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.
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
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 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).
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?
@@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!
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.
@@dontspam7186 nice! That's the key observation!
Amazing sir
Thank you very much!
Whenever the title of a maths video is a question, the answer is no. Same as newspaper headlines.
@@cycklist haha not always! But I would agree in the majority of cases!
Nice problem and very good channel.
Thank you very much, I'm glad you're a fan of the content!
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
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
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.
Are you sure? 2*3*4=4*6
@@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).
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?
@@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!
@@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
@@wjeksababakqabzzhzaab873 Ah interesting, I'll take a look at the problem now - thanks for sharing!