A number n is not a Prime no and lies between 1 to 301,how many such numbers are there which is not divisible by 2,3,5,7. 
A number n is not a prime and lies between $1$ and $301$, how many such numbers are there which are not divisible by $2$, $3$, $5$, $7$?  
a) $6$
  b) $8$
  c) $10$
  d) $12$

My approach:
Multiplication of prime with itself or other primes gives composite numbers. So
11*11=121
11*13=143
11*17=187
11*19=209
11*23=253
11*29=319 which is not in range of 1-301 so we cant conider it
13*13=169
13*17=221
13*19=247
13*23=299
17*17=289
17*19=323 which is not in range of 1-301 so we can't consider it
So total possible numbers are $10$ and option c) is the right one.
Are there any better approaches to this problem?
 A: Your approach is valid and probably the best possible. 
Of course, a composite number could also be constructed from the product of three or more primes, so to be really pedantic one might also look at $11 \cdot 11 \cdot 11$ but since this is way bigger than $301$, we needn't bother with any 3-products (in this case).
A: One way to formalise this a bit is to note that $\sqrt {301}\lt 18$ so there must be at least one prime factor less than $18$. Consider these in turn:
The highest values you need to consider are $$17\times \left\lfloor \frac {301}{17}\right\rfloor =17 \times 17$$$$13\times \left\lfloor \frac {301}{13}\right\rfloor =13 \times 23$$where $23$ is the largest prime (only if the second factor is $\gt 121$ does the question of three factors arise) and $$11\times \left\lfloor \frac {301}{11}\right\rfloor =11 \times 27$$ where $23$ is the largest prime.

Original version had $323$ as the target number

This might save you a little work, because it identifies the highest values first, and once you have done this, you know that all the lower values work without having to do the multiplications.
