RE: Brainsteem Mathematics Challenges: Rectangles in Squares
So as far as I've come now in the evening (22:35 local time) is the following:
M has to be 11, 12 and/or 13 since you would have to find a combination working for these numbers.
These numbers you get by the upper and lower bound.
The smallest number you can achive is:
1x10+2x9+3x8+4x7+5x6=110
sqr(110)=10,...
and the biggest number:
10x9+8x7+6x5+4x3+2x1=190
sar(190)=13,...
since only integers are allowed only 11,12,13 work.
Also there are only only some ways to achive these.
let u=uneven e=even
so
uu=u
ue=e
ee=e
So you can only achive u (11 and 13) with
uu ue ue ue ee
And e (12) only with
uu uu ue ee ee or
ue ue ue ue ue
That's as far as I got right now.
I'll post an update if there is any. As you can see the only thing left to do is to find an eample for each one.
I'll keep you updated.
I thought that since I've seen your post two days after you posted it that it'll surely be answered till now, but maybe I'll be the first with the results (If there really are solutions).
Otherwise maybe my thoughts helped somebody
Yeah, as you've seen, I'm not limited to just one upvote; sometimes different people solve different parts of the puzzle.
I quite like this question :-) Without many stated conditions, the nature of integer arithmetic is enough to really narrow down the field!
The odd-and-even argument is good; one extra insight about the geometry of the rectangles.
And... everyone else is free to join in!
Yes, I've seen that. It was more kind of a personal thing with motivation and so.
So if I see an unsolved or not completly solved question I'm more likely to give the highes effort possible. Anyway now I've time to try finishing it. Thanks by the way for posting all these interesting questions and also for upvoting. Much appreciated!
If I don't get it today I'll probably ask my mother, if she also wants to try. She's a math teacher in Grundschule (elementary school) and also quite enjoys mathematical questions, riddles or similar things.
I didn't have much time yesterday but now I cracked at least the case M=11
Edit: I can proof (at least I think I can) that there is no solution for M=12 and so only M=11 and M=13 work.
And while thinking about the proof I realized that I could have solved it much easier with a little more logical thinking. If somebody wishes a short way to get the numbers I can hand it in just reply to one of my comments. BTW hope there aren't that many spelling or grammar mistakes.