Question

Show that any positive odd integer is in the form $6q+1,6q+3,6q+5$ where $q$ is some integer.

Answer 1

Hint: To solve this question, we should know the concept of Euclid division lemma. It states that for two positive integers a and b for which there exist another two integers q and r such that $a=qb+r$ where $0\le r < b$. We can use this theorem by considering a as the positive odd integer, and $b=6$. We get the possibilities of r as $0\le r < 5$. In those possibilities, we should consider the possibilities for which we get an odd integer.

Complete step by step answer:
The Euclid division lemma is stated for two positive integers a and b for which there exist another two integers q and r such that $a=qb+r$ where $0\le r < b$. Generally, we decode this division lemma as the number b is the divisor and q is the quotient that we get when we divide a by b and r is the remainder.
In our question, we are asked to prove that any odd positive integer can be of the form $6q+1,6q+3,6q+5$. In the three possibilities, we can see that there is a common number which is 6. So, we should use 6 as the divisor in the Euclid division lemma.
Let us consider the positive odd number as a. Let us take the value of b which is the divisor as 6. By applying the Euclid division lemma, we get
$a=6q+r$where $0\le r < 6$.
The possible values of r are $r=0,1,2,3,4,5$.
So, we can write the possible values of a as
\[\begin{align}
  & a=6q+0 \\
 & a=6q+1 \\
 & a=6q+2 \\
 & a=6q+3 \\
 & a=6q+4 \\
 & a=6q+5 \\
\end{align}\]
If we consider the term $6q$, it is always an even number for the integral values of q because the product of an even number and any positive integer is even. We also know that the sum of an even number and an odd number is odd and the sum of an even number and even number is an even number. From this, we can write the possible odd numbers as
\[\begin{align}
  & a=6q+1 \\
 & a=6q+3 \\
 & a=6q+5 \\
\end{align}\]
So, we got the possibilities of odd numbers as $6q+1,6q+3,6q+5$.
$\therefore $Hence we proved the statement.

Note: The key to prove this question is by recognising that 6 should be taken as the divisor and the term b in the Euclid division lemma. If we consider a different divisor in place of 6, we cannot prove what is required in the question. Students know the general odd number term as $2n\pm 1$ where n is a positive number. They might think that the question is wrong but we should consider a different divisor which is 2.