#### Re: Re: Gold ingots of Geriona

There are two combinations of numbers if them to present in two ingots, in the total giving 11. 2+9 = 11 1+10 = 11 the First weighing 1+9 = 10 bag is not torn, we postpone an ingot in 9 1+10 = 11 bag is not torn, we postpone an ingot in 10 2+9 = 11 bag is not torn, we postpone an ingot in 9 2+10 = 12 bag is torn the ingot Means in the first pair 1 kg was powerful. If the ingot in weight in 3 and more  gets to us

#### Re: Re: Gold ingots of Geriona

Hello, - n1l - you wrote: N> There are two combinations of numbers if them to present in two ingots, in the total giving 11. N> 2+9 = 11 N> 1+10 = 11 N> the First weighing N> 1+9 = 10 bag is not torn, we postpone an ingot in 9 N> 1+10 = 11 bag is not torn, we postpone an ingot in 10 N> 2+9 = 11 bag is not torn, we postpone an ingot in 9 N> 2+10 = 12 bag You are torn only showed Gerionu that there are two pairs which do not tear a bag. It is a lot of such pairs... Meanwhile everything that m we can state that among these 4 is not present 11... For example, the same circuit: 3+8 = 10 2+9 = 11 2+8 = 10, we postpone 8 2+9 = 11, we postpone 9 3+9 = 12 and we  2 as 1, if Gerion to us checks it, certainly...

#### Re: Re: Gold ingots of Geriona

You in the decision do the same, you assume that already know weight of any ingot. For example in your case there is still a heap of other combinations in three ingots which do not tear a bag.

#### Re: Re: Gold ingots of Geriona

And and this true assumption as Archimedes learned weight of all ingots and wants to prove Gerionu that the specific ingot weighs 1 kg. It is necessary to wait  behind details to the task.

#### Re: Re: Gold ingots of Geriona

Hello, - n1l - you wrote: N> You in the decision do the same, you assume that already know weight of any ingot. 1) the weight of ingots is known on a condition. Archimedes somehow weighed them, and now it should convince Geriona that is is specific here this ingot - 1... N> For example in your case there is still a heap of other combinations in three ingots which do not tear a bag. Well we suppose that Archimedes . He wrote on ingots of weight from 1 to 11, and actually there others. Show swap of numbers from 1 to 11 which at substitution in my decision, does not tear bags five times, but shows not on 1, instead of something another...

#### Re: Re: Gold ingots of Geriona

Hello, - n1l - you wrote: N> N> Archimedes learned weight of all ingots and wants to prove Gerionu that the specific ingot weighs 1 kg. N> it is necessary to wait  behind details to the task. Something is not clear to you in a condition? There are 11 ingots bar_1, bar_2. bar11 They weigh from 1 to 11 We should find such minimum sequence of inequalities of a type bar_i + bar_j +... + bar_n <12 And, probably, one type bar_i + bar_j +... + bar_n> = 12 That they can be fulfilled simultaneously all only if bar_1 = 1

#### Re: Re: Gold ingots of Geriona

Than the sequence 6+1+2 = 9 will be cause a stir from 1+3+7 = 11 in the given task?

#### Re: Re: Gold ingots of Geriona

Well here I also found.

#### Re: Re: Gold ingots of Geriona

Hello, Kodt, you wrote: And then for a long time became silent in  generally. Perhaps in  is at war or generally in  what. Well I, as involved, before it  preferred not to read E>> I know the decision from 5 weighings... I know the decision from 2 weighings (removed a spoiler - Kodt) Cool! In one, obviously is impossible... About 40-kg the set of weights did not know, . And remaining - well so-so... Well I googled three seconds only updt: about false weights, by the way, if to admit and an error in the form of swap of markings, too it is unusual... Same Wikipedia! Come yes correct. Or I do not understand something? Well at one time there not all resolved, but it can is again possible for all

#### Re: Re: Gold ingots of Geriona

Hello, - n1l - you wrote: N> Than the sequence 6+1+2 = 9 will be cause a stir from 1+3+7 = 11 in the given task? You about my decision? We showed Gerionu two groups on 4 ingots which do not tear a bag. Three general ingots in these groups - 1, 2, 3 those, two which changed - 4,5 we take one of group 6-11, and two of 1-5 further... As we could show three dial-ups 6+1+, 6 cannot be 7...

#### Re: Re: Gold ingots of Geriona

I take two groups: {1. 8} {9,10} 11 at all I do not use. Well and here at me a combination on two weighings. And if I take 9 there it can not appear 8 in any way...

#### Re: Re: Gold ingots of Geriona

Hello, - n1l - you wrote: N> Well here I also found. What found? ? Show? That on what to replace? bar_1 = 1 bar_2 = 2 bar_3 = 3 bar_4 = 4 bar_5 = 5 bar_6 = 6 bar_7 = 7 bar_8 = 8 bar_9 = 9 bar_10 = 10 bar_11 = 11

#### Re: Re: Gold ingots of Geriona

Hello, - n1l - you wrote: N> Ok I take two groups: N> {1. 8} {9,10} N> 11 at all I do not use. N> well and here at me a combination on two weighings. And if I take 9 there it can not appear 8 in any way... Well look. You showed  11 ingots, naming them bar1, bar_2. bar_11 weighed 4 ingots bar_2+bar_9 bar_1+bar_10 bar_1+bar_9 bar_2+bar_10 here a bag it was tore So Further? Look, swap bar_1 = 2; bar_2 = 3; bar_9 = 8; bar_10 = 9 yields the same result of weighings...

#### Re: Re: Gold ingots of Geriona

Hello, Erop, you wrote: {1,2,3,4,5,6,7,8,9,10,11} 1+2+3+4 = 10 1+2+3+5 = 11 {1,2,3,4,5} {6, 7, 8, 9, 10, 11} 6+1+2 = 9 6+1+3 = 10 6+1+4 = 11 {1} {2,3,4} {5} {6} {7, 8, 9, 10, 11} Well look. You showed  11 ingots, naming them bar1, bar_2. bar_11 E> weighed 3 ingots bar_6+bar_1+bar_2 = 9 bar_6+bar_1+bar_3 = 10 bar_6+bar_1+bar_4 = 11 Look Further, swap bar_6 = 1; bar_1 = 2; bar_4 = 3; yields the same result of weighings...

#### Re: Re: Gold ingots of Geriona

Hello, - n1l - you wrote: N> Hello, Erop, you wrote: N> {1,2,3,4,5,6,7,8,9,10,11} N> 1+2+3+4 = 10 N> 1+2+3+5 = 11 N> {1,2,3,4,5} {6, 7, 8, 9, 10, 11} N> 6+1+2 = 9 N> 6+1+3 = 10 N> 6+1+4 = 11 N> {1} {2,3,4} {5} {6} {7, 8, 9, 10, 11} N> Well look. You showed  11 ingots, naming them bar1, bar_2. bar_11 E>> weighed 3 ingots N> bar_6+bar_1+bar_2 = 9 N> bar_6+bar_1+bar_3 = 10 N> bar_6+bar_1+bar_4 = 11 N> Look Further, swap bar_6 = 1; bar_1 = 2; bar_4 = 3; N> yields the same result of weighings... Then on any of bar_1+bar_2+bar3+bar_4 bar_1+bar_2+bar3+bar_5 the bag will be torn...

#### Re: Re: Gold ingots of Geriona

And? As it justifies your decision. Where you precisely know what of ingots how many weighs and does not justify mine where happens too most. In a statement of the problem there is such point.

#### Re: Re: Gold ingots of Geriona

Hello, - n1l - you wrote: N> And? N> As it justifies your decision. Where you precisely know what of ingots how many weighs and does not justify mine where happens too most. N> in a statement of the problem there is such point. Look, it is necessary to find such sequence of weighings which is impossible, if bar_1! = 1. I offered you swap which breaks I not the full resulted, but you remaining do not use, but it is possible also any to result full. For example the such: bar | kg 1 | 2 2 | 3 3 | 1 4 | 4 5 | 5 6 | 6 7 | 7 8 | 10 9 | 8 10 | 9 11 | 11 Try to show same forging bar_1 swap for my decision...

#### Re: Re: Gold ingots of Geriona

You so also did not prove anything. You mark ingots, I do too most.

#### Re: Re: Gold ingots of Geriona

Hello, Erop, you wrote: E> Hello, - n1l - you wrote: N>> Hello, Erop, you wrote: N>> {1,2,3,4,5,6,7,8,9,10,11} N>> 1+2+3+4 = 10 N>> 1+2+3+5 = 11 N>> {1,2,3,4,5} {6, 7, 8, 9, 10, 11} N>> 6+1+2 = 9 N>> 6+1+3 = 10 N>> 6+1+4 = 11 N>> {1} {2,3,4} {5} {6} {7, 8, 9, 10, 11} N>> Well look. You showed  11 ingots, naming them bar1, bar_2. bar_11 E>>> weighed 3 ingots N>> bar_6+bar_1+bar_2 = 9 N>> bar_6+bar_1+bar_3 = 10 N>> bar_6+bar_1+bar_4 = 11 N>> Look Further, swap bar_6 = 1; bar_1 = 2; bar_4 = 3; N>> yields the same result of weighings... E> Then on any of E> bar_1+bar_2+bar3+bar_4 E> bar_1+bar_2+bar3+bar_5 E> the bag will be torn... With ingots 7 8 9 bag too will be torn and what?

#### Re: Re: Gold ingots of Geriona

Hello, - n1l - you wrote: N> You so also did not prove anything. Well it is strict to me to prove laziness I there specified lemmas that type bar1, bar2. bar5 contains swap 1, 2. 5, and remaining - swap 6. I wrote down 11 It so: {1, 2, 3, 4, 5} {6, 7, 8, 9, 10, 11} Lemmas to me to prove laziness, IMHO, they are obvious. After in record of weighings lemma record got, it already is proved. According to it it is possible to use in the further reasonings... N> You mark ingots, I do too most. I state the following. If to do those 5 weighings it does not turn out to pick up such swap of markings on ingots that bar1! =1 If you do not agree - show swap which does not tear a bag in one of those of 5 weighings... But there is a decision from 2. Very class, by the way...

#### Re: Re: Gold ingots of Geriona

Hello, - n1l - you wrote: N> With ingots 7 8 9 bag too will be torn and what? And at true marking, it will not be torn on one of 5... Well and still there are how many markings at which too it will not be torn, but all of them possess that property that bar_1 = 1...

#### Re: Re: Gold ingots of Geriona

Well at me at true marking too all converges.

#### Re: Re: Gold ingots of Geriona

N> Well at me at true marking too all converges. And it is necessary, what at any, where bar_1! = 1 did not converge...

#### Re: Re: Gold ingots of Geriona

Hello, Kodt, you wrote: I know the decision from 2 weighings (UPD. Removed a spoiler) I too found for 2 weighings! If you did not tell that it exists, would not find...