Question

Find the HCF of 28, 42 and 98 using prime factorisation.​

Answer 1

Answer:

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Answer 2

$\large\underline{\sf{Solution-}}$

$\red{ \bf \: Prime \: Factorization \: of \: 28}$

$\begin{gathered}\begin{gathered}\begin{gathered} \:\: \begin{array}{c|c} {\underline{\sf{2}}}&{\underline{\sf{\:\:28\:}}}\\ {\underline{\sf{2}}}& \underline{\sf{\:\:14}} \\\underline{\sf{7}}&\underline{\sf{\:\:7}}\\\underline{\sf{}}&{\sf{\:\:1}} \end{array}\end{gathered}\end{gathered}\end{gathered}$

$\rm :\longmapsto\:28 = 2 \times 2 \times 7$

$\red{ \bf \: Prime \: Factorization \: of \: 42}$

$\begin{gathered}\begin{gathered}\begin{gathered} \:\: \begin{array}{c|c} {\underline{\sf{2}}}&{\underline{\sf{\:\:42\:}}}\\ {\underline{\sf{3}}}& \underline{\sf{\:\:21 \: }} \\\underline{\sf{7}}&\underline{\sf{\:\:7 \: }}\\\underline{\sf{}}&{\sf{\:\:1}} \end{array}\end{gathered}\end{gathered}\end{gathered}$

$\rm :\longmapsto\:42 = 2 \times 3 \times 7$

$\red{ \bf \: Prime \: Factorization \: of \: 98}$

$\begin{gathered}\begin{gathered}\begin{gathered} \:\: \begin{array}{c|c} {\underline{\sf{2}}}&{\underline{\sf{\:\:98\:}}}\\ {\underline{\sf{7}}}& \underline{\sf{\:\:49 \: }} \\\underline{\sf{7}}&\underline{\sf{\:\:7 \: }}\\\underline{\sf{}}&{\sf{\:\:1}} \end{array}\end{gathered}\end{gathered}\end{gathered}$

$\rm :\longmapsto\:98 = 2 \times 7 \times 7$

So,

$\bf\implies \:HCF(28, 42, 98) = 2 \times 7 = 14$

Additional Information :-

1. HCF is a factor of LCM, i.e. HCF always divides LCM.

2. Let a and b are two numbers, then

• HCF(a, b) × LCM(a, b) = a × b

3. HCF of numbers is always asmaller than or equal to smaller of given numbers.

4. LCM of numbers is always greater than or equal to larger of given number.