Question

find the HCF of 64,80 by listing all their factors​

Answer 1

Answer:

16

Step-by-step explanation:

1. 64=2*2*2*2*2*2
2. 80=2*2*2*2*5
3. here 4 times 2 are common so 2*2*2*2=16

Answer 2

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

• Two numbers 64 and 80.

$\large\underline{\sf{To\:Find - }}$

• HCF of 64 and 80 using prime factorization.

$\large\underline{\sf{Basic \: Concept \: Used- }}$

HCF by Prime Factorization

• Find the prime factors of each of the given number.

• Then, identify the common prime factors of the given numbers.

• Then, multiply the common factors.

• The product of these common factors is the HCF of the given numbers

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

Consider,

• First number 64,

• Its prime factorization as follow :-

$\begin{gathered}\begin{gathered}\begin{gathered} \:\: \begin{array}{c|c} {\underline{\sf{2}}}&{\underline{\sf{\:\:64\:\:\:}}}\\ {\underline{\sf{2}}}& \underline{\sf{\:\:32\:\:\:}} \\\underline{\sf{2}}&\underline{\sf{\:\:16\: \:\:}}\\{\underline{\sf{2}}}&{\underline{\sf{\:\:8\:\:\:}}} \\{\underline{\sf{2}}}&{\underline{\sf{\:\:4\:\:\:}}} \\ {\underline{\sf{2}}}&{\underline{\sf{\:\:2\:\:\:}}} \\ \underline{\sf{}}&{\sf{\:\:1\:\:\:}} \end{array}\end{gathered}\end{gathered}\end{gathered}$

$\bf\implies \:64 = 2 \times 2 \times 2 \times 2 \times 2 \times 2 - - (1)$

Now,

Consider

• Second number 80,

• Its prime factorization as follow :-

$\begin{gathered}\begin{gathered}\begin{gathered} \:\: \begin{array}{c|c} {\underline{\sf{2}}}&{\underline{\sf{\:\:80\:\:\:}}}\\ {\underline{\sf{2}}}& \underline{\sf{\:\:40\:\:\:}} \\\underline{\sf{2}}&\underline{\sf{\:\:20\: \:\:}}\\{\underline{\sf{2}}}&{\underline{\sf{\:\:10\:\:\:}}} \\{\underline{\sf{5}}}&{\underline{\sf{\:\:5\:\:\:}}} \\ {\underline{\sf{}}}&{{\sf{\:\:1\:\:\:}}} \\ {\sf{}}&{\sf{\:\:\:\:\:}} \end{array}\end{gathered}\end{gathered}\end{gathered}$

$\bf\implies \:80 = 2 \times 2 \times 2 \times 2 \times 5$

Hence,

$\sf \: \rm :\implies\:Common \: factors \: are \: 2 \times 2 \times 2 \times 2 = 16$

$\bf\implies \:HCF \: of \: 64 \: and \: 80 \: is \: 16.$