[Archive!] Pure mathematics, physics, chemistry, etc.: brain-training problems not related to trade in any way - page 208
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ключей от квартиры (с) в архиве тоже не нашлось?:))))
Did you put it down? Shit!!!! There's a lot of thieves around here. (((((((((((((((((((((((((((
Well, I can write 10 equations with 10 unknowns. Then solve them. Boring... You can do the math in your head. ;)
Yeah, I'm confused by the system of equations too :)
By the way, I have 11 equations.
Да, я тоже запутался в системе уравнений :)
Кстати, уравнений у меня 11.
There's nowhere to get mixed up. Subtract them in pairs from each other and you'll be happy... ;)
I had five, very simple ones - the sum of all five numbers, the sum of the minimum, the sum of the maximum, and the closest to them. And then I checked the solution - it added up.
У меня их было 5, очень простых, - сумма всех 5 чисел, сумма минимальных, сумма максимальных, а также ближайших к ним. А потом решение проверил - сошлось.
I only have 3 equations :))) and the solution comes out unambiguous, the rest is logical reasoning.
I had five, very simple ones - the sum of all five numbers, the sum of the minimum, the sum of the maximum, and the closest to them. And then I checked the solution - it added up.
The sum of all five numbers and I had. It was four times less than the sum of those 10. Four, because five digits minus one.
У меня их было 5, очень простых, - сумма всех 5 чисел, сумма минимальных, сумма максимальных, а также ближайших к ним. А потом решение проверил - сошлось.
And I just thought logically:
1) there can't be two negatives.
2) one pair must be obligatory. (for otherwise there would be two zeros (in the initial ones), and hence two twos (in sums))
3) the negative must have a symmetric positive
4) this pair must be modulo smaller than all the others
5),6),7),8)..... etc.
In fact, I had to choose only ten, the rest appeared by themselves.
;)
We have 10 sums
0, 2, 4, 4, 6, 8, 9, 11, 13, 15
Denote our numbers x1, x2, x3, x4, x5. To be clear, let the first sum of 10 be the sum of x1 and x2
x1+x2=0 (1)
Obviously, if you take all the sums together, each number will occur there exactly 4 times (a total of 20 numbers in the 10 sums). So
(x1+x2+x3+x4+x5)*4 = 0+2+4+4+6+8+9+11+13+15, where
x1+x2+x3+x4+x5 = 18, or given (1), x3+x4+x5=18.
But then (x3+x4)+(x4+x5)+(x5+x3) = 36, i.e. among the 10 sums there must be a triplet of such that the sum gives 36. It is easy to make sure that this triplet must necessarily include the numbers 13 and 15, as if it were not so, then from the remaining numbers would have to make at least one pair, the sum giving no less than 36-15 = 21, which is not satisfied for the largest numbers: 9+11=20<21. So we have two sums: 13 and 15. The third one is easy to find: 36-15-13=8. Now, since we are still free in the order of the variables, we can write
(x3+x4)=8 x3=5
(x4+x5)=13 => x4=3
(x5+x3)=15 x5=10
Three of the five numbers we have found, it is up to the remaining two. To do this in the six sums
2, 4, 4, 6, 9, 11
we need to find three pairs in which the "participants" differ by the same even number:
(x3+x1)-(x3+x2)=(x4+x1)-(x4+x2)=(x5+x1)-(x5+x2) = x1-x2 = 2*x1
Since only 9 is evenly divisible from number 11, it follows that 2*x1 = 2 => x1 = 1, x2=-1
MetaDriver, 1) если не секрет, а как вы архивчик "сделали"? Я в этих вопросах не спец,
2) хочется научится :)
1) I quickly remade the script on mql5 (I had a preparation for opening a safe in CA).
2) So learn! I have shown you the route - go ahead. // the script is commercial, you know....
;)