Which one of the following is a prime number?

### Concept & Strategy To verify if a larger number $N$ is prime, find its approximate square root $\sqrt{N}$. Check if $N$ is divisible by any prime number up to that integer limit. If it isn't divisible by any, it is a prime number. ### Step-by-Step Solution * **Option (a) 161:** The approximate square root is slightly less than 13 ($\sqrt{169} = 13$). Test primes up to 11. Testing 7: $161 = 7 \times 23$. Thus, 161 is composite. * **Option (b) 221:** The approximate square root is slightly less than 15. Test primes up to 13. Testing 13: $221 = 13 \times 17$. Thus, 221 is composite. * **Option (d) 437:** The approximate square root is roughly 21 ($\sqrt{400} = 20, \sqrt{441} = 21$). Test primes up to 19. Testing 19: $437 = 19 \times 23$. Thus, 437 is composite. * **Option (c) 373:** The approximate square root is roughly 19.3. Test primes up to 19 (2, 3, 5, 7, 11, 13, 17, 19). 373 is odd, doesn't end in 5, sum of digits is 13 (not div by 3). $373 \div 7$ leaves remainder 2. $373 \div 11$ leaves remainder 10. $373 \div 13$ leaves remainder 9. $373 \div 17$ leaves remainder 16. $373 \div 19$ leaves remainder 12. It does not divide evenly by any of these primes. Therefore, 373 is prime. ### Exam Strategy & Shortcut Aptitude test setters intentionally use products of mid-range primes (7, 13, 17, 19, 23) as options to trap students. Memorize common combinations like $13 \times 17 = 221$. When checking 161, recognize that $140 + 21 = 161$, making it instantly divisible by 7. By eliminating the easier composites, you can confidently select 373 without having to manually divide it by every single prime up to 19. ### Common Pitfall Giving up too early on division tests. Many students check for 2, 3, 5, 7, and 11, and if none work, they assume the number is prime. You must test primes all the way up to the approximate square root (like 17 and 19 for these larger options) to be accurate. ### Final Answer **Therefore, the correct answer is 373.**