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"per calf - 0.5 0.5 roubles"
How do you understand that?
all figures duplicated
all the numbers duplicated
I take it this branch is about who's cooler at googling?
An old problem. How many bulls, cows and calves can be bought for 100 roubles, if 100 cattle are to be bought, and the price of a bull is 10 roubles, the price of a cow is 5 roubles, and the price of a calf is 0,5 roubles?
Answer
program prim3
uses crt;
var b, k, t: integer;
begin
clrscr;
for b:=0 to 10 do
for k:=0 to 20 do
for t:=0 to 200 do
if(20*b+10*k+t=200) and (b+k+t=100)
then
writeln('bulls ', b, 'cows ', k ' calves ', t);
readln;
end.
Ancient task
There is 100 roubles.
solve with nested cyclesHow many bulls, cows and calves can be bought with all this money,
if the fee for a bull is 10 rubles,
for a cow is 5 rubles,
for a calf is 0.5 rubles
and 100 cattle have to be bought?
1 bull, 9 cows and 90 calves
The Diophantus equation (if prices are counted in fifty cents)
i take it this branch is who's the cooler googler?
how much do you get?
i used python to solve this problem.
Here's more, better to decide without googling, then there's no point
*
How much is it?
i solved this problem in python
Let me better ask my own problem closer to the subject ;)
I finally remembered how I was doing one of my strategies with forecasting ;))))
I wrote somewhere on a forum that a forecast is the simplest school problem ;)
Suppose the price moves along a parabola
Which decision will be rejected - BUY or SELL ?
I wrote about it here somewhere, maybe the picture will correct, I'm reading it because my old nickname was deleted
https://www.mql5.com/ru/forum/37348/page69#comment_1158229
picture:
Ancient task
There is 100 roubles.
solve with nested cyclesHow many bulls, cows and calves can be bought with all this money,
if the fee for a bull is 10 rubles,
for a cow is 5 rubles,
for a calf is 0.5 rubles
and you have to buy 100 cattle?
1+9+90 but that's it, in my mind almost without cycles, even/odd: Calves should divide by 10, others in total too, bulls and cows an odd number
Ancient problem
There is 100 roubles.
solve with nested loopsHow many bulls, cows and calves can be bought with all this money,
if the fee for a bull is 10 roubles,
for a cow is 5 roubles,
for a calf is 0.5 roubles
and 100 cattle have to be bought?
.