Solving This Problem With One Trick

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  • Опубліковано 24 гру 2024

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  • @f5673-t1h
    @f5673-t1h 2 роки тому +26

    This has complex numbers written all over it.
    The equations basically say (a+di)(c+bi) = 19+34i. Take squared norms on both sides, (a²+d²)(c²+b²) = 1517 = 37*41. 37 and 41 are primes that are 1 mod 4, so they can be written as the sum of two squares. The rest of the solution is the same.

    • @zadsar3406
      @zadsar3406 2 роки тому +2

      The question was probably designed with complex numbers in mind. Unfortunately, there are almost always alternative approaches to these competition problems which mask the elegance and usefulness of the complex method.

    • @obrod7080
      @obrod7080 2 роки тому +1

      What does 'taking square norms' mean?

    • @zadsar3406
      @zadsar3406 2 роки тому +1

      @@obrod7080 A norm of a complex number z = x + yi is usually defined as N(z) = sqrt(x^2 + y^2). So taking the square norm would be N^2(x) = x^2 + y^2.
      Although, in the context of Gaussian integers, the norm is commonly defined as this latter function.

    • @f5673-t1h
      @f5673-t1h 2 роки тому +3

      @@obrod7080 If you have a complex number a+bi, then its absolute value (the "norm") is |a+bi| = sqrt(a²+b²). Squaring it gives the square norm, a²+b².
      The absolute value/norm is multiplicative. That means if you have two complex numbers z and w, then |zw| = |z|*|w|. The same thing obviously applies to the square norm, since you're just squaring everythng.

    • @bait6652
      @bait6652 2 роки тому

      @@zadsar3406 but the trick is beautiful in itself.....finding a simplistic trick to stay within Z
      Finding such natural patterns in numbers is more sophisticated and takes more intuition then jumping out of the plane let a cheat code

  • @sarathbabunalam5931
    @sarathbabunalam5931 2 роки тому +2

    Indeed a very tricky problem and the solution is achieved with utmost logic!!! Thanks

  • @yoav613
    @yoav613 2 роки тому +14

    Only the first set. The last set ac-bd=(-1)×(-5)-(-4)×(-6)=-19 and not 19.

    • @healblazer3172
      @healblazer3172 2 роки тому +1

      True, only has one solution.

    • @ubncgexam
      @ubncgexam 2 роки тому +1

      I love it, when there are people whose eyes are working. 😁 😅 😂 🤣 Man... you made my day. 👍🏼😂😂😂

  • @martindedek2885
    @martindedek2885 2 роки тому +3

    I actually thought that the second equation being 19 is pretty big so I assumed that b is probably c+1 and c has to be pretty big because a can't be that big so I tried trial and error and it worked

  • @mcwulf25
    @mcwulf25 2 роки тому +3

    I had to look up the prime factors of 1517 but other than that I did it like you.

    • @willbishop1355
      @willbishop1355 2 роки тому +4

      I checked every prime number up to the square root of 1517 by hand. The very last one, 37, was the answer!

  • @zadsar3406
    @zadsar3406 2 роки тому +4

    Here's another interesting approach. Take z = a + di, w = b + ci, where i^2 = -1. Denote by x' the conjugate of x.
    => z'w = ab + cd + i(ac - bd) = 34 + 19i.
    The norm of a complex number is defined by N(x + yi) = x^2 + y^2. It can be checked that this is a multiplicative function. Taking the norm of both sides gives:
    (a^2 + d^2)(b^2 + c^2) = 1517.
    Now proceed as in the video. If we want a prettier approach utilizing more properties of the Gaussian integers, we can note that 34 + 19i = (1 + 6i)(4 - 5i) and both of these factors are prime in Z[i]. Since none of the a, b, c, d are zero, we are left with the cases where z' and w are both one of these factors multiplied by a unit (that is a number from the set {±1, ±i}) and these can be checked to obtain the solutions from the video.

  • @avinashthakur80
    @avinashthakur80 2 роки тому

    Audio & video are not in sync, audio is lagging a bit.

  • @SuperYoonHo
    @SuperYoonHo 2 роки тому

    nice trick!

  • @chalkfulloffun2866
    @chalkfulloffun2866 2 роки тому

    Also (a,b,c,d)=(4,1,6,5)

    • @2070user
      @2070user 2 роки тому +2

      not at all, since a>b>c>d