WHICH IS THE SMALLEST NUMBER WHICH LEAVES REMAINDER 8 AND 12 WHEN DIVIDED BY 28 AND 32?

Asked by  | 22nd Apr, 2013, 07:07: AM

Expert Answer:

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

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

Answered by  | 22nd Apr, 2013, 10:21: AM

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