Problem Statement
Show that any positive odd integer is of the form $6q + 1$, or $6q + 3$, or $6q + 5$, where $q$ is some integer.
Verified Solution & Marking Scheme
Apply Euclid's Division Lemma with b = 6
Let $a$ be any positive odd integer. By Euclid's lemma with divisor $b = 6$:
$a = 6q + r, \quad \text{where } 0 \le r < 6$
Possible remainders are $r \in \{0, 1, 2, 3, 4, 5\}$.
Examine Even Values of a
- If $r = 0$: $a = 6q = 2(3q)$ (Even)
- If $r = 2$: $a = 6q + 2 = 2(3q + 1)$ (Even)
- If $r = 4$: $a = 6q + 4 = 2(3q + 2)$ (Even)
Conclude for Odd Integers
Since $a$ is odd, $a$ cannot be $6q, 6q+2$, or $6q+4$. Therefore, any positive odd integer must be of the form $6q + 1, 6q + 3$, or $6q + 5$.