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Using prime factorisation, find the HCF of $180, 140$ and $210$
Using prime factorisation, find the HCF of $180, 140$ and $210$
This Question has 1 answers.
We need to find the HCF of $180, 140$ and $210$ using prime factorization.
### Step 1: Prime Factorization of Each Number
Factorizing $180$:
$180 = 2^2 \times 3^2 \times 5$
Factorizing $140$:
$140 = 2^2 \times 5 \times 7$
Factorizing $210$:
$210 = 2 \times 3 \times 5 \times 7$
### Step 2: Identify Common Factors
The common prime factors among all three numbers are:
- $2$ appears in $180$ as $2^2$, in $140$ as $2^2$, and in $210$ as $2$. The minimum power is $2^1 = 2$.
- $5$ appears in all three numbers as $5^1 = 5$.
### Step 3: Compute the HCF
The HCF is the product of the lowest powers of the common prime factors:
$HCF = 2^1 \times 5^1 = 2 \times 5 = 10$
### Final Answer:
The HCF of $180, 140,$ and $210$ is $10$.
### Step 1: Prime Factorization of Each Number
Factorizing $180$:
$180 = 2^2 \times 3^2 \times 5$
Factorizing $140$:
$140 = 2^2 \times 5 \times 7$
Factorizing $210$:
$210 = 2 \times 3 \times 5 \times 7$
### Step 2: Identify Common Factors
The common prime factors among all three numbers are:
- $2$ appears in $180$ as $2^2$, in $140$ as $2^2$, and in $210$ as $2$. The minimum power is $2^1 = 2$.
- $5$ appears in all three numbers as $5^1 = 5$.
### Step 3: Compute the HCF
The HCF is the product of the lowest powers of the common prime factors:
$HCF = 2^1 \times 5^1 = 2 \times 5 = 10$
### Final Answer:
The HCF of $180, 140,$ and $210$ is $10$.
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