[Archive!] Pure mathematics, physics, chemistry, etc.: brain-training problems not related to trade in any way - page 100
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Let the people distract themselves and solve the new one. In principle, it could be solved by a sixth-grader in the old school. The old one will be finished in general form later.
IMHO, it really is simpler than that.
b/c is the tangent of the angle, which is easy to construct: plot c on the Oh axis, then b on the perpendicular.
Now from the same point (from the vertex of the angle) on the O axis we plot a. The reconstructed perpendicular inside the constructed angle will give the segment a*tg(alpha)=ab/c
Well, yes, you can do it without a tangent, with simple proportions.
Next (for those who don't really like geometry, but again for 9th grade): Prove that there are 2000 different natural numbers n_1, n_2, ..., n_2000 such that 1/n_1 +1/n_2 +...+1/n_2000 =1.
I don't know the solution myself yet. Note: for three numbers it's 2, 3, 6. For four - er. 2, 4, 6, 12. I'm too lazy to go any further.
Yep. ab/c = x; Move b to the right.
a/c = x/b
Ага. ab/c = x; Перенесём b вправо.
a/c = x/b
Ahem. Inattentive, however. In the picture, b and x are rearranged. I don't want to redraw it. Please credit. ;)
The principle is clear.
Mathemat писал(а) >>
Next (for those who don't like geometry much, but again for 9th grade): Prove that there exist 2000 different natural numbers n_1, n_2, ..., n_2000 such that 1/n_1 +1/n_2 +...+1/n_2000 =1.
I don't know the solution myself yet. Note: For three numbers it's 2, 3, 6. For four, it's, uh... 2, 4, 6, 12. I'm lazy to go further.
Direct example of existence:
1 = summ(2^n, (where n = 1 ... 1998)) + 3*2^1998 + 3*2^1999
Proved.
PS. I seem to like geometry more - just lose my head sometimes. :-)
Next (for those who don't really like geometry, but again for 9th grade): Prove that there are 2000 different natural numbers n_1, n_2, ..., n_2000 such that 1/n_1 +1/n_2 +...+1/n_2000 =1.
I think you're trying to get us to unlearn simple problems so that when we'll relax... :-)
Any series of powers of two { 2, 4, 8, ..., 2^(N-1), 2^N } when summed back gives a number different from 1 by 1/2^N . It remains to divide this number into two, so that the denominator has different numbers. You can divide it in any way you like, e.g. in the ratio 2:1.
Yurixx писал(а) >>
You can break it up however you like, for example in a 2:1 ratio.
I'm not sure any other way is possible. Then only rational-integer ones seem to work
Credit to both, OK. The decision is not unanimous, apparently.
The next one is a game (a joke, but I'm just shocked to my core):
Ostap Bender played a simultaneous chess match with grandmasters Garry Kasparov and Anatoly Karpov. He played with one of his rivals with white pieces and with the other with black. Despite the fact that this was only the third time Bender had played chess in his life and his previous experience in Vasyuki had been very poor, he managed to get one point in this session. (One point is given for winning a chess game, half a point for a draw, and 0 points for a loss.) How did he manage to achieve this?