Pure maths, physics, logic (braingames.ru): non-trade-related brain games - page 164
You are missing trading opportunities:
- Free trading apps
- Over 8,000 signals for copying
- Economic news for exploring financial markets
Registration
Log in
You agree to website policy and terms of use
If you do not have an account, please register
This option allows you to divide the cake (ingot) into both 9 (obviously) and 8 parts. Do you want to try it further on your own?
This option allows you to divide the cake (ingot) into both 9 (obviously) and 8 pieces. Do you want to try it further on your own?
You need more hints.
:))))
This problem is solved in 3 steps.
2 steps have been shown to you, do you need a hint for step 3?
I think it's elementary. First cut the cake into nine equal pieces. Then, in spite of the cuts (as if it were whole), into 8 more, and then into 7 more as well. Now you can distribute it among 7, 8 and 9 people equally. If you count the number of pieces, it comes to 24 in total. But you can minimize it by making some of the slices overlap. But the point is that the numbers 7, 8, and 9 do not have any common divisor, which definitely says that the matching can be only in the place of the first cut, i.e. where the point 0 (it is also 7/7 and 8/8 and 9/9 in total), i.e. where the first cut, when divided by 9, is also the first for 8 and for 7. So we minimise by 2 pieces. We get 22. Please note that when cutting the cake in a circular pattern, the number of cuts will be strictly equal to the number of slices received. It's also easy to understand the fact that it's not important how to cut the cake (evenly/nearly, perpendicular to the table or diagonally, etc.), since by convention we just need to divide it into any number of parts, each of which may constitute any part of the whole cake (however small or large, but each of them strictly <1), but then everything must be divided equally for all and equally for all. I think it's impossible to argue with that. Suppose we have the restriction that you can cut strictly in a circular pattern from the centre and directly perpendicular to the table without slopes at all (for example, you cut into 2 equal parts through the centre, it is considered that the 2 cuts, which actually 2 pieces and get, as you know). So, for this case, the question is. Will such a problem be equivalent to the given one? Obviously yes, of course. Can we cut it into any number of pieces and make each of them any size we want? Absolutely, it's quite obvious. So it turns out that if this problem has a solution in less than 22 pieces, it can be solved by such cuts. Now let's turn to common sense. There can be 9 people, so no slice can be > 1/9 of the whole cake, otherwise you can't exactly distribute equally to everyone. So, in general, the cake must be cut in such a way that 9 times 1/9 each can be assembled, which means, of course, that the cuts must be made (other cuts may go between them, but don't ignore it) so as to divide 9 times 1/9 each (remember, all cuts are made exactly from the centre to the edge, perfectly straight and perpendicular to the table, so any "tricks" etc. are excluded). Similar should be the cuts that divide into 7 and again into 8 equal fractions. All cuts, due to the lack of common divisors of these numbers, will not coincide, hence we have 24 pieces, hence 24 pieces. Of them 3 can coincide in one place, in zero point (it has already been said about it, see above), so we minimize by 2, and get 22 cuts, and then we get 22 pieces. Again, for want of common divisors, in case of sort of "rotation" of our seven-, eight- or nine-way cuts around the axis, it turns out that the cuts may have just one coincidence, or not at all. It's obvious, obviously. So it can't be less than 22. No way!
WHO IS BRAVE, FIND A MISTAKE IN THE PROOF OF MINIMALITY, AT LEAST SOME. OR AT LEAST A HINT THAT WOULD ALLOW TO DOUBT THE STRICTNESS OF THE EQ. NO, SERIOUSLY, I'M CURIOUS MYSELF)) I'M JUST SURE I CAN BACK IT ALL UP. BUT THERE'S SOME SMART-ASSES OUT THERE WHO SAY IT'S OKAY TO GO UNDER 22. NO, I CAN'T!((
It seemed to me that you had another solution.
So did I, to be frank. As it turned out later, I didn't have a solution for 22.
But I didn't find any universality in Road_king's reasoning either, which proves that it can't be less than 22. There are too many "obviously" that are non-obvious.
What do you think about this? EFFICIENCY=30-50. Bullshit or not?