Question

Find the H.C.F of $\frac{2}{3},\frac{8}{9},\frac{32}{81},\frac{10}{27}$

Answer 1

H.C.F of fractions$=\frac{H.C.Fofnumerators}{L.C.Mofdenominators}$

H.C.F of $(\frac{2}{3},\frac{8}{9},\frac{32}{81},\frac{10}{27})=\frac{H.C.Fof(2,8,32,10)}{L.C.Mof(3,9,81,27)}$

H.C.F of $2,8,32,10$

Factors of $2=2\times 1$

Factors of $8=2\times 2\times 2$

Factors of $32=2\times 2\times 2\times 2\times 2$

Factors of $10=2\times 5$

H.C.F$=2$

L.C.M of $3,9,81,27$

Factors of $3=3\times 1$

Factors of $9=3\times 3$

Factors of $81=3\times 3\times 3\times 3$

Factors of $27=3\times 3\times 3$

L.C.M$=3\times 3\times 3\times 3=81$

$\therefore$ H.C.F of $(\frac{2}{3},\frac{8}{9},\frac{32}{81},\frac{10}{27})=\frac{2}{81}$