Humour - page 123
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
The fifth problem is more interesting. It is immediately obvious that A=0, D=1
E=5; M=2; but B and D could be 3 and 5; 4 and 6; 6 and 8; 7 and 9, respectively.
G=1, A=0,
Т < 5
T+B=9. B=9-T
2*T=M
2*E=10+M=10+2*T. E=5+T
Э=5
cannot = 5, as M cannot be =0, as A=0
G=1, A=0,
Т < 5
T+B=9. B=9-T
2*T=M
2*E=10+M=10+2*T. E=5+T
Thus, passing through the variants T=2, 3, 4 we find
Т=2 - cannot be, because it means B=E=7
Т=3 - cannot be, because it means B=M=6
we are left with
Т=4 - 4940+5940=10880
That's right, I wrote it all down wrong.
Well, I've already made up my mind there.
It's easier to see:
Well, I've already made up my mind there.
mine would fit if M was the only one )). Well almost, there's still a little mistake I have
Well, I've already made up my mind there.
Admit it, you just have this textbook.
at the end are the answers. ))