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**Deveno** both Opalg's and Soroban's method both give that a=8 is not possible. although their reasoning in finding b+c differ, both give the same possible pairs (b,c).

so, given a in {1,2,3,4,5,6,7,8}, if 10a+d+11(b+c)/2 = 100, there can be only one possible choice for (b+c)/2 and d.

for example, if a = 3, then 70 - d is divisible by 11. since d < 10 (in fact, it's actually < 9), (b+c)/2 has to be 6, so b+c = 12, and from

70 - d = 66, we get d = 4.

if we choose a = 4, we do NOT get d = 3. if a = 4, then 60 - d is divisible by 11, so (b+c)/2 = 5, hence b+c = 10. from 60 - d = 55, we get d = 5.

so each pair (a,d) is uniquely defined, we need only count them once.