Pure maths, physics, chemistry, etc.: brain-training tasks that have nothing to do with trade [Part 2] - page 16
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
Yes.
So the trick of the task is to set unnecessary conditions to confuse?
So, the trick of the problem is to set superfluous conditions to confuse?
They're all sadists out there. ;)
It's a simple task: weigh to the left, weigh to the right and calculate the geometric mean. It always helps with scales... ;-)
They're all sadists out there. ;)
And the task is simple: weigh to the left, weigh to the right and take the geometric mean. It always helps with the scales. ;-)
This is an approximate method because the effect of the difference in the shoulders is non-linear with respect to the weight being measured and if you measure on different sides, the effect will be different.
It's easier to put a ruby on one side of the scale. On the other, you put weights or whatever you want to balance it. Take away the ruby and put the right weights in its place. It is also balanced. The total weight of the weights will be the weight of the ruby.
This is an approximate method as the effect of the difference in the shoulders is non-linear with respect to the weight being measured and if you measure on different sides, the effect will be different.
Uh-oh. I don't believe it! ;)
But I'm malleable, I'm willing to admit that your way is more versatile, and is good even if you disguise a couple of hidden springs. As long as the friction doesn't inhibit it.
As for "ideal" lever scales - my method is quite workable. You can't prove otherwise, you can try it. We've got all the non-linearity under control... )))
Uh-oh. I don't believe it! ;)
But I'm malleable, I'm willing to admit that your way is more versatile, and is good even if you disguise a couple of hidden springs. As long as the friction doesn't inhibit it.
As for "ideal" lever scales - my way is quite workable. You can't prove otherwise, you can try it. We have all non-linearity under control... )))
I agree)), for classic scales without any springs - a medium geometric will do as well.
Megamind came up with a ten-digit natural number. The first (left) digit of this number equals the number of zeros in its entry, the second digit equals the number of ones, the third digit equals the number of twos, etc., the last digit equals the number of nines in the entry of this number. Can you repeat Megamind's achievement and find this number?
What about this one? Too-simple?
--
Here, by the way. I found a solution, but I'm not sure it's the only one. It wouldn't hurt to find out, either.
I agree)), for classic scales without any springs - a medium geometric will do as well.
All right, then. Can you find the tricky number?
I think there's only one option: 6210001000.
there seems to be one option: 6210001000
This is an approximate method because the effect of the difference in the shoulders is non-linear with respect to the weight being measured, and if you measure on different sides, the effect will be different.
Put the ruby on one side of the scale. On the other, put weights or anything else to balance it. Remove the ruby and put the right weights in its place. It is also balanced. The total weight of the weights is the weight of the ruby.
Yeah, I see. I did not think in such a direction, although it is really a more universal method. Using only problem conditions ("different shoulders"), this is how I solved it.
2 MD: I don't want to waste my brain on problems with difficulty less than 3 :) It seems that a proof is not required here. But if you want, you may think about uniqueness.
Here's another one (4 points). This one is serious:
Find all natural numbers which, when multiplied by 4, turn into their mirror image. (A mirror image is when the digits in it go in reverse order).