Find the HCF (210, 55) by Euclid's Division algorithm

Answer

To find the HCF of 210 and 55 using Euclid's Division Algorithm, we will follow these steps:

### Step 1: Apply Euclid's Division Lemma
According to Euclid's Division Lemma, we can express the relationship between two numbers \( A \) and \( B \) as:
\[ A = B \cdot Q + R \]
where \( R \) is the remainder when \( A \) is divided by \( B \).

Here, let \( A = 210 \) and \( B = 55 \).

### Step 2: Divide 210 by 55
Now, we perform the division:
\[ 210 \div 55 \]
Calculating this, we find that 55 goes into 210 three times (since \( 55 \times 3 = 165 \)), leaving a remainder:
\[ R = 210 - 165 = 45 \]
So we can write:
\[ 210 = 55 \cdot 3 + 45 \]

### Step 3: Repeat the process with 55 and 45
Next, we take the divisor (55) and the remainder (45) and apply the same process:
\[ 55 = 45 \cdot Q + R \]
Calculating this, we find that 45 goes into 55 once (since \( 45 \times 1 = 45 \)), leaving a remainder:
\[ R = 55 - 45 = 10 \]
So we can write:
\[ 55 = 45 \cdot 1 + 10 \]

### Step 4: Repeat the process with 45 and 10
Now we take 45 and 10:
\[ 45 = 10 \cdot Q + R \]
Calculating this, we find that 10 goes into 45 four times (since \( 10 \times 4 = 40 \)), leaving a remainder:
\[ R = 45 - 40 = 5 \]
So we can write:
\[ 45 = 10 \cdot 4 + 5 \]

### Step 5: Repeat the process with 10 and 5
Next, we take 10 and 5:
\[ 10 = 5 \cdot Q + R \]
Calculating this, we find that 5 goes into 10 two times (since \( 5 \times 2 = 10 \)), leaving a remainder:
\[ R = 10 - 10 = 0 \]
So we can write:
\[ 10 = 5 \cdot 2 + 0 \]

### Step 6: Conclusion
At this point, we have reached a remainder of 0. According to Euclid's Division Algorithm, the last non-zero remainder is the HCF. Thus, the HCF of 210 and 55 is:
\[ \text{HCF} = 5 \]

### Final Answer
The HCF of 210 and 55 is **5**.

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