First weigh-in possible outcomes: both pans of balance are equal, which means A: 8(1), 8(1) or B: 8(0), 8(0); or one pan of balance is heavy, which means A: 8(1), 8(0); or B: 8(2), 8(0). In former one we can measure one of the measured groups (A:8(1) or B:8(0)) with another unmeasured (A:8(0) or B:8(2)). If unmeasured is lighter than the measured then it is case A because 8(1)(measured)=8(1)(measured)>8(0)(unmeasured). Else if unmeasured is heavier than the measured then it is case B because 8(0)(m)=8(0)(m)<8(2)(unm). So far we did 3 measurements.

Case A: 8(1), 8(1), 8(0). In this case we totally disregard 8(0) because there is no fake coin. Now, let's focus on each of 8(1). Next measurement (case A 1st measure) we find 4(1), 4(0). After that (case A 2nd measure) we find 2(1), 2(0). Then (A 3rd m) we find the exact fake coin (1(1), 1(0)) in that group 8(1). In this case it took us 3 weigh-ins to find fake coin. Another 3 weigh-ins to find the second fake coin in the second 8(1). So total of 3 measurements until we reach case A, and then 6 measurements in case A. So total of 9 measurements altogether.

Case B: 8(0), 8(0), 8(2). In this case we totally disregard 8(0) groups, instead we focus on 8(2). First weigh-in of case B: 4(0), 4(2) or 4(1), 4(1). In the latter (4(1), 4(1)) it will take us 2 more measurements to find fake coin in 4(1). So each 4(1) will take 2 more measurements, 4 in total; 1 measurement in case B until 4(1), 4(1) point; and 3 more measurements before case B. 8 measurements altogether in this part (4(1), 4(1)) of case B.

Now, 4(0), 4(2) of case B. We disregard 4(0) and focus on 4(2). Next measure we find either 2(2), 2(0) or 2(1), 2(1). In 2(2), 2(0) we reached the result because both fake coins are there. So far 3 measures until case B; 1 more in case B to reach 4(0), 4(2); and 1 more to reach 2(2), 2(0). Altogether 5 measurements in this way.

Now, 2(1), 2(1) of 4(0), 4(2). It will take 2 measures to find fake coins because it was heavy than the original coins. 3 measure until case B; 1 more until 4(0), 4(2); 1 more until 2(1), 2(1); and 2 more to find exact fake coins. In total of 7 measurements in this case of B.

I tried to detail all the weigh-ins as best as I could. I got the help of my friends: Bazarbay Halmedov and Kerven Durdymyradov, both from Turkmenistan IMO team. Thanks to them and you all for nice questions and support.

First weigh-in possible outcomes: both pans of balance are equal, which means A: 8(1), 8(1) or B: 8(0), 8(0); or one pan of balance is heavy, which means A: 8(1), 8(0); or B: 8(2), 8(0). In former one we can measure one of the measured groups (A:8(1) or B:8(0)) with another unmeasured (A:8(0) or B:8(2)). If unmeasured is lighter than the measured then it is case A because 8(1)(measured)=8(1)(measured)>8(0)(unmeasured). Else if unmeasured is heavier than the measured then it is case B because 8(0)(m)=8(0)(m)<8(2)(unm). So far we did 2 measurements.

Case A: 8(1), 8(1), 8(0). In this case we totally disregard 8(0) because there is no fake coin. Now, let's focus on each of 8(1). Next measurement (case A 1st measure) we find 4(1), 4(0). After that (case A 2nd measure) we find 2(1), 2(0). Then (A 3rd m) we find the exact fake coin (1(1), 1(0)) in that group 8(1). In this case it took us 3 weigh-ins to find fake coin. Another 3 weigh-ins to find the second fake coin in the second 8(1). So total of 2 measurements until we reach case A, and then 6 measurements in case A. So total of 8 measurements altogether.

Case B: 8(0), 8(0), 8(2). In this case we totally disregard 8(0) groups, instead we focus on 8(2). First weigh-in of case B: 4(0), 4(2) or 4(1), 4(1). In the latter (4(1), 4(1)) it will take us 2 more measurements to find fake coin in 4(1). So each 4(1) will take 2 more measurements, 4 in total; 1 measurement in case B until 4(1), 4(1) point; and 2 more measurements before case B. 7 measurements altogether in this part (4(1), 4(1)) of case B.

Now, 4(0), 4(2) of case B. We disregard 4(0) and focus on 4(2). Next measure we find either 2(2), 2(0) or 2(1), 2(1). In 2(2), 2(0) we reached the result because both fake coins are there. So far 2 measures until case B; 1 more in case B to reach 4(0), 4(2); and 1 more to reach 2(2), 2(0). Altogether 4 measurements in this way.

Now, 2(1), 2(1) of 4(0), 4(2). It will take 2 measures to find fake coins because it was heavy than the original coins. 2 measure until case B; 1 more until 4(0), 4(2); 1 more until 2(1), 2(1); and 2 more to find exact fake coins. In total of 6 measurements in this case of B.

I tried to detail all the weigh-ins as best as I could. The most weigh-ins is 8. I got the help of my friends: Bazarbay Halmedov and Kerven Durdymyradov, both from Turkmenistan IMO team. Thanks to them and you all for nice questions and support.