Answer:

Consider x as the required number,

Apply Euclid’s lemma,

x = 28p + 8 and x = 32q + 12    [p and q – quotients]

28p + 8 = 32q + 12

28p = 32q + 4

7p = 8q + 1….. (1)

p = 8n – 1 and q = 7n – 1       [n – natural number]

If n = 1,

p = 8 – 1 = 7 and q = 7 – 1 = 6

x = 28p + 8

x = 28 × 7 + 8

x = 204

The smallest number which when divided by 28 and 32 leaves remainders 8 and 12 is 204.