Are all Odd,Non-Prime numbers compulsorily multiple of 3.If yes,how so?? I just thought that if all odd numbers which are non prime, compulsorily multiples of 3,I tried to get some mathematics but in vain. Kindly help me.
 A: No. For example, $25 = 5 \cdot 5$ or $35 = 5 \cdot 7$ are odd, not prime, and not multiples of 3.
More generally, take the product of at least two odd primes $\neq 3.$
A: No.  Just take
$X = \displaystyle \prod_1^n p_i^{a_i}, \tag 1$
where
$2, 3 \ne p_i \in \Bbb P, \; a_i \in \Bbb N, 1 \le i \le n; \tag 2$
that is, $X \in \Bbb N$ is the product of $n$ prime powers, where no prime in the product is $2$ or $3$; then it is clear that $X$ is odd, since it is a product of odds, and
$3 \not \mid X. \tag 3$
In this way, a myriad of examples my be constructed, viz.
$X = 5 \cdot 7 \cdot 11 = 385; \; X = 11 \cdot 13 \cdot 17 = 2431, \; X = 29 \cdot 31 \cdot 37 = 33263, \; \text{etc. etc. etc;} \tag 4$
the list of such $X$ is in fact quite long . . . 
A: The most relevant OEIS entry to your question is probably A038509, composite numbers congruent to $\pm 1 \bmod 6$.
Given any nonzero integer $k$, it is obvious that $6k$ is composite (let's ignore 0 for now). If $k \neq -1$ or 0, then $6k + 3$ is an odd number, composite and clearly divisible by 3, as you already know. $6k + 2$ and $6k + 4$ are obviously even but neither is divisible by 3.
So that leaves us $6k + 1$ and $6k + 5$. Neither is divisible by 2 or 3. They could both be prime. But you should also know that the primes thin out as you go further out towards infinity.
In fact, given a positive integer $n$, you can always find $n$ consecutive integers such that none of them are prime.
Let's try $n = 7$. Set $k = 15$, then clearly $6k = 90$ is composite, on account of being divisible by 2, 3 and 5. So 91 can't be divisible by 2, 3 or 5, it could even be prime. Nope: $91 = 7 \times 13$. Then 93 is obviously divisible by 3. And 95 is divisible by 5 but not by 2 or 3, and it's obviously not prime.
