HCF Notion #2

Number Theory Level 3

( 1 0. 3 ‾ & 0. 3 ‾ ) \Large \left( \color{skyblue}{\dfrac{1}{0.\overline{3}} \quad \color{#D61F06}{\&} \quad 0.\overline{3}} \right)

Find the highest common factor (H.C.F.) of the above two numbers.

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1 0. 3 ‾ \dfrac{1}{0.\overline{3}} 1 Doesn't exist. None of the given choices. ( 0. 3 ‾ ) 2 (0.\overline{3})^2 1 ( 0. 3 ‾ ) 2 \dfrac{1}{(0.\overline{3})^2} 0. 3 ‾ 0.\overline{3}

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Sandeep Bhardwaj by Sandeep Bhardwaj , 28, India

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1 solution

Sandeep Bhardwaj
May 27, 2015

Since we know that 0. 3 ‾ = 1 3 \boxed{0.\overline{3}=\dfrac{1}{3}} . (How ? : Try to figure it out)

So the given no. are equivalent to :

( 3 & 1 3 ) \Large \left( \color{skyblue}{3 \quad \color{#D61F06}{\&} \quad \dfrac{1}{3}} \right)

H.C.F. of fractions is calculated as : H.C.F. of ( a b & c d ) = H . C . F . ( a , c ) L . C . M . ( c , d ) \boxed{\text{H.C.F. of} \left( \dfrac{a}{b} \ \& \ \dfrac{c}{d} \right)=\dfrac{H.C.F. (a, c)}{L.C.M.(c,d)}}

∴ \therefore

H.C.F. of ( 3 1 & 1 3 ) = H . C . F . ( 3 , 1 ) L . C . M . ( 1 , 3 ) = 1 3 = 0. 3 ‾ \text{H.C.F. of} \left( \dfrac{3}{1} \ \& \ \dfrac{1}{3} \right)=\dfrac{H.C.F. (3, 1)}{L.C.M.(1,3)}=\dfrac{1}{3}=\boxed{0.\overline{3}}

enjoy !

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