BMO: Can the product of 4 consecutive integers equal the product of 2 consecutive integers?

Поділитися
Вставка
  • Опубліковано 24 гру 2024

КОМЕНТАРІ •

  • @dontspam7186
    @dontspam7186 Місяць тому +2

    Very neat solution. When I saw it for the first time I observed multiplying four consecutive integers always gives you one less than a square, which you factorise and show with a bit of heavy algebra and then proceeded as follows in the video with the square sandwiching.

    • @JPiMaths
      @JPiMaths  Місяць тому

      @@dontspam7186 nice! That's the key observation!

  • @Physicslover1879
    @Physicslover1879 Місяць тому +1

    Amazing sir

    • @JPiMaths
      @JPiMaths  Місяць тому

      Thank you very much!

  • @cycklist
    @cycklist Місяць тому +2

    Whenever the title of a maths video is a question, the answer is no. Same as newspaper headlines.

    • @JPiMaths
      @JPiMaths  Місяць тому

      @@cycklist haha not always! But I would agree in the majority of cases!

  • @stefanoguarini4710
    @stefanoguarini4710 Місяць тому

    Nice problem and very good channel.

    • @JPiMaths
      @JPiMaths  Місяць тому

      Thank you very much, I'm glad you're a fan of the content!

  • @user3lskq3
    @user3lskq3 Місяць тому

    what was the intution to get the line of (n^2+3n)(n^2+3n+2)?
    why did you combine the first and 4th term? why not any other combination at first

    • @JPiMaths
      @JPiMaths  Місяць тому

      Good question, I think noticing that then both will be very similar quadratics (both of the form n^2+3n+k), which means that they'll be the product of two integers a fixed distance apart, which seems like a useful observation for this problem

  • @pawebielinski4903
    @pawebielinski4903 Місяць тому

    I'd like to note that you can prove a more general result, namely that a product of integers that differ by 1 is never the product of integers that differ by 2, simply by considering the last digit.

    • @JPiMaths
      @JPiMaths  Місяць тому

      Are you sure? 2*3*4=4*6

    • @pawebielinski4903
      @pawebielinski4903 Місяць тому

      @@JPiMaths No no, I mean two factors, so that it implies the main problem after partly distributing on the left.
      While this result holds, it is not by just the last digit analysis, at least not without additional consideration (so yes, I was partly mistaken).

  • @wjeksababakqabzzhzaab873
    @wjeksababakqabzzhzaab873 Місяць тому

    Amazing explanation! This question is hard in general , but easy for a q5 on a bmo1 paper. Please carry on with the bmo problems! Have you considered some geometry problems?

    • @JPiMaths
      @JPiMaths  Місяць тому +1

      @@wjeksababakqabzzhzaab873 I agree, I've definitely seen harder problems for Q5 BMO. I definitely have a few BMO problems I have in the schedule for recording. I will try some geometry ones, but this is normally a bit trickier to make titles/thumbnails for!

    • @wjeksababakqabzzhzaab873
      @wjeksababakqabzzhzaab873 Місяць тому

      @@JPiMaths i abit late but bmo1 2006/7 q6 is one i did recently that is approachable and the solution is very simple number theory plus some logic

    • @JPiMaths
      @JPiMaths  Місяць тому

      @@wjeksababakqabzzhzaab873 Ah interesting, I'll take a look at the problem now - thanks for sharing!