A Diophantine Equation with 2s | Number Theory

Поделиться
HTML-код
  • Опубликовано: 29 июн 2024
  • 🤩 Hello everyone, I'm very excited to bring you a new channel (SyberMath Shorts).
    Enjoy...and thank you for your support!!! 🧡🥰🎉🥳🧡
    / @sybermath
    / @aplusbi
    ⭐ Join this channel to get access to perks:→ bit.ly/3cBgfR1
    My merch → teespring.com/stores/sybermat...
    Follow me → / sybermath
    Subscribe → ruclips.net/user/SyberMath?sub...
    ⭐ Suggest → forms.gle/A5bGhTyZqYw937W58
    If you need to post a picture of your solution or idea:
    intent/tweet?text...
    #numbertheory #diophantineequations
    via @RUclips @Apple @Desmos @NotabilityApp @googledocs @canva
    PLAYLISTS 🎵 :
    ▶ Trigonometry: • Trigonometry
    ▶ Algebra: • Algebra
    ▶ Complex Numbers: • Complex Numbers
    ▶ Calculus: • Calculus
    ▶ Geometry: • Geometry
    ▶ Sequences And Series: • Sequences And Series

Комментарии • 15

  • @runenorderhaug7646
    @runenorderhaug7646 10 дней назад

    Desmos 3d depection of it is interesting

  • @sanamite
    @sanamite 10 дней назад +1

    1) Odd, so n or k is 0, let's say k = 0 for the next step
    2) 2^m - 2^n = 96 = 2⁵•3 = 2⁵(2²-2⁰) = 2⁷-2⁵
    3) you get k or n is 0 and the other is 5 :
    [7;5;0] or [7;0;5]

  • @mcwulf25
    @mcwulf25 10 дней назад

    Obvs m > n,k
    Assume n >= k
    Factor out 2^k:
    2^k (2^m-k - 2^m-n) = 95
    Obvs k=0 for RHS to be odd.
    So 2^m - 2^n = 96
    2^n (2^m-n - 1) = 96 = 32 * 3
    2^n = 32
    n = 5
    2^m-n - 1 = 3
    2^m - n = 4
    m - n = 2
    m = 2 + 5 = 7

  • @allanmarder456
    @allanmarder456 11 дней назад +1

    Clearly m> n and m>k or else the equation would result in a negative answer. Assume k is the smallest of the integers.
    Then (2^k)*[2^(m-k) + 2^(n-k) -1] =95 Now 95 can be factored as 5*19 or 1*95 and one of the factors has to be a power of 2.
    Thus the only possibility is for k=0 and 2^k=1. Then the second factor is (2^m -2^n -1)=95 or (2^n)*[2^(m-n)-1]=96. We can factor
    96 as 2x48 4x24 8x12 16x6 32x3 96x1, The first factor must be a power of 2 and the second factor must be 1 less than a power of 2.
    The only possibility is for the second factor to be 3 (2^2-1) and the factorization of 96 must be 32x3 so n=5 and m-n=2 or m=7.
    Thus k=0 n=5 m=7. If we had assumed n to be the smallest of the integers we would get k=5 n=0 m=7.

  • @rdspam
    @rdspam 9 дней назад

    N or k must be 0. 2^0 is the only way to get an odd number. Then powers of 2 that are 96 apart must be 128 and 32. So 7, 5, and 0.

  • @neilmccoy9390
    @neilmccoy9390 10 дней назад

    a) k=0. b) lhs is multiple of 2^5 but not 2^6 ; so n=5 c) 2^m = 95+1+32=128=2^7. So, m=7

  • @phill3986
    @phill3986 11 дней назад

    😊😊😊👍👍👍

  • @yoav613
    @yoav613 11 дней назад

    Nice

  • @nasrullahhusnan2289
    @nasrullahhusnan2289 11 дней назад

    Let p in an integer. 2^p =1 if p=0 and is even for p>0.
    As RHS is odd, then either n or k, but not both, is 0. Without lost of generality, let n>k --> k=0
    Thus (2^m)-(2^n)-1=95
    (2^m)-(2^n)=96
    =128-32
    =2⁷-2⁵
    Therefore (m,n,k)={(7,5,0),(7,0,5)}

  • @phill3986
    @phill3986 11 дней назад

    Did it in binary - 96 -> 1100000. Rounded up to 10000000 to get n=7, subtracted 1100000 to get 100000 to get m=5

    • @phill3986
      @phill3986 11 дней назад

      Oh and subtract 1 binary (k=0) to get to 95

  • @icems.a.6645
    @icems.a.6645 10 дней назад

    Por tanteo 7, 5 , 0

  • @FisicTrapella
    @FisicTrapella 11 дней назад

    There is another way to get an odd number, but it doesn't work. That is n = k = -1. (=> 2^m = 96; with no solution if m has to be an integer).