5. Given a numerical riddle: TETA + BETA = GAMMA. (Different letters correspond to different digits, the same letters correspond to the same digits.) Find all its solutions and prove that there are no others.
Answer: $4940+5940=10880$

Solution. Since $\Gamma$ is the result of carrying over to the next digit, $\Gamma=1$. Since $A+A$ ends in $A$, then $A=0$. Therefore, 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 the odd number Э+Э+1 would end in the even $M$. Since there is no carry, $2T<10$. Possible options are 2, 3, 4. If $T=2$, then Э=7, from which $Б=7$ - but 7 is already taken. If $T=3$, then $M=6$, Э=8, from which $Б=6$, but $6=M$. The last option is $T=4$. Then $M=8$, Э=9. From which $Б=5$ - there is no contradiction. Therefore, only one option is possible: $4940+5940=10880$