### Theory:

**1**.

**Show that any positive odd integer is of form \(4q + 1\) or \(4q + 3\), where \(q\) is some integer**.

**Solution**:

**:**

**Recall****Euclid's division lemma**

Given positive integers \(a\) and \(b\), there exist unique integers \(q\) and \(r\) satisfying \(a = bq + r, 0 ≤ r < b\).

Let \(a\) be an odd positive integer, and \(b\) be the divisor.

And, take \(b = 4\).

Now, apply Euclid's division lemma.

\(a = 4q + r\), where \(0 \le r < 4\)

We know that remainder is always lesser than the divisor.

So, \(r\) can be \(0\), \(1\), \(2\) or \(3\).

If \(r = 0\), then the equation becomes \(a = 4q\).

If \(r = 1\), then the equation becomes \(a = 4q + 1\).

If \(r = 2\), then the equation becomes \(a = 4q + 2\).

If \(r = 3\), then the equation becomes \(a = 4q + 3\).

Here, \(4q\) and \(4q + 2\) are even numbers.

Since \(a\) is an odd positive integer, \(a\) cannot be \(4q\) or \(4q + 2\).

Since \(a\) is an odd positive integer, \(a\) can be \(4q + 1\) or \(4q + 3\).

**Thus, any positive odd integer is of form \(4q + 1\) or \(4q + 3\), where \(q\) is some integer**.

**2**.

**Check whether \(6^n\) can end with the digit \(0\) for any natural number \(n\)**.

**Solution**:

The number, which ends with digit \(0\), must be a multiple of \(10\).

The prime factor of \(10\) is \(2 \times 5\).

That means, if a number ends with digit \(0\), then it must contain \(2\) and \(5\) as the factors.

Now, \(6^n = (2 \times 3)^n\).

Here, \(6^n\) contains \(2\) as a factor, but it does not contain the factor \(5\).

So, it does not end with the digit \(0\).

**Therefore, \(6^n\) cannot end with the digit \(0\) for any natural number \(n\)**.

**3**.

**Find the LCM and HCF of \(15\) and \(42\) by the prime factorisation method. Also, find their product**.

**Solution**:

First, let us write the given numbers as a product of primes.

\(15 = 3 \times 5\)

\(42 = 2 \times 3 \times 7\)

LCM \(=\) \(2 \times 3 \times 5 \times 7 = 210\)

HCF \(=\) \(3\)

HCF \(\times\) LCM \(=\) \(210 \times 3 = 630\)

**Note from the above example**:

HCF \(\times\) LCM \(=\) \(630\)

We know, \(15 \times 42 = 630\).

So, HCF \(\times\) LCM \(=\) \(15 \times 42 \)

**That is, HCF**\(\times\)

**LCM**\(=\)

**Product of the given numbers**.

Important!

**For any two positive integers \(a\) and \(b\):**

**HCF \((a, b)\) \(\times\) LCM \((a, b) = a \times b\)**