Question

what's the hcf of 125,75 and 290

do in continued division method

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Answer 1

Answer:

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Answer 2

To Find:-

The HCF of 125,75 and 290.

Solution:-

$\left.\begin{array}{c c c} \LARGE\begin{gathered}\begin{gathered}\begin{gathered}\begin{gathered} \:\: \begin{array}{c|c} {\underline{\sf{5}}}&{\underline{\sf{\:\:125 \:\:}}}\\ {\underline{\sf{5}}}& \underline{\sf{\:\:25 \:\:}} \\\underline{\sf{5}}&\underline{\sf{\:\:5\:\:}} \\ {\underline{\sf{}}}& \underline{\sf{\:\:1\:\:}} \end{array}\end{gathered}\end{gathered}\end{gathered}\end{gathered}\end{array}\right \} \tt{125 = 5 \times \red5 \times 5 = {5}^{3} }$

$\left.\begin{array}{c c c} \LARGE\begin{gathered}\begin{gathered}\begin{gathered}\begin{gathered} \:\: \begin{array}{c|c} {\underline{\sf{5}}}&{\underline{\sf{\:\:75 \:\:}}}\\ {\underline{\sf{5}}}& \underline{\sf{\:\:15 \:\:}} \\\underline{\sf{3}}&\underline{\sf{\:\:3\:\:}} \\ {\underline{\sf{}}}& \underline{\sf{\:\:1\:\:}} \end{array}\end{gathered}\end{gathered}\end{gathered}\end{gathered}\end{array}\right \} \tt{75 = 5 \times \red5 \times 3 = {5}^{2} \times 3 }$

$\left.\begin{array}{c c c} \LARGE\begin{gathered}\begin{gathered}\begin{gathered}\begin{gathered} \:\: \begin{array}{c|c} {\underline{\sf{2}}}&{\underline{\sf{\:\:290 \:\:}}}\\ {\underline{\sf{5}}}& \underline{\sf{\:\:145 \:\:}} \\\underline{\sf{29}}&\underline{\sf{\:\:29\:\:}} \\ {\underline{\sf{}}}& \underline{\sf{\:\:1\:\:}} \end{array}\end{gathered}\end{gathered}\end{gathered}\end{gathered}\end{array}\right \} \tt{290 = 2 \times \red5 \times 29 }$

$\pmb {\tt{∴HCF = 5}}$