[Archive!] Pure mathematics, physics, chemistry, etc.: brain-training problems not related to trade in any way - page 429
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We're going with what we've got.
And there is no need to duplicate pairs of numbers in a loop.
And they are not duplicated. When i=2 in the parent loop, the child ii will occur only once = 2. It means that the combination of numbers 2 and 2 will occur only once. There is no duplication.
X - rod length
Z - length and width of the cell
b=Z*4 - length of rod per cell
c=X/b - number of cells
It remains to calculate the total walls somehow. -1 per row.
A=X/(Z*4)-2
?
In fact, it's like 5th grade when they start learning percentages, maybe they should be screwed in?
And they are not duplicated. When i=2 in the parent loop, then ii will only occur once in the child loop = 2. So the combination of the numbers 2 and 2 will only occur once. There is no duplication.
But (2 and 3) and (3 and 2) ???
One way in which B could say "I know in advance...": sum = 11.
11 = 2+3*3 = 3+2*2*2 = 2*2+7 = 5+2*3 = ...
And there aren't many numbers like that with a sum of less than 100, by the way.
Well, this is where the idea of the program comes in.
A (2 and 3) and (3 and 2) ???
These situations have to be handled by code too. Otherwise we risk missing something. Anyone familiar with combinatorics would immediately say that we have the total number of pairs of two-letter combinations = 98*98 = 9604. He would say that we are facing a tuple of two disks of 98 elements each. The risk of being a fool would increase with each attempt to cross out the extra. You can cross it out, but when the program is going through the options, this risk is not logically justified. Especially since there aren't many elements and CPU time can be neglected.
Anyway, you can't go through a lot of solutions quickly if you "get an earful" of a complex number. A system of two equations with three unknowns on a complex number will not work quickly.
P.S.
Perhaps I should clarify. When you have to calculate the number of choices, it is better to abstract from the concept of numbers and look at the two disks as disks containing letters. In this case, the combination A-B is not the same word as B-A. So it's better to go through all the variations.
These situations must also be handled by code. Otherwise we risk missing something. Anyone familiar with combinatorics would immediately say that we face a total number of pairs of two-letter combinations = 98*98 = 9604. He would say that we are facing a tuple of two disks of 98 elements each. The risk of being a fool would increase with each attempt to cross out the extra. You can cross it out, but when the program is going through the options, this risk is not logically justified. Especially since there aren't many items and CPU time can be neglected.
Anyway, you can't go through a lot of solutions quickly if you "get an earful" of a complex number. a system with three unknowns on a complex number won't work quickly.
Apparently you don't understand me. The key to the solution of the problem is the statements of the sages, and they operate only with products and sums. They were told the product and sum of the two Cheslas. Considering all possible pairs including their permutations won't change anything. Wouldn't it?
Well, I gave the right answer in the first post. 2*2=4 и 2+2 = 4. The answer is exactly the same as the problem!
Well, I gave the right answer in the first post. 2*2=4 и 2+2 = 4. The answer is exactly the same as the problem!
No match!!!
The first wise man wouldn't have said he couldn't find those numbers then!