[Archive!] Pure mathematics, physics, chemistry, etc.: brain-training problems not related to trade in any way - page 506
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
1. weigh 300. then move the weight to the buckwheat and deliver 50g. on the second weighing, 700.
2. how theExpert got two weighings I can't even guess...
1. I got 650 for the second weighing :(
2. neither do i. that's fantastic. spill it andrei!
Let's finish off the hockey players first.
If not, the hockey players can always be beaten by overkill, there aren't many options left, they just want to be beautiful.
2. me too. fantastic. spill it Andrei!
I'm working with WinAPI in MQL5 at the moment, so I start looking for solutions via *an ass :) it's like a hint.
in two!
And indeed. I decided for two too.
Then a correction:
Имеем рычажные весы, 9 кило гречки и две гирьки : 50гр. и 200 гр.
Нужно за 3 взвешивания отмерить ровно 2 кило.for 2!
?
?
1. 4350+300==4650 // 4350+4650 = 9000
2. 2350==2000+350
Crooks . Both . Wallpaper. How for 2 ?
Shouldn't be poured from scratch but from 9
it doesn't seem possible.
Because:
1. An integer must be between 2 and 4. Since the minimum number 123456*5=617280 - does not fit, and the next minimum 132456*5=662280 is already more than the maximum possible (654321)
2. The number 2 as a multiplier does not fit. If to multiply in column as taught at school))), in that position of hockey number where 4 in multiplication will be 8, and to pull up to 1 (also hockey number), in mind d.b. 3 - i.e. in the previous digit in the multiplication shall be obtained over thirty, and this is impossible with the given multiplier and hockey numbers
3. Similarly, 4 cannot be a multiplier. In the position where the first number has a 2, also the multiplication results in 8, and from the previous digit 3 cannot be obtained in any way, because even if the largest hockey number (6) is multiplied by 4, it would only be 24 and 2 would be carried over into the senior digit.
Similarly, the multiplier 3 does not work either. Also, multiplying by 6 results in 8, which in no way will not reach the hockey number 1
P.S. TheXpert already replied while I was writing :)