Question

find the hcf of the given number by prime factorisation method 105,140,175

Answer 1

Answer:

Prime factorisation of given numbers are :

105=3×5×7

140=2×2×5×7

165=5×5×7

Notice that 5 occurs as a common prime factor at least factor one time and 7 one time in given numbers.

∴ H.C.F =5×7=35

Answer 2

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

Consider,

$\red{\rm :\longmapsto\:Prime \: Factorization \: of \: 105}$

$\red{\begin{gathered}\begin{gathered}\begin{gathered} \:\: \begin{array}{c|c} {\underline{\sf{5}}}&{\underline{\sf{\:\:105 \:\:}}}\\ {\underline{\sf{3}}}& \underline{\sf{\:\:21 \:\:}} \\\underline{\sf{7}}&\underline{\sf{\:\:7\:\:}} \\\underline{\sf{}}&{\sf{\:\:1 \:\:}} \end{array}\end{gathered}\end{gathered}\end{gathered}}$

$\red{\rm :\longmapsto\:Prime \: Factorization \: of \: 105 = 5 \times 3 \times 7}$

Now, Consider

$\green{\rm :\longmapsto\:Prime \: Factorization \: of \: 140}$

$\green{\begin{gathered}\begin{gathered}\begin{gathered} \:\: \begin{array}{c|c} {\underline{\sf{2}}}&{\underline{\sf{\:\:140 \:\:}}}\\ {\underline{\sf{2}}}& \underline{\sf{\:\:70 \:\:}} \\\underline{\sf{5}}&\underline{\sf{\:\:35\:\:}} \\ {\underline{\sf{7}}}& \underline{\sf{\:\:7 \:\:}}\\\underline{\sf{}}&{\sf{\:\:1 \:\:}} \end{array}\end{gathered}\end{gathered}\end{gathered}}$

$\green{\rm :\longmapsto\:Prime \: Factorization \: of \: 140 = {2}^{2} \times 5 \times 7}$

Now, Consider

$\blue{\rm :\longmapsto\:Prime \: Factorization \: of \: 175}$

$\blue{\begin{gathered}\begin{gathered}\begin{gathered} \:\: \begin{array}{c|c} {\underline{\sf{5}}}&{\underline{\sf{\:\:175 \:\:}}}\\ {\underline{\sf{5}}}& \underline{\sf{\:\:35 \:\:}} \\\underline{\sf{7}}&\underline{\sf{\:\:7\:\:}} \\\underline{\sf{}}&{\sf{\:\:1 \:\:}} \end{array}\end{gathered}\end{gathered}\end{gathered}}$

$\blue{\rm :\longmapsto\:Prime \: Factorization \: of \: 175 = {5}^{2} \times 7}$

So, we have now

$\red{\rm :\longmapsto\:Prime \: Factorization \: of \: 105 = 5 \times 3 \times 7}$

$\green{\rm :\longmapsto\:Prime \: Factorization \: of \: 140 = {2}^{2} \times 5 \times 7}$

$\blue{\rm :\longmapsto\:Prime \: Factorization \: of \: 175 = {5}^{2} \times 7}$

So,

$\purple{\rm :\longmapsto\:HCF(105, 140, 175) = 5 \times 7 = 35}$

▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬

More to Learn

Let a and b are two positive real numbers having HCF as h and LCM as l then

• 1. HCF × LCM = a × b

• 2. LCM is always divisible by a, b and HCF

• 3. HCF always divides a, b and LCM.