Question

Find the smallest number which when divides 28 and 32, leaving remainders 8 and 12 respectively.

Answer 1

Let the required number be x.
Using Euclid’s lemma,
x = 28p + 8 and x = 32q + 12, where p and q are the quotients
⇒28p + 8 = 32q + 12
⇒28p = 32q + 4
⇒7p = 8q + 1….. (1)
Here p = 8n – 1 and q = 7n – 1 satisfies (1), where n is a natural number
On putting n = 1, we get
p = 8 – 1 = 7 and q = 7 – 1 = 6
Thus, x = 28p + 8
= 28 × 7 + 8

= 204
Hence, the smallest number which when divided by 28 and 32 leaves remainders 8 and 12 is 204.