Question

Find the H.C.F and L.C.M of the following by the fundamental theorem of arithmetic.

(i) 9 and 12
(ii) 60 and 72
(iii) 8624 and 21658
(iv) 12, 30, 144
(v) 17, 13, 29

Answer 1

$$\begin{gathered} \left(\mathrm{i}\right)9=3\times 3=3^2 \\ 12=2\times 2\times 3=2^2\times 3^1 \\ \mathrm{HCF}\mathrm{of}9\mathrm{and}12=3^1=3 \\ \mathrm{LCM}\mathrm{of}9\mathrm{and}12=2^2\times 3^2=4\times 9=36 \end{gathered}$$


$$\begin{gathered} \left(\mathrm{ii}\right)60=2\times 2\times 3\times 5=2^2\times 3^1\times 5^1 \\ 72=2\times 2\times 2\times 3\times 3=2^3\times 3^2 \\ \mathrm{HCF}\mathrm{of}60\mathrm{and}72=2^2\times 3^1=2\times 2\times 3=12 \\ \mathrm{LCM}\mathrm{of}60\mathrm{and}72=2^3\times 3^2\times 5^1=8\times 9\times 5=360 \end{gathered}$$



$$\begin{gathered} \left(\mathrm{iii}\right)8624=2\times 2\times 2\times 2\times 7\times 7\times 11=2^4\times 7^2\times 11 \\ 21658=2\times 7\times 7\times 13\times 17=2^1\times 7^2\times 13\times 17 \\ \mathrm{HCF}\mathrm{of}8624\mathrm{and}21658=2^1\times 7^2=2\times 49=98 \\ \mathrm{LCM}\mathrm{of}8624\mathrm{and}21658=2^4\times 7^2\times 11\times 13\times 17 \\ =16\times 49\times 11\times 13\times 17 \\ =1905904 \end{gathered}$$

$$\begin{gathered} \left(i\mathrm{v}\right)12=2\times 2\times 3=2^2\times 3^1 \\ 30=2\times 3\times 5=2^1\times 3^1\times 5^1 \\ 144=2\times 2\times 2\times 2\times 3\times 3=2^4\times 3^2 \\ \mathrm{HCF}\mathrm{of}12,30\mathrm{and}144=2\times 3=6 \\ \mathrm{LCM}\mathrm{of}12,30\mathrm{and}144=2^4\times 3^2\times 5^1=720 \end{gathered}$$

$$\begin{gathered} \left(\mathrm{v}\right)17={17}^1\times 1 \\ 23={23}^1\times 1 \\ 29={29}^1\times 1 \\ \mathrm{HCF}\mathrm{of}17,23\mathrm{and}29=1 \\ \mathrm{LCM}\mathrm{of}17,23\mathrm{and}29=17\times 23\times 29=11339 \end{gathered}$$