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