Question

Find the HCF of \[960\] and \[432\] is \[................\]

Answer 1

Hint: HCF stands for highest common factor.
So, to find the HCF of any two or more numbers, we need to factorise those numbers into its prime factors first.
And, then we will try to find the highest number of common factors in between those numbers.

Complete step-by-step solution:
Let's say, \[A\] and \[B\] are two non-prime numbers.
Where \[A\] has four prime factors and also,\[B\] has four prime factors.
Also, assume \[A\] and \[B\] as in the following form.
\[A = (p \times q \times r \times s)\] .
And,
\[B = (a \times b \times r \times s)\] .
Where \[a,b,p,q,r,s\] are prime numbers.
Then we can say that \[A\] and \[B\] are two prime factors common in them; and, they are \[r,s\] .
So, HCF of \[A\] and \[B\] should be \[ = (r \times s).\]
So, we have to factorise \[960\] and \[432\] into its least possible prime factors and then we will try to find the maximum prime factors, and we will multiply them to find the required HCF.
So, we can factorise \[960\] in following manner:
\[\begin{array}{*{20}{c}}
  {2\left| \!{\underline {\,
  {960} \,}} \right. } \\
  {2\left| \!{\underline {\,
  {480} \,}} \right. } \\
  {2\left| \!{\underline {\,
  {240} \,}} \right. } \\
  {2\left| \!{\underline {\,
  {120} \,}} \right. } \\
  {2\left| \!{\underline {\,
  {60} \,}} \right. } \\
  {2\left| \!{\underline {\,
  {30} \,}} \right. } \\
  {3\left| \!{\underline {\,
  {15} \,}} \right. } \\
  {\,\,5}
\end{array}\]
So, \[960 = (2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 3 \times 5)\] .
Also, we can factorise \[432\] in the following manner:
\[\begin{array}{*{20}{c}}
  {2\left| \!{\underline {\,
  {432} \,}} \right. } \\
  {2\left| \!{\underline {\,
  {216} \,}} \right. } \\
  {2\left| \!{\underline {\,
  {108} \,}} \right. } \\
  {2\left| \!{\underline {\,
  {54} \,}} \right. } \\
  {3\left| \!{\underline {\,
  {27} \,}} \right. } \\
  {3\left| \!{\underline {\,
  9 \,}} \right. } \\
  {\,\,3}
\end{array}\]
So,\[432 = (2 \times 2 \times 2 \times 2 \times 3 \times 3 \times 3)\] .
So, in \[960\] and \[432\] four \[{2^{'s}}\] and one \[3\] are common.
So, the required HCF of \[960\] and \[432\] is\[ = (2 \times 2 \times 2 \times 2 \times 3) = (16 \times 3) = 48.\]

\[\therefore \] The HCF of \[960\] and \[432\] is \[48\].

Note: Another approach we can use to find HCF of numbers.
We can also use a division method to find out the HCF of two numbers.
In that case we have to divide the larger number by the smaller number in the following manner to find out the HCF:
$\begin{array}{*{20}{c}}
  {\begin{array}{*{20}{c}}
  {432)960(2} \\
  {\,\,\,\,\underline {864} } \\
  {96)432(4} \\
  {\,\,\underline {384} } \\
  {48)96(2} \\
  {\,\,\underline {96} }
\end{array}} \\
  0
\end{array}$
So, using the division method we can find the HCF of two numbers.
One must remember that each and every number is formed by multiplying only prime numbers.