LCM or GCD?

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.


The answer is 4.28.

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1 solution

Chew-Seong Cheong
Apr 21, 2017

Let the smallest fraction be a b \dfrac ab , where a a and b b are positive coprime integers. Then, we have a b ÷ 6 35 = a b × 35 6 = n 1 \dfrac ab \div \dfrac 6{35} = \dfrac ab \times \dfrac {35}6 = n_1 , a b × 21 10 = n 2 \dfrac ab \times \dfrac {21}{10} = n_2 , and a b × 49 15 = n 3 \dfrac ab \times \dfrac {49}{15} = n_3 , where n 1 n_1 , n 2 n_2 , and n 3 n_3 are integers. For a b \dfrac ab to be the smallest, a a should be the smallest, while b b the biggest. For n 1 n_1 , n 2 n_2 , and n 3 n_3 to be integers, a 6 \dfrac a6 , a 10 \dfrac a{10} , and a 15 \dfrac a{15} must be integers and the smallest a a is a m i n = lcm ( 6 , 10 , 15 ) = 30 a_{min} = \text{lcm } (6,10,15) = 30 . Similarly, 35 b \dfrac {35}b , 21 b \dfrac {21}b , and 49 b \dfrac {49}b must also be integers and the biggest b b is b m a x = gcd ( 35 , 21 , 49 ) = 7 b_{max} = \gcd (35,21,49) = 7 . Therefore, a b = 30 7 4.28 \dfrac ab = \dfrac {30}7 \approx \boxed{4.28} .

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