
Hi Abhishek
N = 6a + 3 = 7b + 1 i.e. 6a + 2 = 7b i.e. 7b = 2(3a + 1) i.e. 3a + 1 must be multiple of 7, so smallest value of a which satisfy is 2.
Thus N = c*LCM(6, 7) + 6(2) + 3 = 42c + 15
ALso it is given that N = 11d + 4.
So 42c + 15 = 11d + 4 i.e. 11d = 42c + 11, so smallest value of c which satisfy this is c = 0
Thus N = k*LCM(11, 42) + 42(0) + 15 = 462k + 15.
Kamal Lohia
