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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. ))