B and C will contain three digit numbers of the form 9k+2 and 9k+ℓ respectively. We need to find sum of all elements in the set B∪C effectively. Now, S(B∪C)=S(B)+S(C)−S(B∩C) where S(k) denotes sum of elements of set k . Also B={101,110,......992} ∴S(B)=
100
2
(101+992)=54650
Case-I : If ℓ=2 then B∩C=B ∴S(B∪C)=S(B) which is not possible as given sum is 274 × 400=109600
Case-II : If ℓ≠2 then B∩C=φ ∴S(B∪C)=S(B)+S(C)=400×274