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