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 \((a, b) = a \times b\)