Find the smallest positive fraction, which when individually divided by 6/35, 10/21, and 15/49 yields an integer.
Write answer correct up to 2 decimal places.
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Let the smallest fraction be b a , where a and b are positive coprime integers. Then, we have b a ÷ 3 5 6 = b a × 6 3 5 = n 1 , b a × 1 0 2 1 = n 2 , and b a × 1 5 4 9 = n 3 , where n 1 , n 2 , and n 3 are integers. For b a to be the smallest, a should be the smallest, while b the biggest. For n 1 , n 2 , and n 3 to be integers, 6 a , 1 0 a , and 1 5 a must be integers and the smallest a is a m i n = lcm ( 6 , 1 0 , 1 5 ) = 3 0 . Similarly, b 3 5 , b 2 1 , and b 4 9 must also be integers and the biggest b is b m a x = g cd ( 3 5 , 2 1 , 4 9 ) = 7 . Therefore, b a = 7 3 0 ≈ 4 . 2 8 .