Applied math - using numbers in real life situations

To be clear, if you include decimals from 12-15, I believe it is impossible. However, if you just mean integers, I'd have to think about it

Had I not mentioned the any length greater part, you could say 12 but then 16 would not be possible and 16 is greater than 12.

ease up on the accusations

48 is correct per verification.

I have yet to discover a nice, neat formula. In an algebraic work-up 12(t+1)=15*(t) yields t=4 which you could without a deep understand multiply by 12 but I really think number theory has to be invoked for the general case and I might embark on that route.

Here is an approach shy of rigorous math in finding a solution to a more generalized problem of this nature. Looking at 12 and 15 as the sought lengths:

y/N=[12,15] where y =length of log, N=number of pieces cut into

With N=1, 12≤y≤15

With N=2, 24≤y≤30

With N=3, 36≤y≤45

With N=4, 48≤y≤60

With N=5, 60≤y≤75

With N=6, 72≤y≤90

And so once y gets up to 48, any length above this can be divided up by an integral quantity to yield a number between 12 and 15. There are no more gaps in the range for y at this point relative to the integral divisors.

Try 47: 12,12,12 and left with 11
With 15's, the max, 15,12,12, and left with 9

15,15,12, and left with 6

15,15,15, and left with 3

With 48, obviously 4*12 with zero waste

With 49 you can divide by 4 and have a length of cut just over 12 and definitely less than 15, no waste

24 to 30 inches

48 is the ultimate minimum. Trying for example:

47.7, you see after three cuts of the min length of 12 you have a remainder of 11.7...pretty close to 12 but not quite.

OK but the question asks for the minimum length so that any length greater would also be a solution.
here, 31 would not work.

Let's play around with some small numbers to get a feel for what's going on.
if we were working with the numbers 2 and 3 instead of 12 and 15, you could see 4 would be the minimum length. 3.5, 3.6, 3.7 wouldn't divide up well enough. 4.5 = 2*2.75, fine. 5=2*2.5=fine. 5.5=2.75*3=fine, 6 obviously fine. 6.3=3*2.1, fine.

You can see, the LCM is involved but tempered with a number just lower than it OR it is some multiple of the lower number. Try for more different pairings and you could discover more.

And so as long as you cut logs to a length of 48" or greater, as if you are working down a big tree, conserving cuts to get the major work done and not trying to deal with the minutia at the same time, you'd be assured you could later process the logs to obtain the desired length.

...............

with 47.7 speak of your two cuts.

r8j, putting so much effort into trying to find fault doesn't help in solving the problem.

now tell me, since I mentioned 47.7 go and explain away the cuts.......

you really persist. two cuts is immaterial.  the cuts per the minimum length would depend greatly on the LCM/max(x,y) where x and y are the lengths desired for final processing (or something around/near this).

please give it up.

well well well....

OK, I discovered the formula. I'll post it here after I present elsewhere. It's rather interesting, partially simple, partially complicated. Requires use of ceiling(x) which is identical to round(x+1/2).

Of all places, why do you post on a chess.com forum?