITB
    @FAUXVIKINGIITB 2 หลายเดือนก่อน

    Ans-d
    The given expression is divisible by 2,3,6,17 also, it is not divisible by 4 hence, the only option that satisfies is d.
    It should also be divisible by 34 i suppose.

  • @anirudhpratapsinghchauhan
    @anirudhpratapsinghchauhan 2 หลายเดือนก่อน

    3:01 by factor theorem we can say

  • @saumyatheallrounder6193
    @saumyatheallrounder6193 2 หลายเดือนก่อน

    Congduence modulo se ho jata hai

  • @rajeevpandey6220
    @rajeevpandey6220 2 หลายเดือนก่อน +3

    Consider the polynomial (P(x) = a^n - b^n). Notice that (P(b) = 0), which means (x - b) is a factor of (P(x)). Thus, (a - b) is a factor of (a^n - b^n).
    In more straightforward terms, for any integers (a) and (b), when you subtract (b^n) from (a^n), the result can always be expressed as a product of (a - b) and another polynomial. Hence, (a^n - b^n) is always divisible by (a - b).
    Does this make sense or not 😅
    Please reply ❤ .
    Thank you ❤
    And yes I have a shorter method then this maybe 5to6 lines .

  • @anonyme_3
    @anonyme_3 2 หลายเดือนก่อน +1

    Solve this using modulo concept

  • @ashutosh4189
    @ashutosh4189 2 หลายเดือนก่อน

    for proof you asked : let take common a^2n from expresssion, then divide , and multply term with (1-b/a ) , this create a normal gp(a^2n+.....b^2n) *(1-b/a) , clearly div, by (a-b )

  • @IITIAN-g9m
    @IITIAN-g9m 2 หลายเดือนก่อน

    Sir you can solve using euler & fermat theorem

  • @adityajha2889
    @adityajha2889 2 หลายเดือนก่อน

    3:05 gp k sum se kiya jaaskta h

  • @priyank6356
    @priyank6356 2 หลายเดือนก่อน

    2:34 how to confirm that it is divisible by 34???

    • @manasonly
      @manasonly หลายเดือนก่อน

      Bcz 34 is a multiple of 17

  • @pranavaggarwal7965
    @pranavaggarwal7965 2 หลายเดือนก่อน

    Well you can use modulu as you know many aspirant give exam of ioqm these are the basic question like remainder so if you use moduluit will take around 2 min but it will get the answer but i love your method

  • @jayagarwal6930
    @jayagarwal6930 2 หลายเดือนก่อน

    A piece of cake for oly asp

  • @ghostisalive3962
    @ghostisalive3962 2 หลายเดือนก่อน

    Bhai property nahi bhi aati ho toh we can split 25 and 19 as 17+8 and 17-2, binomial se open karne ke baad the 8^190 and 2^190 automatically get cut so it becomes divisble by 17. Thoda dhyaan se dekhein toh it becomes divisible by 34 also as all terms have a multiple of 2

  • @GleshawnDsouza-x9d
    @GleshawnDsouza-x9d หลายเดือนก่อน

    can put it in the form x^n-1/x-1 where x is a/b and this is gp sum

  • @arpandutta2404
    @arpandutta2404 2 หลายเดือนก่อน

    This theorem was in cengage book but we ignored like our crush used to ignore us 😅 and jee gave question on this concept

  • @hariprakashbachchas2822
    @hariprakashbachchas2822 2 หลายเดือนก่อน

    Bhaiya pls continue concept crunch series

  • @itachi5061
    @itachi5061 หลายเดือนก่อน

    a2n -b2n ÷ a-b is actually g.p sum

  • @shlokdas9940
    @shlokdas9940 2 หลายเดือนก่อน

    mains mei jo sawal tha waha 14 aur 34 mei hi options diye the yaha alag kyu hai?

  • @Khuntapurva
    @Khuntapurva 2 หลายเดือนก่อน

    Proof
    Consider a GP
    x^n-1+(x^n-2)y+(x^n-3)y^2+......+y^n-1
    Do its sum and simplify it result will be magnificient

  • @Tanishk-g6l
    @Tanishk-g6l 2 หลายเดือนก่อน

    Which software are you using.

  • @PranavSharma-u6z
    @PranavSharma-u6z 2 หลายเดือนก่อน

    Bhai maths ke liye kaunsi achi book rhegi??

  • @blurryface602
    @blurryface602 2 หลายเดือนก่อน

    a^2n wala
    is it because after factoring we get a^n-b^n which divides a-b??

  • @zombie190
    @zombie190 2 หลายเดือนก่อน

    Your prove can be proved by mathematical induction

  • @titanfreefire193
    @titanfreefire193 2 หลายเดือนก่อน

    Congruence Modulo

  • @harshdas4061
    @harshdas4061 2 หลายเดือนก่อน

    Ye maine pyqs krte time kiya tha
    First attempt me to nhi hua pr soln dekhke mja ayega

  • @UnknownGhost97
    @UnknownGhost97 หลายเดือนก่อน

    Answer is D

  • @pratyushlokhande1200
    @pratyushlokhande1200 2 หลายเดือนก่อน

    Can we prove it by lmvt ?

  • @monishrules6580
    @monishrules6580 2 หลายเดือนก่อน

    Sir just learn modulo, mod 17 is so easily then look at mod 7 very easy

  • @sirak_s_nt
    @sirak_s_nt 2 หลายเดือนก่อน

    Congruence modulo and ye without pen paper solved

  • @girishchandrasrivastava12thA
    @girishchandrasrivastava12thA หลายเดือนก่อน

    25 ≡ 8 (mod 17)
    19 ≡ 2 (mod 17)
    8 ≡ 8 (mod 17)
    2 ≡ 2 (mod 17)
    Ab isme
    ((8^(190) -2^(190) -8^(190) +2^(190)) (mod 17)
    Simply kr lo
    0(mod 17)
    Matlb hai ab 28 mein bhi kr lo nahi hoga
    I know ki yeh topic OLYMPIAD ka hai but ek true JEE aspirant tabhi hota hai jab woh har approach ko seekhe aur chota sa hai yaar yeh toh tumhaara B. T. easy kr dega seekh lo jaake 😊

  • @NitinKumar-bg4or
    @NitinKumar-bg4or 2 หลายเดือนก่อน

    Gp ke sum se bhi proof ho jayega sum of this series
    a^n-1+a^n-2b+a^n-3b+.....+b^n-1. Is

  • @gamingtecherz3269
    @gamingtecherz3269 2 หลายเดือนก่อน

    a^n-b^n=(a-b)(a^n-1+a^n-2b+a^n-3b^2+........+a.b^n-2+b^n-1) .hence, it is divisible by a-b
    pls like and pin.

  • @uttu2008
    @uttu2008 2 หลายเดือนก่อน

    Sir we can solve it by binomial method by breaking 25 as 17+8, 19 as 17+2, 2^some as (17-1)^some. At the end we only get multiple of 17..
    Is my method correct?

  • @anikchakraborty92
    @anikchakraborty92 2 หลายเดือนก่อน +1

    Sandal bhaiya 2024 cc op

  • @annanay007
    @annanay007 2 หลายเดือนก่อน

    Easier than my dpp.

  • @ALGEO-777
    @ALGEO-777 2 หลายเดือนก่อน

    Bro use cogurence Modulo IT WILL become more easy

  • @anshukartikeyan9146
    @anshukartikeyan9146 2 หลายเดือนก่อน

    no need of this for to prove it's divisible by 17.. remainder of 1st one is 16, second one is 4, third one is 1 and fourth one is 13.. add all and it's divisible by 17...

  • @VedMohod-jw3gb
    @VedMohod-jw3gb 2 หลายเดือนก่อน

    Pta nahi sahi proof hai ya nahi but I just tried
    (a^2n -b^2n)/a-b. Isko a^2-b^2 identify se break kiya toh {(a^n -b^n)(a^n+b^n)}/a-b aise break karte gaye toh iske baad n/2 banega uske baad n/4 till n/n which is equal to 1 toh ekdam last step mein a-b numerator mein banega toh wo cut hojayega that why divisible

  • @SeemaGupta-ob5pe
    @SeemaGupta-ob5pe 2 หลายเดือนก่อน +5

    did it, as a IOQM student

  • @SagarKumar-l3k1d
    @SagarKumar-l3k1d หลายเดือนก่อน

    But I got it correct 😌

  • @shivx3295
    @shivx3295 2 หลายเดือนก่อน +1

    fermats little theorem aur khatam karo isko

  • @shauryavardhansingh9186
    @shauryavardhansingh9186 2 หลายเดือนก่อน

    Euler and fermat laughing in corner

    • @IITIAN-g9m
      @IITIAN-g9m 2 หลายเดือนก่อน

      Yes

  • @Aizen-ll9oc
    @Aizen-ll9oc หลายเดือนก่อน

    Ye wali property aati h. 🌚🐮

  • @TPD_Dev
    @TPD_Dev 2 หลายเดือนก่อน

    Solved your question for the first time just by looking at it (it was done in our class)