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.