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Find the HCF and LCM of 90, 144 by prime factorization method.

Answer Verified Verified
- Hint:First of all, find the LCM by taking both the numbers together and prime factorizing it and then multiplying all the factors. Now, find the HCF by prime factorizing each number separately and then choosing the common factor of the given numbers.

Complete step-by-step solution -

In this question, we have to find the LCM and HCF of 90 and 144 by using the prime factorization method. Before proceeding with this question, let us understand what LCM and HCF are.
LCM: It is the least common multiple or the smallest common multiple of two or more given natural numbers.
For example,
\[\begin{align}
  & 10=5\times 2 \\
 & 100=5\times 5\times 2\times 2 \\
\end{align}\]
So, the LCM of 10 and 100 is
\[5\times 5\times 2\times 2=100\]
HCF: It is the highest common factor of any two or more natural numbers.
For example,
\[\begin{align}
  & 10=5\times 2 \\
 & 100=5\times 5\times 2\times 2 \\
\end{align}\]
So, the HCF of 10 and 100 is
\[5\times 2=10\]
Now, let us find the LCM of 90 and 144. For finding the LCM of the numbers, we do the prime factorization of the numbers together as follows.
\[\begin{align}
  & 2\left| \!{\underline {\,
  90,144 \,}} \right. \\
 & 2\left| \!{\underline {\,
  45,72 \,}} \right. \\
 & 2\left| \!{\underline {\,
  45,36 \,}} \right. \\
 & 2\left| \!{\underline {\,
  45,18 \,}} \right. \\
 & 3\left| \!{\underline {\,
  45,9 \,}} \right. \\
 & 3\left| \!{\underline {\,
  15,3 \,}} \right. \\
 & 5\left| \!{\underline {\,
  5,1 \,}} \right. \\
 & \text{ }\left| \!{\underline {\,
  1,1 \,}} \right. \\
\end{align}\]
So, we get the LCM of 90 and 144 as
\[3\times 3\times 2\times 2\times 2\times 2\times 5=720\]
Now, let us find the HCF of 90 and 144. To find the HCF of the numbers, we do the prime factorization of the numbers separately as follows.
\[\begin{align}
  & 2\left| \!{\underline {\,
  90 \,}} \right. \\
 & 3\left| \!{\underline {\,
  45 \,}} \right. \\
 & 3\left| \!{\underline {\,
  15 \,}} \right. \\
 & 5\left| \!{\underline {\,
  5 \,}} \right. \\
 & \text{ }\left| \!{\underline {\,
  1 \,}} \right. \\
\end{align}\]
\[\begin{align}
  & 2\left| \!{\underline {\,
  144 \,}} \right. \\
 & 2\left| \!{\underline {\,
  72 \,}} \right. \\
 & 2\left| \!{\underline {\,
  36 \,}} \right. \\
 & 2\left| \!{\underline {\,
  18 \,}} \right. \\
 & 3\left| \!{\underline {\,
  9 \,}} \right. \\
 & 3\left| \!{\underline {\,
  3 \,}} \right. \\
 & \text{ }\left| \!{\underline {\,
  1 \,}} \right. \\
\end{align}\]
So, we get,
\[90=3\times 3\times 2\times 5\]
\[144=2\times 2\times 2\times 2\times 3\times 3\]
Factors common to both are \[3\times 3\times 2\].
So, we get, HCF as \[3\times 3\times 2=18\]

Note: Students must use only prime numbers starting from the smallest one in the prime factorization method to get the correct answer. Also, students can check this answer as follows:
For two numbers,
(LCM \[\times \] HCF) = Product of two numbers
So, we get,
\[\left( 720\times 18 \right)=\left( 90\times 144 \right)\]
12960 = 12960
Hence, LHS = RHS
So, our answer is correct.
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