5x7 = 35 is the smallest no. which is divisible by 5 and 7 and has remainder of 2 when divided by 3
3x7 = 21 is the smallest no. which is divisible by 3 and 7 and has a remainder of 1 when divided by 5
therefore 21x3 = 63 is divisible by 3 and 7 and has a remainder of 3 when divided by 5
3x5 = 15 is the smallest no. which is divisible by 3 and 5 and has a remainder of 1 when divided by 7
therefore 15x4 = 60 is divisible by 3 and 5 and has a remainder of 4 when divided by 7
LCM 3x5x7 = 105
N = 35+63+60 + 105t where t is integer
N = 158 + 105t (general solution)
N = 53 is smallest when (t = -1)
It's just an I.Q. math question for a P 5 student. We can get to the answer by simple logical thinking (trial by error), but the approach posted by baekdoosan is obviously beyond primary or even secondary school level.