[Archive!] Pure mathematics, physics, chemistry, etc.: brain-training problems not related to trade in any way - page 214
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
Mathemat писал(а) >>
Which of the polygons inscribed in this circle has the maximum sum of the squares of the sides?
A triangle.
It remains to be proven.
Осталось это доказать.
That's the easy part:))
OK, next.
What is the greatest power of 2 that (2^n) is divisible by! ?
Follow up with a similar one:
How many zeros end in 1000! ?
Это-то как раз и несложно:))
Studio, please. I was relying more on logic than maths :) .
____
Although... you can prove using the cosine theorem and the sum of angles of a polygon that it is less for an n-gon than for an n-1-gon.
Barbaric duck liver.
Goodkat:
A concentration of 10^-400 is what?
smirik:
It means that once near the cure, at a distance of no more than 1,000 km
a barbaric duck flew by.
Goodkat:
There are about 10^80 atoms in the known part of the universe.
10^-400 - the duck flew in the next universe :)
Smirik:
Yes, by the way. Just like that, unobtrusively, we proved the theory of parallel
universes.
Mathemat писал(а) >>
What is the greatest power of 2 that (2^n) is divisible by! ?
How many zeros end in 1000! ?
1) The degree is 2^n - 1, i.e. (2^n)! is divisible by 2^(2^n - 1).
2) 249.
I won't prove it: the degree of a prime in a factorial is calculated by a known and easily deduced formula.
В студию, плз. Я опирался больше на логику, чем на математику :) .
____
Хотя... можно доказать используя теорему косинусов и сумму углов многоугольника, что для n-угольника она меньше чем для n-1-угольника.
exactly so.
1. Any n-gon has at least 1 non-acute angle at n>=4. Proof: sum of angles of an n-gon (n-2)*180=a1+a2+...+an. If all angles are acute, i.e. ai<90 for all i, then
(n-2)*180<n*90,
wherefore it follows that n<4.
2. "Straightening an obtuse angle, by the cosine theorem we obtain a side of an (n-1)-angle whose square is greater than the sum of the squares of the two "old" sides. In the case of a "right angle" we obtain equality by the Pythagorean theorem. Thus for any inscribed polygon it is possible to iteratively construct a triangle with sum of squares of sides at least not less than the given polygon. So the optimal polygon is a triangle. It remains to find out which one.
3. If the radius of the circle is R and the angles of the triangle are a, b and pi-(a+b), then the sum of the squares of the sides S=4R^2(sin^2(a)+sin^2(b)+sin^2(a+b)). By differentiating by a and b and equating the derivatives to zero, and solving the resulting equations (I will not give details, there is nothing complicated there), we obtain that a=b=pi/3. Conclusion: the optimal triangle is equilateral.
For today's warm-up
A bus ticket number consists of six digits (the first digits can be zeros). A ticket is called lucky if the sum of the first three digits equals the sum of the last three. Prove that the sum of all lucky ticket numbers is divisible by 13.
One more thing.
Five traders who trade with one brokerage company have 143, 233, 313, 410 and 413 thousand dollars in their accounts. Each of them can transfer money to the other through the internal transfer system of the DC, but the latter will charge 10% more from the sender's account for each transfer. The traders have agreed that they want to send the money in such a way that everyone gets the same amount and the VC receives as little money as possible. How much money will each trader get in the most economical way and what will be the profit for brokerage company?
)))