Question

If \(130 = 15 \times 8 + 10\) and \(15 = 5 \times 3 + 0\), then \(\mathrm{HCF}(130,\,15)\) will be:

• A. \(8\)
• B. \(5\)
• C. \(130\)
• D. \(15\)

Hint:

In Euclid’s algorithm, the HCF is the last non-zero remainder in the sequence of divisions.

Answer 1

The Correct Option is B

Step 1: Apply Euclid’s division algorithm.
From \(130 = 15 \cdot 8 + 10\), the remainder is \(10\).
Next, divide \(15\) by \(10\): \(15 = 10 \cdot 1 + 5\).
Finally, \(10 = 5 \cdot 2 + 0\). The last non-zero remainder is \(5\).
Step 2: Conclude.
Hence, \(\gcd(130,15)=5\).