The step-by-step solution in form of video solution is attached here.



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Answer: Option E

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We see that the last two digits of numbers do add up to 22 (10 + 12 = 22). Therefore, AMC + AMC = 1234 or 2(AMC) = 1234. So AMC = 1234/2 = 617. That is, A = 6, M = 1 and C = 7. So A + M + C = 6 + 1 + 7 = 14.

Alternate Solution:

Carrying out the addition, we notice that the units digit of C + C is 4; therefore C = 2 or C = 7. Next, we notice that the result from M + M is 3; this is only possible if the units digit of M + M is 2 and 1 carried over from the addition of C + C. Thus C = 7 and M = 1 or M = 6. If M were 6, then 1 would carry over to A + A and the first two digits of the sum would have been an odd number. Since the first two digits of the sum is 12, M can only be 1. Finally, since nothing carried over when we add M + M, A must be 6. Thus, A + M + C = 6 + 1 + 7 = 14.

Answer: E
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