Find the smallest no. which leaves remainder 8 and 12 when divided by 28 and 32 respectively .

Let the required number be z.

∴z = 28x + 8 and z = 32y+12

⇒ 7x+2=8y+3

⇒ 7x = 8y+1 … (1)

Here, x =8n -1, y =7n -1 satisfies the equation (1).

Putting n =1, we get

x = 8-1 = 7 and y =7-1 =6

∴ z = 28 × 7 + 8 = 204

⇒ 7x+2=8y+3

⇒ 7x = 8y+1 … (1)

Here, x =8n -1, y =7n -1 satisfies the equation (1).

Putting n =1, we get

x = 8-1 = 7 and y =7-1 =6

∴ z = 28 × 7 + 8 = 204

Thus, the smallest number which leaves remainder 8 and 12 when divided by 28 and 32 respectively is 204.

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