Find the smallest number which leaves …
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Sahdev Sharma 6 years, 9 months ago
Let the required number be z.
z = 28x + 8 and z = 32y+12
=> 28x+8 = 32y+12
=> 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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