### 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 $$(a, b) = a \times b$$