[Archive!] Pure mathematics, physics, chemistry, etc.: brain-training problems not related to trade in any way - page 431

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If we stipulate that the product of the conceived numbers should also be no more than 100, but more than 9, then the number of options is reduced to 138. A three-digit product would be very difficult to operate with. That's why 138 solutions of the problem is still cool, and it's not the only solution. :)
 
drknn:

Based on these conditions, I got 2352 ways of solving a system of equations

a+c= c

a*b=d

Apparently the pundits were too hasty with their conclusion about the uniqueness of the solution. Here is just a piece of the alerter:


Guys, do you see what the big deal is here? It's that we can get so much information from the conversation of the wise men that we have enough to find a solution. More specifically, their dialogue allows us to cut out all the wrong solutions, leaving the ONLY right one.
 

Anyway, here's some simple code to display these numbers on the screen.

//+------------------------------------------------------------------+
//| script program start function                                    |
//+------------------------------------------------------------------+
int start(){
  int  I[5000],II[5000],S[5000],P[5000];
  int SchPar=0,SchPar2=0;
  for(int i=98;i>=2;i--){
    for(int ii=i;ii>=2;ii--){
      if(i+ii<100){
        SchPar++;
        I[SchPar]=i;
        II[SchPar]=ii;
      }
    }
  }
  //числа внесены в массивы - просчитываем сумму и произведение
  for(int z=1;z<=5000;z++){
    if(I[z]>=2 && II[z]>=2 && I[z]*II[z]<100 && I[z]*II[z]>9){
      SchPar2++;
      S[z]=I[z]+II[z];
      P[z]=I[z]*II[z];
      Alert("Пара чисел: ",I[z]," и ",II[z]," Их сумма = ",S[z]," Произведение = ",P[z]);
    }
  }
  Alert("Общее число пар чисел, сумма и произведение которых менее 100 = ",SchPar2);
  return(0);
}
//+------------------------------------------------------------------+
 
drknn:

Anyway, here's some simple code to display these numbers on the screen.


Why is one of the conditions I[z]*II[z]>9 ?
 
drknn:
I mean explicitly by 2 numbers... then it will narrow down the search.
 
I have figured out and rigorously proved what the sums can be, if B says so. They are all odd numbers of the form 2+component that are less than 100. There are 24 such sums in the first hundred. Thinking further. Это числа 11,17,23,27,29,35,37,41,47,51,53,57,59,65,67,71,77,79,83,87,89,93,95,97.
 
drknn:

Anyway, here's some simple code to display these numbers on the screen.


By the way, there's no mention of the fact that the product is less than a hundred in the problem ))

So now you're getting wise)

 
ValS:

Why is I[z]*II[z]>9 one of the conditions?

Because if the product is not a two-digit number, but a one-digit one, the number of choices is so fast that the results then do not make you say, "I don't know the solution.
 
drknn:

Because if the product is not a two-digit number, but a one-digit one, the number of variants goes through so quickly that the results then do not make you say, "I don't know the solution.

That's a bit of a mouthful, don't you think? Based on what? Calculations?
 

So, what do we have?

Sage A says that the product is decomposable into at least three factors greater than 1 (including, possibly, equal ones). There are exceptions, though. These are 8, 27, 125, 343 etc., i.e. the cube of a prime. Here the decomposition is singular anyway.

Sage B says that the sum of numbers is an odd number of type 2+composite - and he tells exactly that to Sage A. But he knew it before, before A. What new information does he know now?

This information is enough for A to now say he knows the numbers. What would that mean?

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