The smallest number would be the LCM of 28 and 32
LCM of 28 and 32 = 224
Therefore the required smallest number which leaves remainders 8 and 12 when divided by 28 and 32 respectively would be:
Number = LCM – (sum of the remainders)
⇒ 224 – (12 + 8)
⇒ 224 – 20 = 204
