4. Solve the numerical riddle: TETA+BETA=GAMMA. (Different letters - different digits.)
Answer: $4940+5940=10880$

Solution. Since $A+A$ ends in $A$, then $A=0$. Since $\Gamma$ is the result of carrying over to the next digit, then $\Gamma=1$. Since $A+A$ ends in $A$, then $A=0$. This means there is no carry to the tens place, i.e., $T+T$ ends in $M$, and thus $M$ is even. There is also no carry to the hundreds place, otherwise an odd number Э+Э+1 would end in an even $M$. Since there is no carry, $2 T Б<10$. Possible values are 2, 3, 4. If $T=2$, then Э=7, from which $Б=7-$

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the variant $T=4$. Then $M=8$, Э=9. From which $Б=5$ - there is no contradiction. Thus, only one variant is possible: $4940+5940=10880$