NCERT Exemplar Solutions
Class 10 - Mathematics - CHAPTER 1: Real Numbers - Exercise 1.3 - Short Answer Questions
Question 8

Question. 8

Use Euclid’s division algorithm to find the HCF of 441, 567, 693.

Answer:

63

Detailed Answer with Explanation:

Step 1: Recall Euclid’s Division Algorithm.

For any two positive integers \(a\) and \(b\), with \(a > b\), we can write:

\(a = bq + r\), where \(0 \leq r < b\).

If \(r = 0\), then \(b\) is the HCF of \(a\) and \(b\). Otherwise, repeat the process with \(b\) and \(r\).

Step 2: Find HCF of 567 and 441.

Divide 567 by 441:

\(567 = 441 \times 1 + 126\)

Now divide 441 by 126:

\(441 = 126 \times 3 + 63\)

Now divide 126 by 63:

\(126 = 63 \times 2 + 0\)

Since remainder is 0, the HCF of 567 and 441 is 63.

Step 3: Find HCF of 693 and 63.

Divide 693 by 63:

\(693 = 63 \times 11 + 0\)

So, the HCF of 693 and 63 is 63.

Step 4: Combine results.

Hence, the HCF of 441, 567, and 693 is 63.

NCERT Exemplar Solutions Class 10 – Mathematics – CHAPTER 1: Real Numbers – Exercise 1.3 - Short Answer Questions | Detailed Answers