Is 81 the ONLY Two-Digit Number with This Property?

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  • Опубліковано 30 жов 2024

КОМЕНТАРІ • 19

  • @CornerstonesOfMath
    @CornerstonesOfMath  6 місяців тому

    Check out the follow-up video of this video, where I attempt to find 4-digit numbers xyzw satisfying √(xyzw) = xy + zw: ua-cam.com/video/2q1ruls5vU8/v-deo.htmlsi=HR2j6soLsQp3-ke5

  • @Bhrz
    @Bhrz 6 місяців тому +1

    I enjoyed your method of proving the question. Thanks for sharing!

    • @CornerstonesOfMath
      @CornerstonesOfMath  6 місяців тому +1

      Thanks! This is one of the most well-known methods to deal with quadratic equation with integer roots.

  • @empireempire3545
    @empireempire3545 6 місяців тому +3

    You should also mention that this result is in decimal system. More interesting would be to generalize this to any number system, in which case i suspect you'd end up with an equation as a result.

    • @CornerstonesOfMath
      @CornerstonesOfMath  6 місяців тому +1

      Well at least I can say that for an arbitrary base n, two-digit number (n - 2)1 base n (n - 2 as n's digit and 1 as 1's digit, base n) ALWAYS satisfies the given property - that is,
      √(11 base 3) = 1 + 1, √(21 base 4) = 2 + 1, √(31 base 5) = 3 + 1, and so on.
      This is because (2-digit number (n - 2)1 base n) = n(n - 2) + 1 = (n - 1)^2, so its square root is n - 1 = (n - 2) + 1.
      I don't know if there are other possible numbers though, and I don't think I will explore that ever (the equation gets way complicated due to the additional variable for the base).

  • @tonyennis1787
    @tonyennis1787 6 місяців тому +2

    Very good

  • @PrimalPower
    @PrimalPower 6 місяців тому +14

    Or you could have just tested all perfect 2 digit squares (there's only 6 of them)

    • @CornerstonesOfMath
      @CornerstonesOfMath  6 місяців тому +2

      Yeah even we don't get to explore the theory of quadratic equations with integer roots, that is perhaps the fastest way to do it.

    • @tonyennis1787
      @tonyennis1787 6 місяців тому +1

      The answer isn't the point. Learning how to solve the problem is the point.

    • @PrimalPower
      @PrimalPower 6 місяців тому

      @@tonyennis1787 I get what you mean but my solution is also a way to solve it. However, the solution shown in the video becomes increasingly better for larger numbers but this "obvious" method is more efficient for 2 digit squares

  • @konstanty8094
    @konstanty8094 6 місяців тому +3

    Can you do something similar for 3 or 4 digits?
    Because the solution to this problem is too easy to find by just checking in your head.

    • @CornerstonesOfMath
      @CornerstonesOfMath  6 місяців тому +3

      Well, I wasn't going to dive too deep into this, but since you asked:
      I haven't tried cases like √(xyz) = x + y + z or √(xyzw) = x + y + z + w, because they already seem too complicated due to the increased number of variables,
      but I managed to handle √(xyzw) = xy + zw pattern, because this too only requires 2 variables and can be expressed as
      √(100a + b) = a + b, where a and b are now two-digit numbers maximum.
      Squaring both sides and doing similar process, we obtain
      a = - (b - 50) ± √(2500 - 99b)
      so we have
      2500 - 99b = k^2 (k is integer) and 2500 - 99b ≥ 0 (equivalent to b ≤ 25.xxx)
      This time there are slightly more cases, and it is actually beneficial to use b = (2500 - k^2)/99 = (50 + k)(50 - k)/99 instead, testing integer values k that makes b integers.
      Then we can find that only possible values are k = 5 and 49, which gives b = 25 and 1, which in turn gives
      (a, b) = (20, 25), (30, 25), (98, 1)
      Therefore, there are three cases (if you include single-digit number such as 01):
      √2025 = 20 + 25 = 45,
      √3025 = 30 + 25 = 55,
      √9801 = 98 + 01 = 99.
      Now I'm wondering if I should make an additional video with these results.

    • @ShivanshPachnanda
      @ShivanshPachnanda 6 місяців тому

      @@CornerstonesOfMath Can't you show that max face sum of a 5 digit number is 54 and even (54)^2

    • @CornerstonesOfMath
      @CornerstonesOfMath  6 місяців тому +1

      @@ShivanshPachnanda It can be easily shown that it is impossible from 5 digits because for a 5-digit number abcde, √(abcde) ≤ 9+9+9+9+9 = 45 and thus abcde ≤ 45^2 = 2025.
      And I think I can show you that even 4-digit numbers (abcd) are impossible since abcd ≤ (9+9+9+9)^2 = 1296, so the value of a is limited to a = 1 only, but then we have 1bcd ≤ (1+9+9+9)^2 = 784, which is impossible.
      If we use cube root ³√ or 4th root ⁴√ then there can be different stories, but again, I just don't think that this problem is worth expanding.

    • @konstanty8094
      @konstanty8094 6 місяців тому +1

      ​@@CornerstonesOfMath
      edit: I was 1h late with the N>=5 proof
      there are no solutions for N-digit numbers where n > 2. I checked up to 15 with a python script.
      1: [1]
      2: [81]
      3: []
      4: []
      5: []
      6: []
      ...
      Proof for N >= 5 is easy, because
      (N * 9)^2 < 10^N if N >= 5
      But the cases for 3,4 are less trivial.

    • @Bhrz
      @Bhrz 6 місяців тому +1

      I would love to see a video on this conversation thread

  • @embers777
    @embers777 6 місяців тому +4

    very noice